VECTOR CALCULUS Solved Exercises Line Integrals

Solved exercises

Equacions Diferencials Grau en Enginyeria en Tecnologies Industrials (GETI) ETSEIB – UPC

Departament de Matem`atiques ETS d’Enginyeria Industrial de Barcelona (ETSEIB) Universitat Polit`ecnicade Catalunya (UPC) Version: May 2021

https://mat-web.upc.edu/etseib/ed/

Note: All the underlined text provides links to websites which may help you better understand each problem, or obtain further information.

Line integrals

1. Length of the arc of the cycloid σ(t) = (R(t − sin t),R(1 − cos t)), 0 ≤ t ≤ 2π. Resolution: A cycloid is the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slipping. To get the parameterization, we ﬁrst parameterize the rim by (R cos(3π/2 − t),R sin(3π/2 − t)), 0 ≤ t ≤ 2π, since the wheel goes clockwise and we measure the angle from the ﬁrst contact point whose angle is 3π/2. As the center of the wheel travels like (Rt, R), the parameterization of the cycloid turns out to be σ(t) = (R(cos(3π/2 − t) + Rt, sin(3π/2 − t) + R) = R(t − sin t, 1 − cos t) , 0 ≤ t ≤ 2π.

0 p t Therefore σ0(t) = R(1 − cos t, sin t), so that σ (t) = R 2(1 − cos t) = 2R sin and 2 2π 2π t=2π t t t L(cycloid) = 2R sin dt = 2R sin dt = 2R −2 cos = 8R. ˆ0 2 ˆ0 2 2 t=0 2. Length of the epicycloid with k cusps parameterized by σ(t) = (R(k + 1) cos t − R cos((k + 1)t),R(k + 1) sin t − R sin((k + 1)t)), 0 ≤ t ≤ 2π. Resolution: An epicycloid or hypercycloid is a plane curve produced by tracing the path of a chosen point on the circumference of a circlecalled an epicyclewhich rolls without slipping around a fixed circle. It is a particular kind of roulette. To get the length, as we have the parameterization, we first have to calculate its first derivative which is σ0(t) = (−R(k + 1) sin(t) + R sin((k + 1)t)(k + 1),R(k + 1) cos(t) − R cos((k + 1)t)(k + 1)) Now let’s simplify this expression saying that a = k + 1 :

σ0(t) = R(−a sin(t) + sin(at)a, a cos(t) − cos(at)a)

The norm is

0 p σ (t) =R (−a sin(t) + sin(at) a)2 + (a cos(t) − cos(at) a)2 = q =R a2(sin2(t) − 2 sin(t) sin(at) + sin2(at) + cos2(t) − 2 cos(t) cos(at) + cos2(at)) = q =aR sin2(t) − 2 sin(t) sin(at) + sin2(at) + cos2(t) − 2 cos(t) cos(at) + cos2(at) = =aRp2 − 2 sin(t) sin(at) − 2 cos(t) cos(at) = v u =aRu2 − 2 (sin(t) sin(at) + cos(t) cos(at)) = t | {z } | {z } 1 1 2 (cos(t−at)−cos(t+at)) 2 (cos(t−at)+cos(t+at)) √ =aRp2 − 2 cos(t − at) = aR 2p1 − cos(t − at) = √ √ =aR 2p1 − cos((1 − a)t) = (cos x is an even function so) = aR 2p1 − cos((a − 1)t) = √ (k + 1)R 2p1 − cos(kt) (where we have changed again k = a − 1)

Finally let us compute the length 2π √ L(epicycloid) = (k + 1)R 2p1 − cos(kt) dt ˆ0 √ 2π = (k + 1)R 2 p1 − cos(kt) dt ˆ 0 | {z }√ q cos(kt) 1−cos(kt) sin kt = 1 − = √ | ( 2 )| 2 2 2 2π 2π √ √ kt kt = (k + 1)R 2 2 sin dt = 2(k + 1)R sin dt. ˆ0 2 ˆ0 2

kt k 2 Now we apply the change u = 2 , du = 2 dt ⇐⇒ dt = k du and use that |sin u| is π-periodic and that |sin u| = sin u ≥ 0 for 0 ≤ u ≤ π,

πk π k + 1 u=π L(epicycloid) = 4 R |sin u| du = 4(k + 1)R |sin u| du = 4(k + 1)R [− cos u]u=0 k ˆ0 ˆ0 = 8(k + 1)R.

8(k+1)R An epicycloid with k cusps has k archs, each one with length k . You can see a ﬁgure for some k just clicking on it: k = 1, k = 2, k = 3, k = 4...

3. Compute the theoretical time of navigation between Lisbon and New York if we follow a trajectory of constant direction and we sail at speed of 30 knots.

Resolution: In this exercise we need to compute a great-circle distance, which is the shortest distance between two points on the surface of a sphere, measured along the surface of the sphere. We are going to establish that Lisbon is our Point 1 and New York is the Point 2.

Let θ1, ϕ1 and θ2, ϕ2, be the geographical longitude and latitude in radians of the two points, and ∆θ, ∆ϕ be their absolute diﬀerences.

2 Illustration of the central angle α, between two points P1 and P2.

The actual arc length d on a sphere of radius r can be trivially computed as d = Rα, where α is the angle between its two endpoints P1 and P2, measured from the center O of the sphere. If −π ≤ θj ≤ π and −π/2 ≤ ϕj ≤ π/2, then:

hOP~ 1, OP~ 2i cos α = = cos ϕ1 cos ϕ2 (cos θ1 cos θ2 + sin θ1 sin θ2) + sin ϕ1 sin ϕ2 ~ ~ OP1 OP2 = sin ϕ1 sin ϕ2 + cos ϕ1 cos ϕ2 cos(∆θ).

[Obs.: This formula can be directly applying the spherical law of cosines to the spherical triangle whose vertices are P1, P2 and one of the poles an auxiliary third point; if the North Pole is chosen, the angles of this triangle are π/2 − ϕ1, π/2 − ϕ2 and α.] First let’s collect some data that we need to solve the problem, ﬁrst of all we need the mean radius of the Earth which is approximately R ≈ 6371 km, then we need the coordinates of Lisbon which o 0 o 0 are (θ1, ϕ1) = (9 8 W, 38 43 N) ≈ (0.1594067, 0.6757333) rad, and also the coordinates of New York o 0 o 0 which are (θ2, ϕ2) = (74 0 W, 40 43 N) ≈ (1.2915436, 0.7106399) rad. Using such data and the formula obtained above, we can calculate the distance, which is d = Rα = 0.8510653 · 6371 = 5422.14 km. Now that we know the distance, we only need to ﬁnish the problem by calculating the time we need to go from Lisbon to New York at a constant speed of 30 knots, knowing that a knot is 1.852 km/h the theoretical time of navigation t will be: 5422.14 t = = 97.59 h, 30 · 1.852 which is a little bit more than 4 days.

4. A cassette tape with thickness of 16µm (micrometers) is rolled in a spool with internal radius r0 = 1.11cm and external radius r1 = 2.46cm. How much does the tape of the spool measure? [Hint: Approximate the form of the spool by the curve described in polar coordinates through the equation r = g(θ) = cθ,θ0 ≤ θ ≤ θ1, being c > 0 a constant to be determined, so that rj = cθj .] Resolution: To solve this problem we shall look at it as if we were to ﬁnd the length of the curve (C) as an Archimedean spiral. See more information about Archimedean spiral in wikipedia.

3 Archimedean spiral

−2 −2 −6 Our data are r0 = 1.11 · 10 m, r1 = 2.46 · 10 m and the increment per revolution is 16 · 10 m. We are given C in polar coordinates as r = r(θ) for θ0 ≤ θ ≤ θ1, so in cartesian coordinates C can be parameterized as C = σ([θ0, θ1]):

(x, y) = σ(θ) = (r(θ) cos θ, r(θ) sin θ), θ0 ≤ θ ≤ θ1.

To ﬁnd θ0 and θ1 ﬁrst we have to determine c. We will do this imposing that the radius increases 16 · 10−6m for each rotation. So, with any rotation

16 · 10−6 = g(θ + 2π) − g(θ) = 2πc, so that 8 · 10−6 c = . π Now we can calculate: r 0.0111 · π θ = 0 = , 0 c 8 · 10−6 r 0.0246 · π θ = 1 = . 1 c 8 · 10−6 We can deﬁne both r(θ) and r0(θ) because it will be useful for us to work with them during the deﬁnition of the parameterization:

r (θ) = c · θ and r0 (θ) = c

Finally, we know that the arc length in polar coordinates is: s dr 2 d` = r2 + dθ, dθ and so, the total arc length of the tape is:

θ1 θ1 q θ1 p 2 p d` = r(θ)2 + r0(θ)2 dθ = (c · θ) + c2 dθ = c 1 + θ2 dθ ˆC ˆθ0 ˆθ0 ˆθ0 1 p 1 p θ=θ1 = c · · θ · 1 + θ2 + ln θ + 1 + θ2 ≈ 94.6m, 2 2 θ=θ0 t a2 p 2 2 p 2 2 p 2 2 where we have used the primitive t + a dt = t + a + ln t + t + a + ctant, which ˆ 2 2 √ can be found either by the change of variables t = a sinh(x) or the change by parts u = t2 + a2, dv = dt.

4 5. Mass of a spiral of the helix of radius R and height h parameterized by

ht σ(t) = R cos t, R sin t, , 0 ≤ t ≤ 2π, 2π

if the density at each point is proportional to the distance from the origin, being k > 0 the constant.

Resolution: Notice that this helix is a 3D-curve with height h. Its mass M is ρ d`, where we ˆ r C h2t2 h p are said that ρ(x, y, z) = kpx2 + y2 + z2. Along C, kσ(t)k = R2 + = t2 + a2, where 4π2 2π 2πR khp 0 h 0 a = , so that ρ(σ(t)) = t2 + a2. Moreover, σ (t) = −R sin t, R cos t, so that σ (t) = h 2π 2π r h2 h p R2 + = 1 + a2. Finally 4π2 2π

t=2π kh2 p 2π p kh2 p t p a2 p M = 1 + a2 t2 + a2 dt = 1 + a2 t2 + a2 + ln t + t2 + a2 2 2 4π ˆ0 4π 2 2 t=0 kh2 p p p 2πR = 1 + a2 π 4π2 + a2 + 2π2 ln 2π + 4π2 + a2 − a2 ln a , a = , 4π2 h t a2 p 2 2 p 2 2 p 2 2 where we have used that t + a dt = t + a + ln t + t + a + ctant, which can be ˆ 2 2 √ found either by the change of variables t = a sinh x or the change by parts u = t2 + a2, dv = dt.

6. Mass of the cardioid deﬁned in polar coordinates by r = R (1 + cos θ) if the density at each point is proportional to the square root of the distance from the origin, being k the constant.

Resolution: First of all we should deﬁne what a cardioid is. A cardioid is a plane curve traced by a point on the perimeter of a circle that is rolling around a ﬁxed circle of the same radius (R). See more information about the cardiod in wikipedia.

Cardiod with R= 0.5 (polar coordinates)

We deﬁne both r(θ) and r0(θ) because it will be useful for us to work with them during the deﬁnition of the parameterization. r(θ) = R(1 + cos θ) , r0(θ) = −R sin θ.

5 Polar coordinates are defined to be: (x, y) = (r cos θ, r sin θ), but in this exercise we are given r as a function of θ, so we have to substitute r = r(θ). As we can see in the figure, θ goes from 0 to 2π. We can define the curve in the following way:

C = σ([0, 2π]) with σ(θ) = (r(θ) cos θ, r(θ) sin θ) .

We know that the line integral along a piecewise smooth curve is deﬁned as:

b 0 f d` = f(σ(t)) · σ (t) dt. ˆC ˆa

So, now we have to compute the derivative and norm of σ(θ):

cos θ − sin θ r0(θ) σ0(θ) = r0(θ) cos θ − r(θ) sin θ, r0(θ) sin θ + r(θ) cos θ = . sin θ cos θ r(θ) If we compute the determinant of the matrix, we can see that it is equal to 1. So, the norm of the derivative of the parameterization is: q 0 p p σ (θ) = r(θ)2 + r0(θ)2 = R2(1 + cos θ)2 + R2 sin2 θ = R 1 + cos2 θ + 2 cos θ + sin2 θ √ √ √ = R 2 + 2 cos θ = R 2 1 + cos θ.

We deﬁne the density of the cardiod as ρ. In the statement of the problem it is said that ρ is proportional to the square root of the distance from the origin, which can be written as: qp ρ(x, y) = k x2 + y2.

If we apply the parameterization rq ρ(σ(θ)) = k r(θ)2 cos2 θ + r(θ)2 sin2 θ = kpr(θ) = kpR(1 + cos θ),

the mass is deﬁned to be 2π π 0 0 M = ρ d` = ρ(σ(θ)) · σ (θ) dθ = 2 ρ(σ(θ)) · σ (θ) dθ ˆC ˆ ˆ 0 0 π π √ √ √ = 23/2kR R 1 + cos θ · 1 + cos θ dθ = k(2R)3/2 (1 + cos θ) dθ ˆ ˆ 0 0 h iπ = k(2R)3/2 θ + sin θ = πk(2R)3/2. 0 The mass of the cardioid if the density at each point is proportional to the square root of the distance from the origin, being k the constant, is πk(2R)3/2. 7. Mass of the arc of the conic helix parameterized by

σ(t) = Ret cos t, Ret sin t, Ret

that joins the points A = (R, 0,R) and B = (0,Reπ/2,Reπ/2) if its linear density is given by ρ(σ(t)) = ket, being k > 0.

Resolution: In this case we see that 0 ≤ t ≤ π/2, as σ(0) = (R, 0,R) = A and σ(π/2) = (0,Reπ/2,Reπ/2) = B.

6 Arc of the conic helix between A and B for R= 1.

The conic helix is a 3D curve and its mass is: π/2 0 M = ρ dl = ρ(σ(t)) σ (t) dt. ˆ ˆ C 0 We are already given ρ(σ(t)) so now we must ﬁnd kσ0(t)k: σ0(t) = (Ret cos t − Ret sin t, Ret sin t + Ret cos t, Ret), q 0 (( (( σ (t) = R2e2t(cos2 t − (2 cos((t sin t + sin2 t + sin2 t + (2 sin((t cos t + cos2 t + 1) √ = Ret 3. Now we can ﬁnally compute its mass:

π/2 π/2 √ 0 2t M = ρ dl = ρ(σ(t)) σ (t) dt = kR 3 e dt ˆ ˆ ˆ C 0 0 √ π/2 √ √ kR 3 kR 3 t=π/2 kR 3 = 2e2t dt = e2t = (eπ − 1) . 2 ˆ 2 t=0 2 0 8. Mass of the semi-circle of radius R if its linear density is proportional to the cube of the distance from the line that divides the circle into two halves, being k > 0 the constant.

Resolution: To compute the mass of the semicircle we compute the following integral ρ dl, where ˆC ρ is the linear density of the semicircle C. The statements says that it is proportional to the cube of the distance that divides the semicircle into two halves, so, we have to ﬁnd that distance d to compute the density. d = R sin θ so that ρ = kR3 sin3 θ. We parameterize our semicircle in polar coordinates, knowing that 0 ≤ θ ≤ π: σ(θ) = (R cos θ, R sin θ), σ0(θ) = (−R sin θ, R cos θ), √ 0 p σ (θ) = (−R sin θ)2 + (R cos θ)2 = R2 = R.

7 We now proceed to the computation of the mass of the semi-circle π π 3 1 ρ dl = kR3 sin3 θ R dθ = kR4 sin θ − sin 3θ dθ ˆC ˆ0 ˆ0 4 4 θ=π! 4 3 θ=π 1 cos 3θ = kR [− cos θ]θ=0 + = 4 4 3 θ=0 3 1 4 = kR4 − = kR4. 2 6 3 4 Finally, we get that the mass of the semicircle is kR4, being R the radius and k > 0 the constant. 3 9. Area and mean height of a fence with a base on a quarter of an astroid inside a circle of radius 30 parameterized by σ(t) = (30 cos3 t, 30 sin3 t), 0 ≤ t ≤ π/2, and whose height is given by the function h(x, y) = 1 + y/3.

Resolution: An astroid is a hypocycloid with 4 cusps: the trace of a ﬁxed point on a small circle of radius r that rolls within a larger circle of radius 4r.

Astroid x2/3 + y2/3 = 302/3

It satisﬁes the equation x2/3 + y2/3 = a2/3 or (x1/3)2 + (y1/3)2 = (a1/3)2, so, similarly to the case of a circle, it is readily parameterized by x1/3 = a1/3 cos t, y1/3 = a1/3 sin t or x = a cos3 t, y = a sin3 t. In this exercise, a = 30, and we deal only with a quarter of an astroid, so we use the parameterization σ(t) = (30 cos3 t, 30 sin3 t), 0 ≤ t ≤ π/2. From σ0(t) = (3 · 30 cos2 t (− sin t), 3 · 30 · sin2 t (cos t)) = 90 sin t cos t(− cos t, sin t), we have that kσ0(t)k = 90 |sin t cos t| = 90 sin t cos t, since 0 ≤ t ≤ π/2, and h(σ(t)) = 1 + 10 sin3 t. The area of the fence is simply the integral of h along C = {σ(t), t ∈ [0, π/2]}, that is π/2 π/2 0 3 A(C) := h d` = h(σ(t)) σ (t) dt = (1 + 10 sin t) 90 sin t cos t dt ˆC ˆ0 ˆ0 π/2 π/2 ! = 90 sin t cos t dt + 10 sin4 t cos t dt ˆ0 ˆ0 ! sin2 tt=π/2 sin5 tt=π/2 1 = 90 + 10 = 90 + 2 = 225. 2 t=0 5 t=0 2

8 h d` ˆ A(C) 225 The mean height h = C = = = 5, since L(C) 45 d` ˆC

π/2 L(C) = d` = 90 sin t cos t dt = 45. ˆC ˆ0

10. Compute the circulation x dy − y dx, being C the positively oriented (traveled counter-clockwise) ˛C boundary of a square D with side 2L and center (x0, y0).[ Obs.: This problem allows us to motivate Green’s theorem.]

4 [ Resolution: We have a piecewise smooth curve C = C ∪ C ∪ C ∪ C , so that = . 1 2 3 4 ˛ ˆ C i=1 Ci

x0+L On C1, x0 − L ≤ x ≤ x0 + L, y = y0 − L, so x dy − y dx = − y0 − L dx = −2L(y0 − L), ˆC1 ˆx0−L y0+L on C2, x = x0 + L, y0 − L ≤ y ≤ y0 + L, so x dy − y dx = x0 + L dy = 2L(x0 + L), ˆC2 ˆy0−L x0−L on C3, x0 − L ≤ x ≤ x0 + L, y = y0 + L, so x dy − y dx = − y0 + L dx = 2L(y0 + L), ˆC3 ˆx0+L y0+L on C4, x = x0 + L, y0 − L ≤ y ≤ y0 + L, so x dy − y dx = x0 + L dy = −2L(x0 − L). ˆC4 ˆy0−L

Domain D with positive oriented boundary

2 1 Summing up, x dy − y dx = 8L , which is just the double of the area of the square: 2 x dy − ˛C ˛∂D y dx = dx dy, which is a consequence of Green’s theorem. ¨D

9 11. Circulation of the vector ﬁeld F (x, y, z) = (z, x, y) along the curve C parameterized by σ(t) = (t, t2, t3), 0 ≤ t ≤ 1.

Resolution: In this exercise, we will apply just the definition of the circulation of a vector field along a curve: 1 hF , dì = hF (σ(t)), σ0(t)i dt. ˆC ˆ0 We need to compute the scalar product of the vector field evaluated on the parameterized curve and the derivative of the parameterization

F (σ(t)) = (t3, t, t2), σ0(t) = (1, 2t, 3t2), hF (σ(t)), σ0(t)i = t3 + 2t2 + 3t4.

Let’s do the computations to get the ﬁnal result.

1 t4 2t3 3t5 t=1 1 2 3 91 hF , d`i = t3 + 2t2 + 3t4 dt = + + = + + = . ˆC ˆ0 4 3 5 t=0 4 3 5 60 As we are given the parameterization, we are also given the orientation of the curve C, which does need to be checked.

The integral theorems I

12. Compute the circulation of the vector ﬁeld: (a) F (x, y, z) = (2xy + z3, x2, 3xz2) along the helix parameterized by √ σ(t) = (cos t2, sin t2, t2), 0 ≤ t ≤ π.

(b) F (x, y, z) = (ey, xey, 2z) along the curve parameterized by

t 2 σ(t) = 1 + t eu du, t, t2 , 0 ≤ t ≤ 1. ˆ1

Resolution: (a) In order to compute this circulation, ﬁrst we will check if F is conservative. This happens when ∇ ∧ F = 0 and F ∈ C1.

i j k ∂ ∂ ∂ 2 2 ∇ ∧ F = = (0 − 0), (3z − 3z ), (2x − 2x) = 0. ∂x ∂y ∂z 2xy + z3 x2 3xz2

Also, F ∈ C1 because all of its components are continuously differentiable (polynomials and exponentials). Since both conditions meet, we can find a potential f for F such that F = ∇f. We will use Newton-Leibnitz Theorem to find the result:

b b b hF , dli = hF (σ(t)), σ0(t)idt = h∇f(σ(t)), σ0(t)idt = (f ◦ σ)0(t)dt = f(σ(b)) − f(σ(a)). ˆσ â â â This shows that the circulation of F doesn’t depend on the path, but only on the initial and final points. Now, we will search for a function f that satisfies: ∂f ∂f ∂f = 2xy + z3, = x2, = 3xz2. ∂x ∂y ∂z

10 For this, we will integrate one by one the components to ﬁnd the common and non-common results:

I = 2xy + z3dx = x2y + z3x + K , x ˆ 1 I = x2dy = x2y + K , y ˆ 2 I = 3xz2dz = z3x + K . z ˆ 3 Out of all these results, we can conclude that f(x, y, z) = x2y + z3x + K. We will now substitute in our original formula: √ hF , dli = f(σ( π)) − f(σ(0)) = f(−1, 0, π) − f(1, 0, 0) = (−π3) − (0) = −π3. ˆσ (b) First, we check if F is conservative, that is if ∇ ∧ F = 0 and F ∈ C1.

i j k ∂ ∂ ∂ y y ∇ ∧ F = = (0 − 0, 0 − 0, e − e ) = (0, 0, 0) = 0. ∂x ∂y ∂z ey xey 2z

And F ∈ C1 because all the components of F are continously diferenciable (since they are polinomials and exponentials which do not cancel out at any point). As F is conservative, we will look for its potential function f so that F = ∇f, and by using Newton-Leibnitz Theorem we know:

b b b hF , dì = hF (σ(t)), σ0(t)i dt = h∇f(σ(t)), σ0(t)i dt = (f ◦ σ)0(t) dt = f(σ(b)) − f(σ(a)), ˆσ â â â this means that the circulation of F does not depend on the path, but only on the initial and final points. Then, we look for a function f(x, y, z) which satisfies: ∂f ∂f ∂f = ey, = xey, = 2z. ∂x ∂y ∂z We begin by integrating each of the components by the different variables and getting the common and non-common results from each of them but just once (non repeated):

I = ey dx = xey + K x ˆ 1 I = xey dy = xey + K y ˆ 2 I = 2z dz = z2 + K z ˆ 3 So we have that F = ∇f with f(x, y, z) = xey + z2 + K, where the K’s are constants. Therefore,

hF , d`i = f (σ (1))) − f (σ(0)) = f(1, 0, 1) − f(1, 0, 0) = (e + 1 + k) − (1 + k) = e, ˆσ as 1 2 σ(1) = 1 + eu du, 1, 1 = (1, 1, 1) , ˆ1 0 2 σ(0) = 1 + 0 eu du, 0, 1 = (1, 0, 0) . ˆ1

11 13. Prove, using Green’s theorem, the area of the cycloid of radius R is 3 times the area of the circle of also radius R.

Resolution: The cycloid Ccyc is parameterized by σ(t) = (R(t − sin t),R(1 − cos t)), 0 ≤ t ≤ 2π, but

with an opposite orientation to the curve C1 of the ﬁgure. To compute A(D) = dx dy, we look ¨D ﬁrst for functions P , Q, such that Qx − Py = 1, for instance P (x, y) = −y, Q(x, y) = 0, and we apply Green’s theorem to the domain D and its boundary C = ∂D = C0 ∪ C1, counter-clockwise oriented with respect to D:

2π A(D) = −y dx = − y dx − y d¨¨x = y dx = R(1 − cos t) R(1 − cos t) dt ‰ ˆ ˆ ¨ ˆ ˆ C=∂D C1 C0⊂{y=0} Ccyc 0 | {z } | {z } y(t) x0(t) dt 2π 2π 2π 3 cos 2t = R2 (1 − cos t)2 dt = R2 1 − 2 cos t + cos2 t dt = R2 − 2 cos t + dt ˆ ˆ ˆ 2 2 0 0 1| {zcos} 2t 0 2 + 2 " #t=2π 3 sin 2 t = R2 t − 2 sin t + = 3πR2. 2 4 t=0

14. Prove, using Green’s theorem, that the area of the rose given in polar coordinates by the equation r = R cos kθ is the half (respectively, the quarter) of the area of the circle of radius R when k is an even number (respectively, when k is odd). [Obs.: There are two interpretations of the meaning of the equation r = g(θ). In the ﬁrst one, the distance from the origin cannot be negative, so the curve is formed by the points that in polar coordinates are those that meet r = g(θ) > 0. In the second one, we can parameterize the curve through the application σ(θ) = (g(θ) cos θ, g(θ) sin θ) and consider it well deﬁned even when g(θ) < 0.] Resolution: We use the Green Theorem to show that we can compute the area computing F through the length of a curve. Using F (x, y) = (−y, x):

1 A(D) = x dy − y dx. 2 ‰∂D

We will parameterize as follows:

σ(θ) = (x(θ), y(θ)) = (R cos kθ cos θ, R cos kθ sin θ) ,

which is in principle 2π-periodic: σ(θ + 2π) = σ(θ) for any θ, and therefore in principle one considers integrals from 0 to 2π to compute lengths, etc. So now we put our parameterization in the vector ﬁeld and derive the parameterization:

F (σ(θ)) = (−R cos(kθ) sin(θ),R cos(kθ) cos(θ))

σ0(θ) = (−Rk sin(kθ) cos(kθ) − R cos(kθ) sin(θ), −Rk sin(kθ) sin(θ) + R cos(θ)R cos(θ)) From rose in Wikipedia:

How the parameter k affects shapes: In the form k = n, for integer n, the shape will appear similar to a flower. If n is odd, half of these will overlap, forming a flower with n petals. However, if n is even, the petals will not overlap, forming a flower with 2n petals.

12 This simply happens because, although the parameterization in cartesian coordinates is in principle 2π-periodic, in some cases it is actually π-periodic. Indeed, using that for any θ

cos k(θ + π) = cos(kθ + kπ)) = cos kθ cos kπ − sin kθsinkπ = (−1)k cos kθ as well as cos(θ + π) = − cos θ, sin θ + π = sin θ cos π + cos θsinπ = − sin θ, we get

x(θ + π) = cos k(θ + π) cos θ + π = (−1)k+1 cos kθ cos θ, y(θ + π) = cos k(θ + π) sin θ + π = (−1)k+1 cos kθ sin θ, so that σ(θ + π) = (−1)k+1σ(θ). Therefore for odd k σ is already π-periodic, and we have to consider it only deﬁned for 0 ≤ θ ≤ π, because it overlaps itself from π to 2π. Therefore, when k is even and k 6= 0 (for k = 0, we have a circle, and the formula is diﬀerent),

D = {σ(θ): 0 ≤ θ ≤ 2π}, so that 1 1 2π A(D) := hF , d`i = −R cos(kθ) sin(θ)(−Rk sin(kθ) cos(kθ) − R cos(kθ) sin(θ)) 2 ‰C 2 ˆ0 + R cos(kθ) cos(θ)(−Rk sin(kθ) sin(θ) + R cos(θ)R cos(θ)) dθ 1 2π = R2 cos2(kθ) sin2(θ) + R2 cos2(kθ) cos2(θ) dθ. 2 ˆ0 Taking common factor and the constants out and knowing that cos2(θ) + sin2(θ) = 1 we obtain this integral: R2 2π A(D) = cos2(kθ) dθ. 2 ˆ0 1 cos(2θ) Using that cos2(θ) = + , 2 2

R2 2π 1 cos(2kθ) R2 θ sin(2kθ)θ=2π πR2 A(D) := + dθ = + = 2 ˆ0 2 2 2 2 4k θ=0 2

Polar plot of one rose with 8 petals.

13 When k is odd and k 6= 0: D = {σ(θ): 0 ≤ θ ≤ 2π} The procedure it’s the same, however, the limits of integration have being changed:

R2 π 1 cos(2kθ) R2 θ sin(2kθ)θ=π πR2 A(D) := + dθ = + = 2 ˆ0 2 2 2 2 4k θ=0 4

Polar plot of one rose with 7 petals.

We have proved that the area of the rose given in polar coordinates by the equation r = R cos(kθ) is the half (respectively, the quarter) of the area of the circle of radius R when k is an even number (respectively, when k is odd). 15. We consider the vector ﬁeld F = (y, 0) and the circle C of radius 1 centered in the origin, oriented counter-clockwise. (a) Deduce, by graphical reasoning, what is the sign of circulation hF , dli. ˛C (b) Calculate the circulation using Green’s Theorem. Resolution: (a) First, let us write the circle C equation, centered at (0, 0) and of radius 1 and oriented counter-clockwise: C = {x2 + y2 = 1}.

The circulation of a vector field around a curve is equal to the line integral of the vector field around the curve. We can see in the graph that the vector field points in the opposite direction of the orientation of the curve C. Consequently, the sign of the circulation is negative.

Graphic representation of the circulation hF , dli along ˛C C: the blue arrows indicate the vector ﬁeld and the circulation is in black.

14 (b) Now, let us compute the circulation. On the one hand, we can compute the circulation using Green’s theorem since the original integral was a line integral ∂Q ∂P hF , d`i = P dx + Q dy = − dx dy, ‰C ‰C ¨D ∂x ∂y In other words, we will change the integral into a double integral considering that the partial derivatives of (P (x, y),Q(x, y)) = F (x, y) = (y, 0) are:

∂Q ∂P ∂Q ∂P = 0, = 1 =⇒ − = −1. ∂x ∂y ∂x ∂y Therefore, hF , d`i = (−1) dx dy. ‰C ¨D

We notice that dx dy is the area of the disk D of radius r = 1, which is A(D) = πr2 = π, therefore ¨D

hF , d`i = − dx dy = −π. ‰C ¨D On the other hand, we can compute the circulation by deﬁnition:

b hF , d`i = hF (σ(t)), σ0(t)i dt. ‰C ˆa We parameterize the circle C with the standard counter-clockwise parameterization

(x, y) = σ(t) = (cos t, sin t), 0 ≤ θ ≤ 2π,

Its derivative is σ0(t) = (− sin t, cos t) If we evaluate the vector ﬁeld in this concrete parameterization,

F (σ(t)) = (sin t, 0).

So,

2π 2π hF , d`i = h(sin t, 0), (− sin t, cos t)i dt = − sin2 t dt ‰C ˆ0 ˆ0 2π 2π 1 cos 2t = − cos2 t − cos 2t dt = − − − cos 2t dt ˆ0 ˆ0 2 2 2π 1 cos 2t sin 2t − 2t2π = − − = = −π. ˆ0 2 2 4 0

16. Verify Greens theorem with the vector ﬁeld F (x, y) = (2x3 − y3, x3 + y3) and the annulus

2 2 2 2 2 D = (x, y) ∈ R : a ≤ x + y ≤ b .

15 Domain D of Exercise 16, with the boundary oriented for Green’s theorem.

Resolution: Green’s theorem gives the relationship between a double integral over a planar region D and the line integral around its boundary ∂D. If Q and P are functions of (x, y) deﬁned on an open region containing D and having continuous partial derivatives there, then

P dx + Q dy = Qx − Py dx dy, ‰∂D ¨D where the path of integration along the boundary ∂D is positive, that is, counter-clockwise with respect to D (leaves the region D on the left side). 2 2 2 Left hand of the formula: Consider ﬁrst a circle CR = (x, y): x + y = R parameterized by (x, y) = σ(θ) = (R cos θ, R sin θ), (x0, y0) = σ0(θ) = (−R sin θ, R cos θ), 0 ≤ θ ≤ 2π, which is a counter-clockwise parameterization. Then for P (x, y) = 2x3 − y3, Q(x, y) = x3 + y3 2π P dx + Q dy = P (x(θ), y(θ))x0(θ) + Q(x(θ), y(θ))y0(θ) dθ ‰CR ˆ0 2π = 2R3 cos3 θ − R3 sin3 θ (−R sin θ) + R3 cos3 θ + R3 sin3 θ R cos θ dθ ˆ0 2π (( (( 4 ((3 ( 4 4 4 4 4 3(( = (−2(R(cos θ sin θ + R sin θ + R cos θ + (R(sin( θ cos θ dθ ˆ0 2π 3π = R4 1 − 2 sin2 θ cos2 θ dθ = R4. ˆ0 2

To compute now the circulation of F = (P,Q) along the boundary of D, we notice that ∂D = Cb ∪ Ca, where Cb must have a counter-clockwise orientation whereas Ca must have a clockwise orientation. Therefore 3π P dx + Q dy = P dx + Q dy − P dx + Q dy = (b4 − a4). ‰ ‰ ‰ ∂D Cb Ca 2 Right hand of the formula: If we introduce polar coordinates (x, y) = T (r, θ) = (r cos θ, r sin θ} which satisfy |det DT (r, θ)| = r, then D = T (R), where R = {(r, θ): 0 ≤ θ ≤ 2π, a ≤ r ≤ b}. Therefore 2π b 4 4 2 2 2 b − a 3π 4 4 Qx − Py dx dy = 3x + 3y dx dy = 3 dθ r · r dr = 6π = (b − a ). ¨D ¨D ˆ0 ˆa 4 2

16 3π Result: We have checked that both sides of Green’s theorem give the same result (b4 − a4). Hence 2 Green’s theorem has been veriﬁed.

2 17. Let C be the boundary of the domain with holes D ⊂ R obtained removing three disks of radii 1 with centers at (−3, 0), (0, 0) and (3, 0) from the disk of radius 10 centered at (1, 0). Compute the circulation 2 y + x3 cos x2 dx + ey + 2x dy ˛C orienting clockwise the largest circle of C and counter-clockwise the three smallest ones. Resolution: For the domain D with the boundary C = ∂D with positive orientation with respect to D, that is, leaving region D on the left side, we can apply Green’s theorem

2 y + x3 cos x2 dx + ey + 2x dy = 2 − 1 dx dy = A(D) = 100π − 3π = 97π. ‰∂D ¨D But we are given a circulation of the four circles of the boundary just in the opposite direction; 2 therefore y + x3 cos x2 dx + ey + 2x dy = −97π. ˛C 18. The polar planimeter is a mechanical instrument that allows us to measure two areas of planar domains (see Wikipedia and a Java app). It has the shape of a ruler with two arms (see the ﬁgure). One of them connects the ﬁxed extreme O = (0, 0) with a mobile point E = (a, b), and the other one connects the point E with another mobile point M = (x, y). In the extreme M there is a small wheel which is perpendicular to the arm EM. We suppose that the arms OE and EM have lengths l and L, respectively. So the point M determines the point E as the intersection between the circle of radius l centered in O and the one with radius L centered in M. This intersection is unique if the angle between both arms is lower to π rad. Therefore, we can consider that a = a(x, y) and b = b(x, y).

Sketch of a polar planimeter.

2 The method to determine the area of the planar simply connected domain S ⊂ R consists in moving the point M over the boundary C = ∂S in a counter-clockwise direction. The wheel in M measures the displacement in the orthogonal direction with respect to the arm. In this problem, we will see that the searched area is L times the total displacement of the wheel.

2 2 2 2 (a) Prove that ax + by ≡ 1 computing the derivatives of the equations a + b ≡ l and (x − a) + (y − b)2 ≡ L2. (b) F (x, y) = (P (x, y),Q(x, y)) is the vector of norm L which is orthogonal to the arm EM in the extreme M = (x, y), given by P (x, y) = b(x, y) − y and Q(x, y) = x − a(x, y). Prove that

hF , d`i = Area(D). ˛C+

17 (c) d is the covered distance by the wheel at the moment it has completed the itinerary along the curve C = ∂D counter-clockwise. Argue that

hF , d`i = Ld. ˛C+

Resolution: (a) We start the proof by diﬀerentiating the equations with respect to x and also with respect to y. From a2 + b2 ≡ l2 we get ∂ ∂ (a2 + b2) = (l2) =⇒ aa + bb = 0, ∂x ∂x x x ∂ ∂ (a2 + b2) = (l2) =⇒ aa + bb = 0. ∂y ∂y y y

From (x − a)2 + (y − b)2 ≡ L2 we get

∂ ∂ (x − a)2 + (y − b)2 = (L2) =⇒ (x − a)(1 − a ) + (y − b)(−b ) = 0, ∂x ∂x x x ∂ ∂ (x − a)2 + (y − b)2 = (L2) =⇒ (x − a)(−a ) + (y − b)(1 − b ) = 0. ∂y ∂y y y

Isolating the term (x − a) in the last two equations, we can equalize them and we get −(y − b)(−bx) −(y − b)(1 − by) (−ay) −(y− b)(1 − by) = =⇒ = =⇒ (1 − ax) (−ay) (1 − ax) −(y− b)(−bx)

=⇒ aybx = (1 − ax)(1 − by) =⇒ aybx = 1 − ax − by + axby =⇒ ax + by = 1 + axby − aybx.

Now that we have isolated the term ax + by, we can substitute ay and bx with the expressions we find −aa −bb from isolating those terms in the first two equations we found previously: b = x and a = y . x b y a So, we get bb aa ¡ y ¡ x ¨¨ ¨¨ ax + by = 1 + axby − = 1 + äxby − äxby = 1. a¡ ¡b Indeed, we have proven that ax + by = 1.

(b) According to Green’s theorem,

hF , d`i = P dx + Q dy = (Qx − Py) dx dy. ˛C+ ˛C+ ¨D

Since Area(D) = dx dy, that means that necessarily (Qx − Py) = 1. Let us check this: ¨D ∂Q ∂P Q = = 1 − a and P = = b − 1, x ∂x x y ∂y y so (Qx − Py) = 1 − ax − by + 1 = 2 − (ax + by) = 2 − 1 = 1, where we have used that ax + by = 1. Therefore, since (Qx − Py) = 1, we have proven that

hF , d`i = dx dy = Area(D). ˛C+ ¨D

18 (c) We note that the vector ﬁeld F is parallel to the tangent vector to the curve and we are told that it has constant norm L. Then,

hF , d`i = FT d` = L d` = Ld, ˛C+ ˛C+ ˛C+

where we have used that d` = d, since we are told that d is the distance traveled by the wheel ˛C+ along the curve C = ∂D (counter-clockwise).

Surface integrals

19. Area of the helicoid parameterized by

ϕ(r, θ) = (r cos θ, r sin θ, hθ), (r, θ) ∈ D = [0,R] ∧ [0, 2π].

Resolution: We start by computing the associated normal vector to this parameterization of the helicoid:       cos θ −r sin θ i j k h sin θ

ϕr = sin θ , ϕθ =  r cos θ  , ϕr ∧ ϕθ = cos θ sin θ 0 = −h cos θ ,

0 h −r sin θ r cos θ h r √ 2 2 from which it follows that kϕr ∧ ϕθk = h + r . Once we have the norm of this normal vector, we can now substitute it in the integral in order to ﬁnd the area:

p 2π R A = h2 + r2 dr dθ = dθ h2 + r2 dr ¨D ˆ0 ˆ0 √ √ r=R  2 2 2 2 2  R p r h + r + h ln h + r + r = 2π h2 + r2 dr = 2π   ˆ0 2 r=0 p 2 2 2 p 2 2 p2 2 2 p 2 2 = π R h + R + h ln h + R + R − 0 · h + 0 + h ln h + 0 + 0 p p √ = π R h2 + R2 + h2 ln h2 + R2 + R − h2 ln( h2) √ !! p h2 + R2 + R = π R h2 + R2 + h2 ln . h

20. Area of the surface S formed by the piece of the plane x + y + z = 1 that is over the ellipse 2 2 2 D = {(x, y) ∈ R : x + 2y ≤ 1}.

Resolution: We can isolate locally one variable from x + y + z = 1, for instance z, to get z = f(x, y), and use the cartesian parameterization ϕ(x, y) = (x, y, z(x, y)), with z = f(x, y) = 1−x−y and (x, y) 2 in a domain D ∈ R such that S = ϕ(D). Computing,       1 0 i j k −(−1)

ϕx =  0  , ϕy =  1  , ϕx ∧ ϕy = 1 0 −1 = −(−1) ,

−1 −1 0 1 −1 1 √ p 2 2 ϕx ∧ ϕy = 1 + (−1) + (−1) = 3.

19 √ In this case we have√ that the interior of an ellipse with semi-axes a = 1 and b = 1/ 2 and and therefore A(D) = πab = π/ 2, so that r √ √ π 3 A(S) = ϕx ∧ ϕy dx dy = 3A(D) = 3√ = π . ¨D 2 2

21. The solid angle of a portion S of the sphere of radius R is Ω(S) := Area(S)/R2, measured in sr ( steradians). Particularly, the complete sphere has 4πsr. (a) SR,α is the portion of the sphere with radius R which is inside the cone with vertex in the centre of the sphere and semi-angle α. For what values of the semi-angle α is Ω(SR,α) = 1? (b) Prove that the area of the spherical cap of radius R and height h < R deﬁned by

3 2 2 2 2 CR,h = {(x, y, z) ∈ R : x + y + z = R , z ≤ R − h} is the same that the area of the lateral of a radius R and height h cylinder. Resolution: (a) We know that the solid angle of a portion a portion S of the sphere is

Area(S) Ω(S) := , R2 and we want to ﬁnd α such that Ω(SR,α) = 1, so we have Area(S ) R,α = 1. R2 First of all, we have to parameterize the sphere with the restrictions of the cone. Therefore, we will use the parameterization

ϕ(u, v) = (R cos u cos v, R sin u cos v, R sin v).

We will calculate the area of the portion SR,α = ϕ(DR,α) of the sphere as

Area(SR,α) = kϕu ∧ ϕvk du dv. ¨DR,α Let us compute the partial derivatives of the parameterization. ∂ϕ ϕ = (u, v) = (−R sin u cos v, R cos u cos v, 0) = R cos v(− sin u, cos u, 0), u ∂u ∂ϕ ϕ = (u, v) = (−R cos u sin v, −R sin u sin v, R cos v) = R(− cos u sin v, − sin u sin v, cos v). v ∂v Now we calculate the cross product of the partial derivatives and its norm

2 ϕu ∧ ϕv = R cos v(cos u cos v, sin u cos v, sin v), q 2 2 2 2 2 2 p 2 2 2 kϕu ∧ ϕvk = R cos v cos v(sin u + cos u) + sin v = R cos v cos v + sin v = R cos v.

Once we have found the norm of the normal vector, we have to ﬁnd the domain DR,α to have SR,α = ϕ(DR,α). Following the spherical change, we know that 0 ≤ u ≤ 2π, and regarding at the restrictions π π of the cone, we can conclude that − α ≤ v ≤ so that 2 2 n π π o D = (u, v): 0 ≤ u ≤ 2π, − α ≤ v ≤ . R,α 2 2

20 Finally, we can calculate our integral 2π π 2 2 2 v=π/2 Area(SR,α) = kϕu ∧ ϕvk dv du = du R cos v dv = 2πR [sin v]v=π/2−α ¨ ˆ ˆ π S 0 2 −α π = 2πR2 1 − sin − α = 2π(1 − cos α)R2. 2 Once we have found the area of our surface in function of the angle α, let us compare with the equation Area(S ) we had for the solid angle. From R,α = 1 we get R2 2π(1 − cos α)R2 1 1 1 = 1 ⇐⇒ 1 − cos α = ⇐⇒ cos α = 1 − ⇐⇒ α = arccos 1 − . R2 2π 2π 2π

In conclusion, when α = arccos(1 − 1/2π), the solid angle of the portion of the sphere SR,α is 1. (b)

It is enough to observe that the spherical cap CR,h equals to a portion of SR,h being h = (1 − cosα)R. It is easier to understand it trhough an image of a spherical cap as a portion of a sphere, as the one on the right. Spherical cap of radius R and height h. 22. Area and volume of the intersection of two cylinders with radius R whose axis are cut perpendicularly (plot done taking R=1). (This is the Steinmetz solid: see Wikipedia and MathWorld.)

Resolution:

Intersection of two cylinders with radius R whose axis are Steinmetz solid (source Wikipedia), from exer- cut perpendicularly, from exercise 22. cise 22. To do this exercise, we are going to split it in two parts: part 1 to solve the area and part 2 to solve the volume. But ﬁrst, we have to evaluate the data given. We have two cylinders represented which create a solid (the Steinmetz solid) represented in the ﬁgure on the right side. These cylinders are: 2 2 2 C1 : x + y = R 2 2 2 C2 : x + z = R

21 Area: To calculate the surface area of a curve, we know that A(W ) = kϕx ∧ ϕyk dx dy, and to do ˆW this, we need to parameterize one of the two equations, using cartesian parameterization, and calculate the vector product of its derivatives, and then ﬁnd the boundaries of the other equation. First, we compute the parameterization and the vector product: p x2 + z2 = R2 =⇒ z = ± R2 − x2, p ϕ(x, y) = x, y, ± R2 − x2 , −2x ϕ = 1, 0, ± √ and ϕ = (0, 1, 0), x 2 R2 − x2 y x ϕ ∧ ϕ = ∓√ , 0, 1 . x y R2 − x2

Now, we calculate the integration limits, by looking at the boundaries. Taking C2 as the surface area calculated, C1 has to be the frontier in which C2 starts and end. From the ﬁgures below, ﬁxing the variable x, the limits for C2 are:

−R ≤ x ≤ R p p − R2 − x2 ≤ y ≤ R2 − x2

Now, we can compute the surface area of the intersection of the two cylinders. Below, we take into account only z > 0. √ s R R∗2−x2 2 x 2 2 A(C2 > 0) = kϕx ∧ ϕyk dx dy = √ −√ + 0 + 1 dy dx = 2 2 2 ˆC2>0 ˆ−R ˆ− R∗2−x R − x √ √ 2 r 2 r R R∗2−x x2 R R∗2−x x2 + R2 − x2 = √ 2 2 + 1 dy dx = √ 2 2 dy dx = ˆ−R ˆ− R∗2−x2 R − x ˆ−R ˆ− R∗2−x2 R − x √ 2 R R∗2−x R R R p R = √ dy dx = √ 2 R2 − x2 dx = 2R dx = √ 2 2 2 2 ˆ−R ˆ− R∗2−x2 R − x ˆ−R R − x ˆ−R = 2R · 2R = 4R2.

2 4R is the surface area of the cylinder C2 for z > 0. So the total surface area for the cylinder C2 is twice this area, for positive and negative z: 2 · 4R2 = 8R2. However, this is not the total surface area of the Steinmetz solid. Because of the symmetry of the solid, the total surface area would be twice the area of one cylinder, that is:

A(Steinmetz solid) = 2 · 8R2 = 16R2

Volume: For the volume, we can find the limits of integration by splitting the plot in two planes: Fixing the x variable, and isolating the y and the z variables in each circumference equation from figures above, we can find the integration limits:

R ≤ x ≤ R p p − R2 − x2 ≤ y ≤ R2 − x2 p p − R2 − x2 ≤ z ≤ R2 − x2

So, the volume of the intersection of two cylinders with radius R whose axis are cut perpendicularly

22 Plane xy of the intersection of the Steinmetz solid (Tak- Plane xz of the intersection of the Steinmetz solid (Tak- 2 ing R = 1), from exercise 22. ing R2 = 1), from exercise 22.

is: √ √ R R2−x2 R2−x2 R p 2 2 p 2 2 V (Steinmetz solid) = √ √ 1 dz dy dx = 2 R − x · 2 R − x dx = ˆ−R ˆ− R2−x2 ˆ− R2−x2 ˆ−R R x3 R R3 −R3 = 4(R2 − x2) dx = 4 · R2x − = 4 · R3 − − −R3 − = ˆ−R 3 −R 3 3 2 4 = 4 · 2R3 − R3 = 4 · R3 = 3 3 16 = R3 3

Another easier method to compute the volume is considering that the cross-section Wx for −R ≤ x ≤ R is a square of area A(x) = 4(R2 − x2) so by Cavalieri’s principle,

R R 16 V (Steinmetz solid) = A(x) dx = 4(R2 − x2) dx = R3. ˆ−R ˆ−R 3

From this exercise, we can say that the surface area and the volume of the Steinmetz solid are 16R2 16 3 and 3 R , respectively. 23. Applying the Pappus’ centroid theorem, compute the centroids of the semi-circle and the semi-disk of radius R.

Resolution: The ﬁrst Pappus’ centroid theorem states that the surface area A of a surface of revolution generated by rotating a plane curve C about an axis external e to C and on the same plane is equal to the product of the length L(C) of the curve C and the distance d = 2πdist(c, e) traveled by the geometric centroid c of C: A = L(C)d = 2πdist(c, e)L(C).

23 Therefore for the semi-circle of radius R and center at the origin, its centroid takes the form c = (r, 0). When we rotate around the z-axis, by the ﬁrst Pappus’ centroid theorem, 4πR2 = dπR, where d = 2πr, 2R so that the centroid of the semi-circle is c = , 0 . π The second Pappus’ centroid theorem states that the volume V of a solid of revolution generated by rotating a plane domain D about an external axis e is equal to the product of the area A(D) and the distance d = 2πdist(c, e) traveled by the geometric centroid c of D. (The centroid of D is usually diﬀerent from the centroid of its boundary curve C.) That is

V = A(D)d = 2πdist(c, e)A(D).

For the semi-disk of radius R and center at the origin, its centroid takes the form c = (r, 0). When 4πR3 πR2 we rotate around the z-axis, by the second Pappus’ centroid theorem, = d , where d = 2πr, 3 2 4R so that the centroid of the semi-disk is c = , 0 . 3π 24. Compute the volume V and the area A of the band which is obtained after drilling a ball of radius R with a drill bit (a cylinder) of radius r < R that goes through its center. Check that if h is the height of the band, then V only depends on h (V = πh3/6), whereas A is the same as the sum of the areas of the sides of two cylinders with radius r and R, both with height h (A = 2πh(R + r)) (“ napkin ring problem”).

Resolution:

Side view of the napkin ring for napkin ring problem Cross-section top view for napkin ring problem First we compute the volume of the napkin ring by means of Cavalieri’s principle. To ﬁnd the area A(y) of the horizontal cross-section we subtract the area of the smaller circle with radius r from the area of the bigger circle with radius rb r !2 p 2 h2 A(y) = πr2 − πr2 = π R2 − y2 − π (R2 − b 2 h2 = π R2 − y2 − R2 + 2 h2 = π − y2 2 It can be clearly seen that this area A(y) does not depend on the radius of the sphere R. If we put

24 the axis at the center of the sphere at (0, 0, 0) and we apply Cavalieri’s principle, we can say that

h/2 h/2 h2 V = A(y) dy = π − y2 dy ˆ−h/2 ˆ−h/2 2 h2 y3 h/2 πh3 1 = π y − = 1 − 2 2 −h/2 4 3 πh3 = . 6 To ﬁnd the area of the napkin ring, we will add the area of the sphere and of the cylinder and subtract the area of the spherical caps at the top and bottom.

Diﬀerent areas of the napkin ring for napkin ring problem

The area for the sphere and cylinder are

2 Asphere = 4πR

Acylinder = 2πrh

For the area of the spherical cap we will use the formula related to Exercise V34 s b dr 2 A(S) = 2π r 1 + dz ˆa dz which is valid for any surface of revolution around the z-axis. Indeed, given a surface of revolution S = {(x, y, z): r = g(z) ≥ 0, a ≤ z ≤ b}, where r = px2 + y2, we can parameterize it as S = ϕ(D), 2 by means of the parameterization ϕ(θ, z) = (g(z) cos θ, g(z) sin θ, z), D = {(θ, z) ∈ R : a ≤ z ≤ b, 0 ≤ θ ≤ 2π} introduced in Exercise 34. Writing the formula for the area of this surface, we obtain the dr 0 formula above (writing r = g(z) and dz = g (z)) as

b p 0 2 A(S) = dS = kϕθ ∧ ϕzk dθ dz = 2π g(z) 1 + g (z) dz ¨S ¨D ˆa where  cos θ  ϕθ ∧ ϕz = g(z))  sin θ  −g0(z)

25 and p 0 2 kϕθ ∧ ϕzk = g(z) 1 + g (z) √ For a sphere we have that r = g(z) = R2 − z2 and so

dr z z = −√ = − . dz R2 − z2 r Also r r √ dr 2 z 2 r2 + z2 R 1 + = 1 + − = = dz r r r and so then s b dr 2 R R A(S) = 2π r 1 + dz = 2π r dz ˆa dz ˆh/2 r R = 2πR z h/2 = πR(2R − h) Adding the area of the cylinder and sphere, and subtracting the area of the spherical cap twice (as there are two caps) we get

A = 4πR2 + 2πrh − 2(π2R2 − πRh)) = 2πh(R + r)

25. We consider a sphere with radius 2R and a cylinder which is tangent to the sphere in a point and goes through the center of the sphere. In coordinates, we can describe the sphere as E = x2 + y2 + z2 = 4R2 and the cylinder as Q = (x − R)2 + y2 = R2. (a) The intersection E ∩ Q results in the Viviani’s curve, that has length cR for a constant c. Provide an integral for c (no need to compute it). (b) We consider a sphere with radius 2R and a cylinder which is tangent to the sphere in a point and goes through the center of the sphere. In coordinates, we can describe the sphere as E = x2 + y2 + z2 = 2 2 2 2 + − 4R and the cylinder as Q = (x − R) + y = R . Compute the area of the portions S1 and S1 of the sphere which are contained in the interior of the cylinder (Viviani’s vault). (c) We consider a sphere with radius 2R and a cylinder which is tangent to the sphere in a point and goes through the center of the sphere. In coordinates, we can describe the sphere as E = x2 + y2 + z2 = 4R2 and the cylinder as Q = (x − R)2 + y2 = R2 . Compute the area of the portion S2 of the cylinder contained in the interior of the sphere.

Resolution: (a) We know that the formula to compute the length of a curve C = σ ([a, b]) is

b 0 d` = σ (t) dt ˆC ˆa We have to parameterize the Viviani’s curve E ∩ Q, and there are at least two ways to do it.

26 Viviani’s curve. A cylinder going through the sphere.

The ﬁrst way of parameterization, based on a cylindrical parameterization for Q, is σ(θ) = (R + R cos θ, R sin θ, z), where p p z = ± 4R2 − x2 − y2 = ± 4R2 − R2 − 2R2 cos θ − R2 cos2 θ − R2 sin2 θ q p = ± R2(4 − 1 − 2 cos θ − (cos2 θ + sin2 θ) = ± R2(2 − 2 cos θ) = ±Rp2(1 − cos θ), so that E ∩ Q is a concatenation of two paths σ(θ) = R + R cos θ, R sin θ, ±Rp2(1 − cos θ) , 0 ≤ θ ≤ 2π, with ! s 2 0 R sin θ 0 sin θ σ (θ) = −R sin θ, R cos θ, ± , σ (θ) = R 1 + . p2(1 − cos θ) 2(1 − cos θ) Since kσ0(θ)k is the same for both paths, we can consider only one and multiply by 2 to compute its length: 2π 2π s 2 0 sin θ d` = 2 σ (θ) dθ = 2R 1 + dθ = cR, ˆE∩Q ˆ0 ˆ0 2(1 − cos θ) where s 2π sin2 θ c = 2 1 + dθ. ˆ0 2(1 − cos θ)

The second way of parameterization is based on ﬁrst parameterizing the sphere E by (x, y, z) = (r cos ϕ cos θ, r cos ϕ sin θ, r sin ϕ), for 0 ≤ θ ≤ 2π, −π/2 ≤ φ ≤ π/2, and then substituting in the equation x2 + y2 − 2Rx = 0 of Q, to get cos φ(cos φ − cos θ) = 0 which implies cos φ = cos θ which implies θ = ±φ, because of the trigonometric identity

θ + φ θ − φ cos φ − cos θ = 2 sin sin . 2 2

Again, E ∩ Q a concatenation of two paths θ = ±φ. Choosing for instance θ = φ, we get a parameterization of half Viviani’s curve:

(x, y, z) = σ(φ) = (r cos φ cos φ, r cos φ sin φ, r sin φ), −π/2 ≤ φ ≤ π/2,

27 with 0 2 2 0 p σ (φ) = −4R sin φ cos φ, 2R cos φ sin φ, 2R cos φ , σ (φ) = 2R 1 + cos2 φ. Since kσ0(φ)k is the same for θ = −φ, we can consider only the path θ = φ and multiply by 2 to compute the length of Viviani’s curve E ∩ Q:

2π π 0 2 p d` = 2 σ (φ) dφ = 2 · 2R 1 + cos2 φ dφ = cR, ˆ ˆ ˆ π E∩Q 0 − 2 where π 2 p c = 4 1 + cos2 φ dφ. ˆ π − 2 Computing both integrals with MATLAB, we get a value of c ≈ 15, 28 . Using both parameterizatons is a good way to check whether our result is right or not:

2π s 2 π sin θ 2 p c = 2 1 + dθ = 4 1 + cos2 φ dφ. ˆ 2(1 − cos θ) ˆ π 0 − 2

(b) To compute both surfaces we will use the following formula for a parameterized surface S = ϕ(D):

A(S) = kϕu ∧ ϕvk du dv ¨D First, we need to parameterize the sphere E. To do so, we will use spherical coordinates and the data given in the statement (r = 2R), such that π π (x, y, z) = ϕ(θ, φ) = (2R cos θ cos φ, 2R sin θ cos φ, 2R sin φ), where − ≤ φ ≤ , −π ≤ θ ≤ π. 2 2 Then, we have to diﬀerentiate the parameterization with respect to θ and φ and calculate the norm of their cross products:

ϕθ = (−2R cos φ sin θ, 2R cos θ cos φ, 0),

ϕφ = (−2R cos θ sin φ, −2R sin φ sin θ, 2R cos φ), q 4 4 4 2 2 p 4 2 2 ϕθ ∧ ϕφ = 16R cos φ + 16R cos φ sin φ = 16R cos φ = 4R cos φ. Before calculating the area of the sphere surface we need to know that, when checking if the spherical curve fulﬁlls the cylinder, we get that cos θ = cos φ and therefore, φ = ±θ. therefore, we are going to take the domain of the parameterization ϕ to compute the area as n π π o D = (θ, φ): − ≤ φ ≤ , 0 ≤ θ ≤ φ , 2 2 + which will parameterize only half of the upper part S1 of the sphere’s surface contained in the cylinder, and therefore the area computed with this parameterization will be one half of the upper area of the sphere’s surface contained in the cylinder, what is the same, one quarter of the total surface area inside the cylinder. To understand it better, one can look at the front view in the ﬁgure below.

Viviani’s window seen from diﬀerent perspectives

28 π φ π + 2 2 2 2 2 θ=φ A(S1 ) = 2 4R cos φ dφ dθ = 2 4R cos φ dθ dφ = 8R [cos φθ]θ=0 dφ ¨D ˆ0 ˆ0 ˆ0 π φ= π 2 2 π 2 2 2 φ= 2 = 8R φ cos φ dφ = 8R φ sin φ − sin φ dφ = 8R [φ sin φ + cos φ]φ=0 ˆ0 ˆ φ=0 hπ i π = 8R2 − 0 + 0 − 1 = 8R2 − 1 = 4πR2 − 8R2. 2 2 + − 2 2 By symmetry, we can see that A(S1 ) = A(S1 ) = 4πR − 8R . Hence, the total area of the sphere surface contained in the interior of the cylinder is:

2 2 2 2 A(S1) = 2(4πR − 8R ) = 8πR − 16R .

(c) First of all we must visualize how the sphere and the cylinder intersect with each other. It’s better an image than a thousand words.

Viviani’s curve of Exercise 25c.

Secondly, we shall use the formula to compute the area of a surface S = ϕ(D), that is:

A(S) = kϕu ∧ ϕvk du dv. ¨D We must ﬁrst parameterize the cylinder Q with cylindrical coordinates that will depend on θ and z

(x, y, z) = ϕ(θ, z) = (R + R cos θ, R sin θ, z).

Then, as we made a parameterization that depends on z and θ, we’ll ﬁnd the limits of integration of those. We substitute the parameterization in the equation of the sphere E:

(R + R cos θ)2 + (R sin θ)2 + z2 = 4R2 ⇐⇒ R2 + 2R2 cos θ + R2 cos2 θ + R2 sin2 θ + z2 = 4R2 p ⇐⇒ z2 = 2R2 − 2R2 cos θ ⇐⇒ z = ± 2R2 − 2R2 cos θ.

In the end we have the following domain D of deﬁnition of the parameterization ϕ such that S2 = ϕ(D): n p p o D = (θ, z): 0 ≤ θ ≤ 2π, − 2R2 − 2R2 cos θ ≤ z ≤ 2R2 − 2R2 cos θ .

The next step is to diﬀerentiate the parameterization with respect to θ and z, make their respective cross product and compute its norm.

ϕθ = R(− sin θ, cos θ, 0), ϕz = (0, 0, 1), ϕθ ∧ ϕz = R(cos θ, − sin θ, 0), kϕθ ∧ ϕzk = R.

29 Now we are ready to compute the area of the portion of the cylinder in the interior of the sphere. √ 2π 2R2−2R2 cos θ 2π 2π p √ √ A(S ) = R dz dθ = R 2 2R2 − 2R2 cos θ dθ = R 2 2R 1 − cos θ dθ 2 ˆ ˆ ˆ ˆ √ 0 − 2R2−2R2 cos θ 0 0 2π 2π √ √ θ θ θ θ=2π = R 2 2R 2 sin dθ = 4R2 sin dθ = −8R2 cos dθ = 16R2 , ˆ 2 ˆ 2 2 θ=0 0 0 α 1 − cos α where we used the trigonometric relation sin2 = , which can also be expressed as 2 2 √ √ α 1 − cos α = 2 sin . 2 26. Compute the centroid of the hemisphere (surface) and the semi-ball (solid) of radius R. Resolution: The hemisphere is deﬁned as

3 2 2 2 2 S = (x, y, z) ∈ R : x + y + z = R , z ≥ 0 . In order to compute the centroid of this surface we have to use the formula and the procedure from vector calculus. By looking at the ﬁgure, we can say thatx ¯ andy ¯ will be zero by symmetry, so we only have to calculate thez ¯ coordinate of the centroid:

zρ dS z (Ψ(θ, ϕ)) ρ (Ψ(θ, ϕ)) kΨ ∧ Ψ k dθ dϕ ˆ ¨ θ ϕ z¯ = S = D , ρ dS ρ (Ψ(θ, ϕ)) kΨθ ∧ Ψϕk dθ dϕ ˆS ¨D where Ψ(θ, ϕ) is the parameterization of S, that is, S = Ψ(D). First, let us write the parameterization of the hemisphere in spherical coordinates and ﬁnd its domain:

(x, y, z) = Ψ(θ, ϕ) = (R cos θ sin ϕ, R sin θ cos ϕ, R sin ϕ), (θ, ϕ) ∈ D,

where we need to ﬁnd the domain D to have S = Ψ(D). As the ﬁgure gives a total turn in the xy π axis, θ goes from 0 to 2π, while it only gives a quarter of a turn in the zy axis, so ϕ goes from 0 to 2 : n π o D = 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ . 2

Then, let us compute k Ψθ ∧ Ψϕ k:

Ψθ = R cos ϕ(− sin θ, cos θ, 0), Ψϕ = R(− cos θ sin ϕ, − sin θ sin ϕ, cos ϕ) 2 2 Ψθ ∧ Ψϕ = R cos ϕ(cos θ cos ϕ, sin θ cos ϕ, sin ϕ), kΨθ ∧ Ψϕk = R cos ϕ.

Since there is no density given in the statement, we will assume ρ = 1, so we can apply the formula to ﬁndz ¯: π 2π 2 z(Ψ(θ, ϕ))ρ(Ψ(θ, ϕ)) k Ψ ∧ Ψ k dθ dϕ R sin ϕ R2 cos ϕ dϕ dθ ¨ θ ϕ ˆ ˆ z¯ = D = 0 0 2π π ρ(Ψ(θ, ϕ)) k Ψ ∧ Ψ k dθ dϕ 2 2 ¨ θ ϕ R cos ϕ dϕ dθ D ˆ0 ˆ0 π 2π π 2 h 2 iϕ= R3 dθ sin ϕ cos ϕ dϕ R32π sin ϕ 2 ˆ ˆ 2 R¡3π R = 0 0 = ϕ=0 = = . 2 ϕ=1 2πR2 2 2 R 2π[sin ϕ]ϕ=0 2πR

30 R Therefore the centroid of the hemisphere is 0, 0, . 2 3 The semi-ball of R is deﬁned as 3 2 2 2 2 W = (x, y, z) ∈ R : x + y + z ≤ R , z ≥ 0 In order to compute the centroid of this solid we have to use the formula and the procedure from triple integrals. Like in the previous case, we can see by symmetry thatx ¯ andy ¯ are 0, so we only have to calculatez ¯: ρz dx dy dz ˚ z¯ = W . V (W ) Let us use the change T to spherical coordinates and specify the domain B where W = T (B):

(x, y, z) = T (r, θ, ϕ) = (r cos θ sin ϕ, r sin θ cos ϕ, r sin ϕ), (r, θ, ϕ) ∈ B, |det DT (r, θ, ϕ)| = r2 cos ϕ,

where n π o B = 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ , 0 ≤ r ≤ R . 2 We also need the volume of the semi-ball W . Since it is a half of a ball of radius R, we will use the 2 3 formula for the volume of a ball of radius R divided by 2: V (W ) = 3 πR . Since there is no density given in the statement, we will assume that ρ = 1, so we can now apply the formula to ﬁndz ¯:

π π 2π 2 R ϕ= r=R 2 h sin2 ϕ i 2 h r4 i ρz dx dy dz r sin ϕ r cos ϕ dr dϕ dθ 2π 4 ˚ ˆ ˆ ˆ 2 4 2π 1 R z¯ = W = 0 0 0 = ϕ=0 r=0 = 2 4 = 2 3 2 3 2 3 V (W ) 3 πR 3 πR 3 πR 6R4π 3R = = . 16 8 3R Therefore the centroid of the semi-ball is 0, 0, . 8 27. Total charge of the lateral side of the cone with height h and radius R if the charge density is proportional to the distance from the base, being k > 0 the constant.

Resolution: To calculate the total charge, we have to integrate the charge density and so, we must compute the following integral:

Q = ρ dS = kz dS = kz kϕθ ∧ ϕzk dz dθ, ˆS ˆS ˆD where ρ is the charge density on each point and S = ϕ(D).

Firstly, we obtain the cone by turning, respect to the z axis, the red line in the ﬁgure which is parameterized by

Rz σ(z) = , 0, z , for 0 ≤ z ≤ h. h

31 The parameterization of the cone S is

Rz Rz (x, y, z) = ϕ(θ, z) = cos θ, sin θ, z , (θ, z) ∈ D, h h

where D is the rectangle D = {(θ, z): 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h} .

We proceed now to compute kϕθ ∧ ϕzk: Rz ϕ = (− sin θ, cos θ, 0), θ h R R ϕ = cos θ, sin θ, 1 , z h h Rz ϕ ∧ ϕ = (h cos θ, h sin θ, −R), θ z h2 Rz p kϕ ∧ ϕ k = h2 + R2. θ z h2 We ﬁnally have all the ingredients to compute the total charge:

2π h Rz p kRp h Q = kzkϕ ∧ ϕ k dz dθ = dθ kz R2 + h2 dz = 2π R2 + h2 z2 dz ˆ θ z ˆ ˆ h2 h2 ˆ D √ 0 0 0 2πkRh R2 + h2 = . 3

28. Total charge of the surface √ n 3 2 2 2 2 o S = (x, y, z) ∈ R : z = x + y + a , a ≤ z ≤ a 2

(piece of the hyperboloid of two sheets) if the charge density is e(x, y, z) = kz, with k > 0.

Resolution: Notice that a hyperboloid of two sheets is a surface of revolution. Therefore, we 2 2 2 will parameterize√ it with cylindrical√ coordinates. That is, r = x + y . Then, S is given by 2 2 r = g(z) := z − a for a ≤ z ≤ a 2. So, S can be parameterized√ as S = ϕ(D) for ϕ(θ, z) = 2 (g(z) cos θ, g(z) sin θ, z),D = {(θ, z) ∈ R : 0 ≤ θ ≤ 2π, a ≤ z ≤ a 2}.

We now compute ϕθ and ϕz in order to obtain the normal vector associated to this parameterization.

> ϕθ = (−g(z) sin θ, g(z) cos θ, 0) 0 0 > ϕz = (g (z) cos θ, g (z) sin θ, 1)

Now we compute ϕθ ∧ ϕz:   i j k g(z) cos θ

ϕθ ∧ ϕz = −g(z) sin θ g(z) cos θ 0 =  g(z) sin θ 

g0(z) cos θ g0(z) sin θ 1 −g(z)g0(z) sin2 θ − g(z)g0(z) cos2 θ  g(z) cos θ   cos θ  =  g(z) sin θ  = g(z)  sin θ  . −g(z)g0(z) −g0(z)

32 Therefore, the normal vector is q 2 2 0 2 p 0 2 kϕθ ∧ ϕzk = g(z) cos θ + sin θ + (−g (z)) = g(z) 1 + (g (z)) s 2 r p z p z2 = z2 − a2 · 1 + √ = z2 − a2 · 1 + z2 − a2 z2 − a2 s 2 2 2 2 2 (z − a ) · (z − a + z ) p 2 2 = = 2z − a . z2− a2

We now have everything we need to compute the total charge surface, which is:

Q(S) = e dS = e (ϕ(θ, z)) · kϕθ ∧ ϕzk dθ dz ¨S ¨D √ √ 2π a 2 p a 2 p = dθ kz · 2z2 − a2 dz = 2πk z · 2z2 − a2 dz. ˆ0 â â We will separately solve the last integral for practical reasons.We will do so by doing a change of variables. √ 2 2 2 2 2 p 2 2 2z − a = u → u = 2z − a 2πk z · 2z − a dz = u du ˆ 4z dz = 2u du → z dz = 2 √ u du u3 = 2¡πk u2 · = πk u2 du = πk + C. ˆ 2¡ ˆ 3 Now we undo the change of variable writing the final expressions in terms of z and thus obtining the total charge of S. √  √ 3 z=a 2 √ 2 2 a 2 p 2z − a Q(S) = 2πk z · 2z2 − a2 dz = πk   â  3  z=a  3  q √ √ 3 √ 3 √ 3  2 · (a 2)2 − a2 2 2 2 2 2  2 · a − a  2 · 2a − a a     = πk  −  = πk  −   3 3  3 3

√ 3  2 " √ # " √ # 3a a3 a3 33 a3 3¡ 3 1 √ 1 = πk  −  = πk − = πka3 − = πka3 3 − .  3 3  3 3 3¡ 3 3

So, the total charge of the surface S is: √ Q(S) = πka3 3 − 1/3 .

29. Let us consider the ellipsoid

3 2 2 2 2 2 2 S = {(x, y, z) ∈ R : x /a + y /b + z /c = 1}

3 2 2 oriented by its unit and outward-pointing normal vector N to the solid W = {(x, y, z) ∈ R : x /a + 2 2 2 2 y /b + z /c ≤ 1} enclosed by S = ∂W . Consider the function f : S → R+ given by f(p) = dist(O,TpS), where O = (0, 0, 0) is the origin and TpS is the tangent plane to the ellipsoid in the point

33 p. (a) Compute f dS. ¨S x2 y2 z2 (b) Prove that the normal component of the vector field F (x, y, z) = , , is F := hF, Ni = a2 b2 c2 N 1/f. Compute the flux hF , dSi. Notice the symbol typically for the flux along the boundary S of a ‹S closed solid W . x y z (c) Compute the flux hG, dSi, where G(x, y, z) = zF (x, y, z), and F (x, y, z) = 2 , 2 , 2 . ‹S a b c Resolution: (a) First, we will parameterize S = ψ(D) with an ellipsoidal parameterization

p = (x, y, z) = ψ(θ, ϕ) = (a cos θ cos ϕ, b sin θ cos ϕ, c sin ϕ) where, as we are considering the whole ellipsoid, D = {(θ, ϕ): 0 ≤ θ ≤ 2π, −π/2 ≤ ϕ ≤ π/2}. For any point p = ψ(θ, ϕ) ∈ S, we have the two independent tangent vectors:

> > ψθ = (−a sin θ cos ϕ, b cos θ cos ϕ, 0) ψϕ = (−a cos θ sin ϕ, −b sin θ sin ϕ, c cos ϕ) , and the associated normal vector, which turns out to be outward-pointing:

 2  i j k bc cos ϕ cos θ 2 ψθ ∧ ψϕ = −a sin θ cos ϕ b cos θ cos ϕ 0 = ac cos ϕ sin θ

−a cos θ sin ϕ −b sin θ sin ϕ c cos ϕ ab cos ϕ sin ϕ  cos θ cos ϕ a cos θ cos ϕ  a   a2       sin θ cos ϕ  b sin θ cos ϕ = abc cos ϕ   = abc cos ϕ   .  b   b2   sin ϕ   c sin ϕ  c c2

Therefore, the tangent plane TpS for p = (x, y, z) = ψ(θ, ϕ) has as unit outward-pointing normal ψθ ∧ ψϕ vector N(p) := . ψθ ∧ ψϕ

Ellipsoid S and the tangent plane TpS to S = ∂W in a point p ∈ S. The red line is the segment between the origin and the closest point of TpS. This segment has the direction of N(p) and length f(p) = |hN(p), pi|.

The distance between the origin and a point p of the surface can be expressed by |hN(p), pi|, where N(p) is the unit normal vector to S at p, so that f(p) = |hN(p), pi|.

For p = (x, y, z) = ψ(θ, ϕ) we have f(ψ(θ, ϕ)) ψθ ∧ ψϕ = ψθ ∧ ψϕ, ψ , where * ! + a cos θ cos ϕ b sin θ cos ϕ c sin ϕ ψ ∧ ψ , ψ = abc cos ϕ , , , (a cos θ cos ϕ, b sin θ cos ϕ, c sin ϕ) θ ϕ 2 2 2 a b c ! a2 cos2 θ cos2 ϕ b2 sin2 θ cos2 ϕ c2 sin2 ϕ = abc cos ϕ + + a2 b2 c2 ! x2 y2 z2 = abc cos ϕ + + = abc cos ϕ, a2 b2 c2 so that

f(ψ(θ, ϕ)) ψθ ∧ ψϕ = abc cos ϕ. We are now ready to compute the integral of f on S = ψ(D):

2π π/2

f dS = f(ψ(θ, ϕ)) ψθ ∧ ψϕ dϕ dθ = abc cos ϕ dϕ dθ ¨S ¨D ˆ0 ˆ−π/2 2π π/2 ϕ=π/2 = abc dθ cos ϕ dϕ = 2πabc [sin ϕ]ϕ=−π/2 = 2πabc · 2 = 4πabc. ˆ0 ˆ−π/2

(b)

The ellipsoid of Exercise 29b with p located at the North pole.

For any point of the surface of the ellipsoid p = (x, y, z) ∈ S ⇐⇒ g(x, y, z) = 0 where g(x, y, z) = x2/a2 + y2/b2 + z2/c2 − 1. According to the resolution of Exercise V27, ∇g(p) = 2(x/a2, y/b2, z/c2) = 2F (p) is a normal vector to S in p, moreover pointing outward. Therefore the outward-pointing unitary normal vector is ∇g(p) F (p) N(p) = = . k∇g(p)k kF (p)k It is easy to check that hF (p), pi = 1 for any p = (x, y, z) ∈ S. The distance from a point q to a plane Π with unitary normal vector N passing though a point p is |hN, p − qi|. Therefore the distance from the origin (q = 0) to the tangent plane TpS to S in p is

|hF (p), pi| 1 f(p) = |hN(p), pi| = = . kF (p)k kF (p)k

We can now compute FN:

hF (p), F (p)i kF (p)k2 1 F = hF (p), N(p)i = = = kF (p)k = . N kF (p)k kF (p)k f(p)

35 To compute the ﬂux we can use Gauss theorem

hF , dSi = div F dx dy dz = h∇, F i dV ‹S ˚W ˚W where the ellipsoid S is oriented by its unit and outward-pointing normal vector N and W is the solid 1 1 1 4π enclosed by the ellipsoid S = ∂W . As h∇, F i = + + V (W ) = abc, we get a2 b2 c2 3 1 1 1 hF , dSi = div F dx dy dz = 2 + 2 + 2 dx dy dz ‹S ˚W a b c ˚W 1 1 1 1 1 1 4π = + + V (W ) = + + · abc a2 b2 c2 a2 b2 c2 3 4π bc ac ab = + + . 3 a b c

(c) First of all we compute G: xz yz z2 G(x, y, z) = , , = (P (x, y, z),Q(x, y, z),R(x, y, z)) , a2 b2 c2 with xz yz z2 P = ,Q = ,R = . a2 b2 c2 To apply Gauss theorem hG, dSi = div G dx dy dz ˆS+ ˚W we compute the divergence ∂P ∂Q ∂R z z 2z 1 1 2 div G = + + = + + = z + + ∂x ∂y ∂z a2 b2 c2 a2 b2 c2 and make a change W = T (B) to adapted spherical coordinates

x = ar cos φ cos θ − π ≤ φ ≤ π     2 2  y = br cos φ sin θ ,B : 0 ≤ θ ≤ 2π , |det DT (r, θ, φ)| = abcr2 cos φ.  z = cr sin φ   0 ≤ r ≤ 1  We can now solve the integral:

2π π 1 2 1 1 2 div G dx dy dz = dθ dφ abcr2 cos φ · z + + dr ˚ ˆ ˆ π ˆ a2 b2 c2 W 0 − 2 0 2π π 1 1 1 2 2 = abc + + dθ cos φ sin φ dφ r2 dr a2 b2 c2 ˆ ˆ π ˆ 0 − 2 0 π 1 1 2 2 1 = abc + + 2π cos φ sin φ a2 b2 c2 ˆ π 3 − 2 φ= π 1 1 2 2π sin2 φ 2 = abc 2 + 2 + 2 = 0. a b c 3 2 π φ=− 2

We could expect this result. As div G is odd in z and the solid W is symmetric with respect to z, it is clear that the integral I of div G on W is zero. Indeed, under the change of variables (x, y, z) = (u, v, −w) one can see that I = −I.

36 30. When a water tap is opened slowly, water ﬂowing out from the opening becomes thinner as it ﬂowed down, and a water jet is formed, with a radius that decreases with the distance from the tap. Compute the form of the water jet as a function of the initial radius r0, the initial speed v0 and the gravity g, assuming that the vertical component of the speed is constant in each horizontal section.

Resolution: To analyze this problem, we ﬁrst have to take four considerations about the ﬂuid:

(a) No viscosity: there is no friction inside it. (b) Stationary: the velocity at a given point is constant with respect to time. (c) Incompressible: the density remains constant. (d) Irrotational: there cannot be whirlpools, there are not angular movement with respect any point.

Under these assumptions, we can continue our analysis. The ﬁrst step is to know the Continuity equation which is synthesized in the following formula, where v1 is the ﬂuid velocity in the point A with a section S1 and v2 is the velocity at the point B with section S2:

v1 · S1 = v2 · S2

Then, let us calculate the velocity of a section of the jet using physics concepts:

∆Em = 0 thus Em0 = Emf .

Then, knowing the two components of the mechanical energy, we can reach the following formula. We will take the ﬁnal height equal to zero (h = 0): 1 1 mv2 + mgh = mv2 + mg · 0. 2 0 0 2 Now, the mass vanishes, and reorganizing all the terms, we obtain the formula just below, where v is the section velocity, v0 the initial section velocity, g is the gravity that we will consider constant and ﬁnally h0 is the initial height:

2 2 v = v0 + 2gh0.

From the two previous equations, we can obtain the radius r in function of the initial radius r0, the initial section velocity v0 and the height h.

s 2 4 v0 r = r0 2 . v0 + 2gh0

31. Flux of the vector ﬁeld F (x, y, z) = (x2, −y2, z2) through the following surfaces: (a)The boundary of:

n 3 2 2 2 2 p o W = (x, y, z) ∈ R : x + y + z ≤ 3R , 0 ≤ z ≤ x2 + y2 − R2 ,

oriented by an outward-pointing normal.

37 W on the (r, z) plane and on the (x, y, z) space with outward-pointing normal vectors for R = 1.

(b) The octant of sphere: 3 2 2 2 2 S = (x, y, z) ∈ R : x + y + z = R , x, y, z ≥ 0 , oriented by an outward-pointing normal.

First octant of the sphere W for R = 1 with outward-pointing normal N. √ n 3 2 2 2 o 2 2 2 Resolution: (a) W = (x, y, z) ∈ R : r + z ≤ 3R , 0 ≤ z ≤ r2 − R2 , where r = x + y , so in cylindrical coordinates (x, y, z) = T (r, θ, z) = (r cos θ, r sin θ, z), W = T (B) with p B = {(r, θ, z): r2 + z2 ≤ 3R2, 0 ≤ z ≤ r2 − R2, 0 ≤ θ ≤ 2π} = {(r, θ, z): r2 + z2 ≤ 3R2, z2 ≤ r2 − R2, z ≥ 0, 0 ≤ θ ≤ 2π}, thus W is a solid of revolution around the z-axis, and its boundary consists of 3 parts: ∂W = S1 ∪ S2 ∪ S3, where these surfaces are deﬁned by the equations √ S0 = {z = 0,R ≤ r ≤ 3R}, √ √ √ p 2 2 p 2 2 S1 = {z = r − R ,R ≤ r ≤ 2R},S2 = {z = 3R − r , 2R ≤ r ≤ 3R},

38 or more precisely, they can be parameterized as √ S0 = ϕ0(D0), ϕ0(r, θ) = (r cos θ, r sin θ, 0),D0 = {(r, θ): R ≤ r ≤ 3R, 0 ≤ θ ≤ 2π}, √ p 2 2 S1 = ϕ1(D1), ϕ1(r, θ) = (r cos θ, r sin θ, r − R ),D1 = {(r, θ): R ≤ r ≤ 2R, 0 ≤ θ ≤ 2π}, √ √ p 2 2 S2 = ϕ2(D2), ϕ2(r, θ) = (r cos θ, r sin θ, 3R − r ),D2 = {(r, θ): 2R ≤ r ≤ 3R, 0 ≤ θ ≤ 2π}.

For a parameterization S = ϕ(D) with ϕ(r, θ) = (r cos θ, r sin θ, f(r)), D = {(r, θ): rm ≤ r ≤ rM, 0 ≤ θ ≤ 2π} of a curve C given by z = f(r), the tangent vectors and the normal vector are given by

     0  cos θ −r sin θ i j k −f (r) cos θ 0 0 ϕr = sin θ  , ϕθ =  r cos θ  , ϕr ∧ ϕθ = cos θ sin θ f (r) = r −f (r) sin θ ,

f 0(r) 0 −r sin θ r cos θ 0 1 where ϕr ∧ ϕθ is always an upward-pointing vector, since its z-component is positive: r > 0. 2 2 2 2 2 2 2 2 As F (x, y, z) = (x , −y , z ), (F ◦ ϕ)(r, θ) = (r cos θ, −r sin θ, f(r) ) and hF ◦ ϕ, ϕr ∧ ϕθi = −r3f 0(r) cos3 θ − sin3 θ + rf(r)2, so that the ﬂux of F through S pointing outward S is given by

hF , dSi = hF , N ext dSi = hF ◦ ϕ, ϕr ∧ ϕθi dr dθ ¨S ¨S ¨D 2π r r M M = − dθ cos3θ − sin3 θ r3f 0(r) dr + 2π rf(r)2 dr ˆ0 ˆrm ˆrm rM = 2π rf(r)2 dr, ˆrm 2π 2π since cos3 θ dθ = sin3 θ dθ = 0 ˆ0 ˆ0 We now apply the formula above to the 3 surfaces S0, S1, S2 with outward-pointing normal vectors to W : on S0 with downward-pointing normal vector (negative induced orientation), and on S1 and S2 with upward-pointing normal vectors: √ 3R 2 On S0, hF , dSi = −2π r 0 dr = 0, ¨S0 ˆR √ √ 2R 4 2 2 r= 2R 2 2 r R r π 4 on S1, hF , dSi = 2π r r − R dr = 2π − = R , ¨S1 ˆR 4 2 r=R 2 √ √ 3R 2 2 4 r= 3R 2 2 3R r r π 4 on S2, hF , dSi = 2π √ r 3R − r dr = 2π − √ = R . ¨S2 ˆ 2R 2 4 r= 2R 2

Summing up, hF , dSi = πR4. ‹∂W Alternatively, we could use the parameterization S = ψ(D) with ψ(θ, z) = (g(z)√ cos θ, g(z) sin θ, z) for 2 2 S = S1 or S√= S2 with D = {(θ, z): 0 ≤ z ≤ R, 0 ≤ θ ≤ 2π} and r = g(z) = R + z on S1, with 2 2 r = g(z) = 3R − z on S2. Then the tangent vectors and the normal vector are given by    0    −g(z) sin θ g (z) cos θ i j k cos θ 0 ψθ=  g(z) cos θ, ψz= g (z) sin θ, ψθ ∧ ψz= −g(z) sin θ g(z) cos θ 0 =g(z)  sin θ , (1)

0 1 g0(z) cos θ g0(z) sin θ 1 −g0(z)

0 where now ψθ ∧ ψz is an outward-pointing vector if g(z)g (z) < 0, and an inward-pointing vector if g(z)g0(z) > 0.

39 0 z z 0 On S1, g (z) = √ = , so g (z)g(z) = z on S1, R2 + z2 g(z) 0 z z 0 on S2, g (z) = √ = − , so g (z)g(z) = −z on S2. 3R2 − z2 g(z)

Therefore ψθ ∧ ψz points inward in S1 and outward in S2,

(F ◦ ψ)(θ, z) = (g(z)2 cos2 θ, −g(z)2 sin2 θ, z2), 3 3 3 2 0 hF ◦ ψ, ψθ ∧ ψzi = g(z) cos θ − sin θ − z g(z)g (z) and

hF , dSi = hF , N ext dSi = − hF ◦ ψ, ψr ∧ ψθi dr dθ ¨S1 ¨S1 ¨D ! 2π R R = − dθ cos3θ − sin3 θ g(z)3 dz − 2π z2g(z)g0(z) dz ˆ0 ˆ0 ˆ0 R R R4 πR4 = 2π z2g(z)g0(z) dz = 2π z3 dz = 2π = , ˆ0 ˆ0 4 2 whereas

hF , dSi = hF , N ext dSi = hF ◦ ψ, ψr ∧ ψθi dr dθ ¨S2 ¨S2 ¨D 2π R R = dθ cos3θ − sin3 θ g(z)3 dz − 2π z2g(z)g0(z) dz ˆ0 ˆ0 ˆ0 R R πR4 = −2π z2g(z)g0(z) dz = 2π z3 dz = . ˆ0 ˆ0 2

On S0 we cannot apply this parameterization r = g(z) since z ≡ 0, but directly F (x, y, 0) = 2 2 > (x , −y , 0), N ext = (0, 0, −1) , so that FN = hF , N exti = 0 and hF , dSi = hF , N exti dS = 0. ¨S0 ¨S0 Summing up again, hF , dSi = πR4. ‹∂W (b)We use a parameterization S = ψ(D) based on spherical coordinates taking ρ = R with (x, y, z) = ψ(θ, φ) = (R cos φ cos θ, R cos φ sin θ, R sin φ) and D = {(θ, φ): 0 ≤ θ ≤ π/2, 0 ≤ φ ≤ π/2} for the octant of the sphere. The tangent vectors and the normal vector are given by

−R cos φ sin θ −R sin φ cos θ 2 2 ψθ =  R cos φ cos θ , ψφ = −R sin φ sin θ , ψθ ∧ ψφ = R |cos φ| = R cos φ, 0 R cos φ with ψθ ∧ ψφ pointing outward the sphere since   i j k cos φ cos θ 2 ψθ ∧ ψφ = −R cos φ sin θ R cos φ cos θ 0 = R cos φ cos φ sin θ .

−R sin φ cos θ −R sin φ sin θ R cos φ sin φ

As F (x, y, z) = (x2, −y2, z2), (F ◦ ψ)(θ, φ) = R2 cos2 φ cos2 θ, − cos2 φ sin2 θ, sin2 φ. We have all the ingredients to compute the ﬂux of the vector ﬁeld F through the surfaces S oriented by an outward-

40 pointing normal:

hF , dSi = hF ◦ ψ, ψθ ∧ ψφi dθ dφ ¨S ¨D = R4 cos4 φ cos3 θ − cos4 φ sin3 θ + cos φ sin3 φ dθ dφ ¨D π/2 π/2 π/2 π/2 = R4 cos3θ− sin3 θ dθ cos4 φ dφ + dθ sin3 φ cos φ dφ ˆ0 ˆ0 ˆ0 ˆ0 π sin4 φφ=π/2 πR4 = R4 = , 2 4 φ=0 8 where we have used that

π/2 π/2 sin3 θ θ=π/2 cos3 θ dθ = (1 − sin2 θ) cos θ dθ = sin θ − = 2/3. ˆ0 ˆ0 3 θ=0

π/2 π/2 Analogously, sin3 θ dθ = 2/3 so that cos3 θ − sin3 θ dθ = 0. ˆ0 ˆ0 Alternatively, we could use the cartesian parameterization ϕ(x, y) = (x, y, f(x, y)), where z = f(x, y) = pR2 − x2 − y2, and D = {(x, y): x2 + y2 ≤ R2, x ≥ 0, y ≥ 0} chosen such that S = ϕ(D). The > −x x associated normal vector is ϕx ∧ ϕy = (−fx, −fy, 1) , where fx = = − , and fy = pR2 − x2 − y2 f y − , which is a vertical vector and therefore points outward the sphere. Recalling that F (x, y, z) = f (x2, −y2, z2), we can compute again the ﬂux of F through the piece S of the sphere oriented by an outward-pointing normal vector as

hF , dSi = hF ◦ ϕ, ϕx ∧ ϕyi dx dy ¨S ¨D 2 2 2 = −x fx + y fy + f dx dy ¨D 3 3 4 x −y 2 πR = dx dy + f dx dy = , ¨D f ¨D 8 where 3 3 π/2 R 4 x − y x = r cos θ 3 3 r dr dx dy = = cosθ − sin θ dθ √ = 0, 2 2 ¨D f y = r sin θ ˆ0 ˆ0 R − r and

π/2 R π R2r2 r4 r=R π R4 R4 πR4 f 2 dx dy = dθ R2 − r2 r dr = − = − = . ¨D ˆ0 ˆ0 2 2 4 r=0 2 2 4 8

32. Flux of the vector ﬁeld F (x, y, z) = (2x, −y, 0) through the surface

3 2 2 2 S = (x, y, z) ∈ R : x + y = R , x, y ≥ 0, 0 ≤ z ≤ h

(a quarter of cylinder) oriented by its outward-pointing normal.

41 Surface and corresponding vectors.

Resolution: We use a parameterization S = ϕ(D) based on cylindrical coordinates, in which r = R, and: ϕ(θ) = (R cos θ, R sin θ, z). The domain is: D = {(θ, z): 0 ≤ θ ≤ π/2, 0 ≤ z ≤ h}, since x and y are positive, and the maximum height is h for the quarter of the cylinder. The tangent vectors are given by the respective partial derivatives as follows:

−R sin θ 0 ϕθ =  R cos θ  , ϕz = 0 , 0 1

We have to take into account that the normal vector points outward the sphere since the cross product indicates a positive direction in the ﬁrst quadrant:

ϕθ ∧ ϕz = (R cos θ, R sin θ, 0)

Now that we have all the necessary information, we can proceed to integrate the ﬁeld using our parameterization. If we develop the calculations:

π/2 h hF , dSi = hF ◦ ϕ, ϕθ ∧ ϕzi dz dθ = h(2R cos θ, −R sin θ, 0), (R cos θ, R sin θ, 0)i dz dθ ¨S ¨D ˆ0 ˆ0 π/2 h π/2 h = 2R2 cos θ − R2 sin θ dz dθ = 2R2 cos θ − R2 sin θ · z dθ ˆ0 ˆ0 ˆ0 0 π/2 π/2 1 1 1 1 = h 2R2 cos θ − R2 sin θ dθ = h 2R2 + cos 2θ − R2 − cos 2θ dθ ˆ0 ˆ0 2 2 2 2 π/2 R2 3R2 hR2 π/2 hR2 3 π/2 R2hπ = h + cos 2θ dθ = 1 + 3 cos 2θ dθ = θ + sin 2θ dθ = . ˆ0 2 2 2 ˆ0 2 2 0 4

33. Flux of the vector ﬁeld F (x, y, z) = (x2, y2, z2) through the surface

3 2 2 2 S = (x, y, z) ∈ R : z = h(x + y )/R , z ≤ h ,

(piece of the elliptic paraboloid) oriented by its interior normal.

42 Resolution: We use a parameterization S = ϕ(D) based on cylindrical parameters with ϕ(r, θ) = r2 r cos θ, r sin θ, h and D = {(r, θ): 0 ≤ r ≤ R, 0 ≤ θ ≤ 2π} for the elliptic paraboloid. R2

cos θ −r sin θ ϕr =  sin θ  , ϕθ =  r cos θ  , h R2 2r 0

 2 h  i j k −2r R2 cos θ h 2 h ϕr ∧ ϕθ = cos θ sin θ 2 2r = −2r 2 sin θ . R R −r sin θ r cos θ 0 r Also r4 F ◦ ϕ(r, θ) = r2 cos2 θ, r2 sin2 θ, h2 . R4 Therefore:

hF , dSi = hF ◦ ϕ, ϕr ∧ ϕθi dr dθ ¨S ¨D 2 4 3 h 4 3 h 5 h = 2r cos θ 2 + 2r sin θ 2 − r 4 dr dθ ¨D R R R 2 h h h = 2r4 cos3 θ + 2r4 sin 3 θ − r5 dr dθ 2 2 4 ¨D R ¨D R ¨D R R 2 2 6 R 5 h h r = 2π r 4 dr dθ = 2π 4 ˆ R R 6 0 0 πR2h2 = . 3

34. Flux of the vector ﬁeld r(x, y, z) = (x, y, z) through the portion S of the hyperbolic paraboloid z = h(x2 − y2)/L2, cut by the plains x = 0, x = L and z = 0, oriented in the point O = (0, 0, 0) by the vector k = (0, 0, 1). Resolution: First, S is the piece of the hyperbolic paraboloid z = h(x2 − y2)/L2, cut by the plains x = 0, x = L and z = 0. So, if we plot the hyperboloid (which will be plotted lately), this means that

3 2 2 2 S = {(x, y, z) ∈ R : z = h(x − y )/L , 0 ≤ x ≤ L, z ≥ 0}.

Imposing z ≥ 0 is equivalent to impose x2 − y2 ≥ 0 (we assume that h and L are positive) or, equivalently y2 ≤ x2 or −x ≤ y ≤ x, so we can write

3 2 2 2 S = {(x, y, z) ∈ R : z = h(x − y )/L , 0 ≤ x ≤ L, −x ≤ y ≤ x} = ϕ(D),

where ϕ(x, y) = (x, y, f(x, y) := h(x2 − y2)/L2 is a Cartesian parameterization of S, deﬁned on

2 D = (x, y) ∈ R : 0 ≤ x ≤ L, −x ≤ y ≤ x .

We do a plot of the hyperbolic paraboloid in order to be easier to understand the problem.

43 Hyperbolic paraboloid and surface S.

By equation  1   0  i j k −f  x q 2 2 ϕx =  0  , ϕy =  1  , ϕx ∧ ϕy = 1 0 fx = −fy , ϕx ∧ ϕy = 1 + fx + fy ,

fx fy 0 1 fy 1 > we know that the associated normal vector to this parameterization is ϕx ∧ ϕy = (−fx, −fy, 1) = 2hx 2hy > − , , 1 , which is an upward-pointing normal vector, and for (x, y, z) = (0, 0, 0) coincides L2 L2 with the normal vector provided in the statement of the problem for the orientation of S, which is (0, 0, 1). We are given the vector field r(x, y, z) = (x, y, z), so that writing z = f(x, y) = h(x2 − y2)/L2, we have that hr ◦ ϕ, ϕx ∧ ϕyi = −f(x, y), and finally after some computations we get the result for the flux: L x h 2 2 hr, dSi = hr ◦ ϕ, ϕx ∧ ϕyi dS = − f(x, y) dx dy = − 2 dx x − y dy ¨S ¨D ¨D L ˆ0 ˆ−x h L 4x3 h x4 L hL4 hL2 = − 2 dx = − 2 = − 2 = − . L ˆ0 3 L 3 0 3L 3

To check this result, we could also use Gauss’ theorem, adding two ﬂat covers S0 = {(x, y, z): 0 ≤ x ≤ > L, −x ≤ y ≤ x, z = 0}, with outward-pointing normal vector N 0 = (0, 0, 1) and SL = (x, y, z): x = > L, −L ≤ y ≤ L, z = f(x, y)}, with outward-pointing normal vector (1, 0, 0) , and area A(SL) = 4hL/3, which enclose a solid 3 2 2 2 W = {(x, y, z) ∈ R : 0 ≤ x ≤ L, −x ≤ y ≤ x, 0 ≤ z ≤ h(x − y )/L }, 2 with volume V (W ) = hL /3, such that ∂W = S ∪ S0 ∪ SL. Gauss’s theorem gives

div r dx dy dz = hr, dSi = hr, dSi + hr, dSi + hr, dSi. ˚ ‹ + ¨ + ¨ + ¨ + W ∂W S S0 SL

2 On the one hand, as div r = 3, div r dx dy dz = 3V (W ) = hL . On the other hand, rN = ˚W hr, N 0i = 0 on S0 so that hr, dSi = 0, and rN = hr, N Li = L on SL so that hr, dSi = ¨ + ¨ + S0 SL 2 hr, N L dSi = hr, N Li dS = LA(SL) = L · 4hL/3 = 4hL /3. Putting all this together, ¨ + ¨ + SL SL 4hL2 hL2 hr, dSi = div r dx dy dz − hr, dSi − hr, dSi = hL2 − 0 − = − . ¨ + ˚ ¨ + ¨ + 3 3 S W S0 SL

44 35. Flux of the vector ﬁeld F (x, y, z) = (2x2, 3y2, z2) through the boundary of the following solid of revolution: n 3 p p o W = (x, y, z) ∈ R : x2 + y2 ≤ z ≤ 2R2 − x2 − y2 , oriented by its outward-pointing normal.

Resolution: W in cylindrical coordinates takes the form

2 2 p 2 2 B = {(r, θ, z): r ≤ z , 0 ≤ z ≤ 2R − r } = B1 ∪ B2

with B1 = {(r, θ, z): r ≤ z/, z ≤ R}, cone of base a√ circle of radius R and height R, and B2 = {(r, θ, z): z2 + r2 ≤ 2R2, z ≥ R}, hemisphere of radius 2R. p 2 2 2 2 2 p 2 2 2 2 2 2 Let S1 = {(x, y, z): z = x + y , x +y ≤ R }, S2 = {(x, y, z): z = 2R − x − y , x +y ≤ R }.

On S1 we use a cylindrical parameterization S1 = ϕ(D1), with ϕ(θ, z) = (g(z) cos θ, g(z) sin θ, z) and 0 g(z) = z, g (z) = 1 on the domain D1 = {(θ, z): 0 ≤ z ≤ R, 0 ≤ θ ≤ 2π}. By equation (1), the 0 > > associated normal vector is ψθ ∧ ψz = g(z)(cos θ, sin θ, −g (z)) = z(cos θ, sin θ, −1) , which points outward W along S1, since has a negative z-component. Therefore the ﬂux of F through S1 oriented by an outward-pointing normal vector is

hF , dSi = hF ◦ ϕ, ϕθ ∧ ϕzi dθ dz ¨S−1 ¨D1 = 2g(z)3 cos3 θ + 3g(z)3 sin3 θ − zg(z)g0(z) dθ dz ¨D−1 2π R 2π R = 2 cos3 θ+ 3 sin3 θ dθ z3 dz − dθ z3 dz ˆ0 ˆ0 ˆ0 ˆ0 πR4 = − , 2

2π 2π since cos3 θ dθ = sin3 θ dθ = 0 ˆ ˆ 0 0 √ 2 2 0 On S2 we also use a cylindrical parameterization S2 = ϕ(D2), with g(z) = 2R − z , g (z) = −z z √ √ = − on the domain D2 = {(θ, z): R ≤ z ≤ 2R, 0 ≤ θ ≤ 2π}. The associated normal 2R2 − z2 g(z) > vector is ψθ ∧ ψz = (g(z) cos θ, g(z) sin θ, z) , which also points outward W along S2, since has a positive z-component. Therefore the ﬂux of F through S2 oriented by an outward-pointing normal vector is

hF , dSi = hF ◦ ϕ, ϕθ ∧ ϕzi dθ dz ¨S2 ¨D2 = 2g(z)3 cos3 θ + 3g(z)3 sin3 θ + z3 dθ dz ¨D2 √ √ 2π 2R 2R 3 3 3 3 = 2 cosθ + 3 sin θ dθ g(z) dz + z dz ˆ0 ˆR ˆR √ 2R 2π 3πR4 = z3 dz = 4R4 − R4 = . ˆR 4 2 3πR4 πR4 Summing up, hF , dSi = − = πR4. ‹∂W 2 2

45 The integral theorems II

36. Given the vector ﬁeld F (x, y, z) = (4xz, −y2, yz), compute its ﬂux through the boundary of the unit cube W = [0, 1]3 oriented by its outward-pointing normal.

Resolution: The cube W has 6 diﬀerent faces, so the direct computation of the ﬂux of F through ∂W involves the computation of 6 integral surfaces. Instead, we are going to apply Gauss’ theorem to compute one single triple integral using that div F = Px + Qy + Rz = 4z − 2y + y = 4z − y

hF , dSi = div F dx dy dz = (4z − y) dx dy dz ‹S+ ˚W ˚W 1 1 1 1 1 1 1 1 = dx dy 4z dz − dx y dy dz = 4z dz − y dy ˆ0 ˆ0 ˆ0 ˆ0 ˆ0 ˆ0 ˆ0 ˆ0 2 y=1 2z=1 y 1 3 = 2z z=0 − = 2 − = . 2 y=0 2 2

37. Let F be a central force without singularities, so F = h(r)r, where r(x, y, z) = (x, y, z), r2(x, y, z) = 2 2 2 1 (x + y + z ) and h(r) is a function of class C in the interval [0, ∞). SR is the sphere with radius R centered in the origen and oriented by its outward-pointing normal. Compute the exit ﬂux hF , dSi: (a) Integrating the normal component of the vector ﬁeld over the sphere. ¨SR (b) Applying Gauss theorem. Compare the result with Gauss Law.

Resolution: (a) First of all, notice that a sphere of radius R centered in the origin has the following equation: x2 + y2 + z2 = R2 and that in the statement of the problem we are given that r2(x, y, z) = (x2 + y2 + z2). Hence, on the sphere we have that r2 = R2. Then, we use a parameterization SR = ψ(D) based on spherical coordinates with (x, y, z) = ψ(θ, φ) = (R cos φ cos θ, R cos φ sin θ, R sin φ) and D = {(θ, φ): 0 ≤ θ ≤ 2π, −π/2 ≤ φ ≤ π/2}. The tangent vectors and the normal vector are given by

−R cos φ sin θ −R sin φ cos θ 2 2 ψθ =  R cos φ cos θ , ψφ = −R sin φ sin θ , ψθ ∧ ψφ = R |cos φ| = R cos φ, 0 R cos φ

with ψθ ∧ ψφ pointing outward the sphere since

 2 2    i j k R cos φ cos θ cos φ cos θ 2 2 ψθ ∧ ψφ = −R cos φ sin θ R cos φ cos θ 0 =  R cos φ sin θ  = R cos φ cos φ sin θ .

−R sin φ cos θ −R sin φ sin θ R cos φ R2 cos φ sin φ sin φ

As F (x, y, z) = h(r)r(x, y, z), (F ◦ ψ)(θ, φ) = h(R)(R cos φ cos θ, R cos φ sin θ, R sin φ). We now have all the necessary data to compute the ﬂux of the central force F through the surface SR oriented by

46 its outward-pointing normal:

hF , dSi = hF ◦ ψ, ψθ ∧ ψφi dθ dφ ¨SR ¨D = h(R) R3 cos3 φ cos2 θ + R3 cos3 φ sin2 θ + R3 sin2 φ cos φ dθ dφ ¨D π/2 2π = h(R)R3 cos3 φ(cos2 θ + sin2 θ) + sin2 φ cos φ dθ dφ ˆ−π/2 ˆ0 π/2 2π = h(R)R3 cos3 φ + sin2 φ cos φ dθ dφ ˆ−π/2 ˆ0 π/2 2π = h(R)R3 cos φ(cos2 φ + sin2 φ) dθ dφ ˆ−π/2 ˆ0 π/2 2π = h(R)R3 (cos φ) dθ dφ ˆ−π/2 ˆ0 π/2 π/2 3 θ=2π 3 = h(R)R (cos φ)[θ]θ=0 dφ = 2πh(R)R (cos φ) dφ ˆ−π/2 ˆ−π/2 3 π/2 3 3 = 2πh(R)R [sin φ]−π/2 = 2πh(R)R (1 − (−1) = 4πh(R)R .

Therefore the exit flux hF , dSi = 4πh(R)R3. ¨SR (b) We must apply Gauss’ theorem on the solid W = {(x, y, z): x2 + y2 + z2 = R2}. Gauss’ theorem tells us that: div F dx dy dz = hF , dSi ˚W ¨SR We must first find the divergence of the vector field F (x, y, z) = (P (x, y, z),Q(x, y, z),R(x, y, z)),

∂P ∂Q ∂R div F = + + ∂x ∂y ∂z

∂P d(h(r)x) dh dr dh 1 dh 1 = = · x + h(r) · 1 = · 2x · x + h(r) = · x2 + h(r). ∂x dx dr dx dr 2px2 + y2 + z2 dr r Similarly, ∂Q dh 1 ∂R dh 1 = · y2 + h(r), = · z2 + h(r). ∂y dr r ∂z dr r Therefore, the divergence is dh 1 dh 1 div F = 3h(r) + (x2 + y2 + z2) = 3h(r) + r2 = 3h(r) + h0(r)r dr r dr r Knowing that the sphere is oriented by its outward-pointing normal we can compute

hF , dSi = div F dx dy dz = 3h(r) dx dy dz + h0(r)r dx dy dz. ¨SR ˚W ˚W ˚W To solve the triple integral we can use spherical coordinates. Let’s take B such that T (B) = W :

−π π B = (r, θ, ρ): 0 ≤ θ ≤ 2π, ≤ ρ ≤ , 0 ≤ r ≤ R . 2 2

47 Then

hF , dSi = 3 h(r)r2 cos ρ dr dρ dθ + h0(r)r3 cos ρ dr dρ dθ ¨SR ˚ ˚ B B π π 2π 2 R 2π 2 R = dθ cos ρ dρ h(r)r2 dr + dθ cos ρ dρ h0(r)r3 dr ˆ ˆ π ˆ ˆ ˆ π ˆ 0 − 2 0 0 − 2 0 R R = 12π h(r)r2 dr + 4π h0(r)r3 dr, ˆ0 ˆ0 ( u = h(r) du = h0(r) dr, and ﬁnally integrating by parts using 2 r3 dv = r v = 3 , " # r3 R R r3 R hF , dSi = 12π h(r) − h0(r) dr + 4π h0(r)r3 dr ¨ ˆ ˆ SR 3 0 0 3 0 R R 3 0 3 0 3 = 4πR h(R) − 4π h (r)r dr + 4π h (r)r dr ˆ0 ˆ0 = 4πR3h(R).

Therefore the solution to this exercise is that hF , dSi = 4πR3h(R). One can also check the ¨SR solution calculating the flow directly through sphere, to obtain the same result. Additionally, if we compare the result with Gauss’ law we can see that the flux (using, for instance, r the expression for the electric field E = can be computed with an integral over the surfaces of the r3 r 1 1 sphere which gives , dS = dS = 4πR2 = 4π. It is important to note that in this 3 2 2 ˆSR r R ˆ R example there is a singularity at r = 0, and to treat it properly you have to take out a small ball of radius , apply Gauss’ theorem with two boundaries, and take a limit when tends to zero. The important information to know is that this result is a generalization of the one we have found using Gauss’ theorem. Furthermore, Gauss’ theorem can be applied to spheres centered at the origin whereas Gauss’ law is valid for any sphere.

38. Compute the ﬂux of the vector ﬁeld F (x, y, z) = (4x, −2y2, z2) through the cylinder

3 2 2 S = (x, y, z) ∈ R : x + y = 4, 0 ≤ z ≤ 3

oriented by its outward-pointing normal. [ Hint: In this problem and from now on, you can cover the surface in order to be able to apply Gauss’ theorem.]

Resolution: Notice that S is a cylinder of equation x2 + y2 = 4 bounded by 0 ≤ z ≤ 3. So, apart from the cylinder we are going to add 2 horizontal ﬂat covers S0 and S3, where the subindex refer to the value of z at the covers. This solid will be called W :

3 2 2 W = (x, y, z) ∈ R : x + y = 4, 0 ≤ z ≤ 3

and its boundary ∂W will consist on 3 pieces, ∂W = S ∪ S0 ∪ S3.

48 Solid W of Exercise V48.

As we have to apply Gauss’ theorem, ﬁrst of all we are going to recall it:

hF , dSi = div F dx dy dz, ‹S+ ˚W where S+ means that we must use a normal vector pointing outward W . First, we calculate the triple integral. Since the vector ﬁeld is F (x, y, z) = (4x, −2y2, z2), div F = 4 − 4y + 2z. Since W is the interior of a cylinder, we are going to parameterize W with cylindrical coordinates. Notice 3 2 2 2 2 that W = (x, y, z) ∈ R : r = 4, 0 ≤ z ≤ 3 , where r = x + y , so in cylindrical coordinates (x, y, z) = T (r, θ, z) = (r cos θ, r sin θ, z), we have W = T (B) with

B = {(r, θ, z): 0 ≤ r ≤ 2, 0 ≤ z ≤ 3, 0 ≤ θ ≤ 2π}.

We now calculate the divergence integral:

2π 3 2 2π 3 2 div F dx dy dz = dθ dz r (4 − 4r sin θ + 2z) dr = 2 dθ dz 2r − 2r2 sin θ + rz dr ˚W ˆ ˆ ˆ ˆ ˆ ˆ 0 0 0 0 0 0 2π 2 2π 3 2r3 sin θ r2z r=2 16 = 2 dθ r2 − + dz = 2 dθ 4 − sin θ + 2z dz ˆ ˆ 3 2 r=0 ˆ ˆ 3 0 0 0 0 2π 2π 2π z=3 16 2 = 2 4z − sin θ z + z dθ = 2 21 dθ − 16 sin θ dθ = 84π. ˆ 3 z=0 ˆ ˆ 0 0 0 Now, it is time to calculate the ﬂux through the covers.

49 Notice the unitary outward-pointing normal vector is N 0 = (0, 0, −1) in S0 and N 3 = (0, 0, 1) in S3. So, this makes very easy and short the computation of the ﬂux through these covers, because, in S3, 2 FN = hF , Ni = z = 9 and

2 hF , dSi = hF , Ni dS = FN dS = 9A(S3) = 9 · π2 = 36π. ¨S3 ¨S3 ¨S3

And in S0, FN = hF , Ni = 0 so

hF , dSi = hF , Ni dS = FN dS = 0. ¨S0 ¨S0 ¨S0

Finally,

hF , dSi = div F dx dy dz − hF , dSi − hF , dSi = 84π − 0 − 36π = 48π. + ‹S ˚W ¨S0 ¨S3

We are asked about the ﬂux through the lateral surface S of the cylinder oriented by its outward- pointing normal, and to apply the Gauss’ theorem, we also use S oriented by its outward-pointing normal, so we don’t have to change sign.

Therefore, the flux of F through the whole boundary ∂W = S ∪ S0 ∪ S3 is 48π. 39. Compute the flux of the vector field F (x, y, z) = (−x, 0, x + z) through the piece of the sphere given 3 2 2 2 by S = (x, y, z) ∈ R : x + (y − 5) + (z − 5) = 25, y ≤ 9, z ≥ 1 oriented by the normal vector N(x, y, z) = (x, y − 5, z − 5).

Sphere S without covers of Exercise 39.

Resolution: We first notice that S is the sphere with center (0, 5, 5) and radius 5 without two caps. We will apply Gauss’ theorem on a solid W , which states that the integral of a vector field over a solid W is equal to the flux of the vector field through its closed boundary surface ∂W . So we have to compute one single triple integral using div F = Px + Qy + Rz = −1 + 0 + 1 = 0, so that

hF , dSi = div F dx dy dz = 0. ‹∂W + ˚W

50 We take W as the interior of S:

3 2 2 2 W = (x, y, z) ∈ R : x + (y − 5) + (z − 5) ≤ 25, y ≤ 9, z ≥ 1 ,

and its boundary ∂W consists of S plus two ﬂat disks S1 and S9 of radius 3 and thus area 9π:

3 2 2 2 2 2 S1 = (x, y, z) ∈ R : x + (y − 5) + (z − 5) ≤ 25, y ≤ 9, z = 1 = (x, y, 1): x + (y − 5) ≤ 9 , 3 2 2 2 2 2 S9 = (x, y, z) ∈ R : x + (y − 5) + (z − 5) ≤ 25, y = 9, z ≤ 1 = (x, 9, z): x + (z − 5) ≤ 9 .

So we get 0 = hF , dSi = hF , dSi + hF , dSi + hF , dSi. ‹ + ‹ + ¨ + ¨ + ∂W S S1 S9 As what we want is the ﬂux of the ﬁeld on S:

hF , dSi = − hF , dSi − hF , dSi. ‹ + ¨ + ¨ + S S1 S9

In order to compute the ﬂux through the ﬂat covers Sj, j = 1, 9,

+ + hF , dSi = hF , N j dSi = hF , N j i dS, ¨ + ¨ + ¨ Sj S1 S1

+ + we may specify the unit outward-pointing normal vectors to W . They are N 9 = (0, 1, 0) and N 1 = (0, 0, −1) (notice that S1 is in the lower part of W ). Recalling that F (x, y, z) = (−x, 0, x + z), we have + + thus hF , N 9 i = 0 on S9 and hF , N 1 i = −x − z = −x − 1 on S1. Therefore

+ hF , dSi = hF , N 9 i dS = 0, ¨ + ¨ S9 S1 ¨ + ¨¨ hF , dSi = hF , N 1 i dS = −x − 1 dS = − ¨x dx dy − dx dy = −A(S1) = −9π, ¨ + ¨ ¨ ¨¨ ¨ S1 S1 S1 ¨S1 S1

where the ﬁrst integral is zero because x is an odd function integrated on the symmetric disk S1. Finally hF , dSi = −(−9π) − 0 = 9π, ‹S+ and we only have to check that the sign is correct, that is, that the normal vector N(x, y, z) = (x, y−5, z−5) of the statement is an outward-pointing vector to W . As the equation of S is g(x, y, z) = 0, with g(x, y, z) := x2 + (y − 5)2 + (z − 5)2 − 25, we ﬁrst recall that any vector proportional to ∇g(x, y, z) = 2(x, y − 5, z − 5) is a normal vector vector to S (see Exercise V27), as is the case of N(x, y, z) = (x, y − 5, z − 5), which clearly points out of the origin and threfore outward W . For instance for p = (5, 5, 5) ∈ S, N(p) = (5, 0, 0) points in the increasing direction of the x axis.

40. Let us consider the piece of the ellipsoid given by

3 2 2 2 S = (x, y, z) ∈ R : x + 2y /3 + z = 1, y ≤ 1, z ≤ 2/3 .

(a) Compute the circulation z dx + y dy − x dz for C = S ∩ {y = 1} oriented by the tangent vector ˛C T (x, 1, z) = (−z, 0, x).

(b) Compute the ﬂux dz dx orienting S in a compatible way with (a). ¨S

51 Domain W = S ∪ S1 ∪ S2/3 of Exercise 40

Resolution: √ (a) The curve C = {(x, y, z): x2 + z2 = 1/3, y = 1} is a circle of radius 1/ 3 centered at the origin of the plane {y = 1} that we parameterize as C = σ(I), with (x, y, z) = σ(t) = √1 cos t, 1, √1 sin t 3 3 and I = [0, 2π]. To check its orientation, we compute its tangent vector (x0, y0, z0) = σ0(t) = − √1 sin t, 0, √1 cos t = (−z, 0, x) = T (x, 1, z), so C with these parameterization is oriented as 3 3 we were asked for. We now can proceed directly to compute the circulation of the vector ﬁeld F (x, y, z) = (z, y, −x) along C:

2π 0 0 0 2 2 2π z dx + y dy − x dz = zx + yy − xz dt = − (z + x ) dt = − . ˛C ˆI ˆ0 | {z } 3 1/3 √ 2 2 Alternatively,√ we could apply Green’s theorem to the disk S1 = {(x, y = 1, z): x + y ≤ 1/ 3} of radius 1 3 contained in the plane {y = 1} with the boundary C = ∂S1 oriented counter-clockwise with respect to S1, or in other words, leaving the region S1 on the left side, as is the case with the parameterization σ, and the vector ﬁeld F (x, y, z) = (z, y, −x), taking into account that rot F =   i j k 0

∂x ∂y ∂z = 2:

z y −x 0 2π z dx − x dz = (−1 − 1) dx dz = −2A(D) = − . ‰C=∂S1 ˆD 3

Even more, we can apply Stokes’ theorem to the surface S1. For this, we have to orient S1 by the + unitary normal vector −N 1 = (0, −1, 0), since in this way its boundary curve C = ∂S1 is oriented by the vector −N 1 leaving S1 on the left side: 2π hF , d`i. = hrot F , dSi = hrot F , −N 1i dS = −2A(S1) = − . ˛ + ¨ + ¨ + 3 C S1 S1

(b) By deﬁnition, if G = (P, Q, R), hG, dSi = P dy dz + Q dz dx + R dx dy, so we are ¨ ¨ ¨ ¨ S S S S zu zv given the vector ﬁeld G = (0, 1, 0). (Recall that dz dx = du dv, if we parameterize ¨S ¨D xu xv S = ϕ(D) and ϕ(u, v) = (x(u, v), y(u, v), z(u, v)).) 3 2 2 2 Take W = (x, y, z) ∈ R : x + 2y /3 + z ≤ 1, y ≤ 1, z ≤ 2/3 , W = S ∪ S1 ∪ S2/3, where S2/3 = {(x, y, z = 2/3): x2 + 2y2/3 ≤ 5/9} is a domain enclosed by an ellipse in the plane z = 2/3, with unitary normal vectors N, N 1 = (0, 1, 0), N 2/3 = (0, 0, 1) pointing outward W . In particular, the

52 unitary normal vector N on S points also outward W , which is compatible with the orientation of C, which is oriented leaving S on the left side. By Gauss’ theorem,

0 = div G dx dy dz = hG, dSi = hG, dSi + hG, dSi + hG, dSi, ˚ ‹ + ¨ + ¨ + ¨ + W ∂W S S1 S2/3 and 1 1 1 π hG, dSi = hrot F , dSi = hrot F , N 1i dS = 2 dS = A(S1) = , ¨ + 2 ¨ + 2 ¨ + 2 ¨ + 3 S1 S1 S1 S1 1 1 hG, dSi = hrot F , dSi = hrot F , N 2/3i dS = 0. ¨ + 2 ¨ + 2 ¨ + S2/3 S2/3 S2/3 π Therefore dz dx = hG, dSi = − hG, dSi − hG, dSi = − . ¨ ¨ + ¨ + ¨ + 3 S S S1 S2/3 41. Flux of the vector ﬁeld F (x, y, z) = 2e−xy, e−xy2, 1 through the piece of the sphere given by S = 3 2 2 4 (x, y, z) ∈ R :(x − 2) + (y − 2) = z , 1 ≤ z ≤ 2 oriented by the normal vector N(x, y, z) = (x − 2, y − 2, −2z3).

Resolution:

Surface S of exercise 41.

We have to apply Gauss’ theorem on a solid W , that states that the integral of a vector field over a solid W is equal to the flux of the vector field through its closed boundary surface. So we have to −x −x compute one triple integral using div F = Px + Qy + Rz = −2e y + 2ye + 0 = 0:

hF , dSi = div F dx dy dz = 0. ‹S+ ˚W In our case the solid will be the interior of the surface S:

3 2 2 4 W = (x, y, z) ∈ R :(x − 2) + (y − 2) ≤ z , 1 ≤ z ≤ 2 ,

and its boundary ∂W will consists on three pieces, the surface S and two ﬂat covers S1 and S2 which happen to be disks of radius 1 and 2, respectively:

3 2 2 4 3 2 2 S1 = (x, y, z) ∈ R :(x − 2) + (y − 2) ≤ z , z = 1 = (x, y, 1) ∈ R :(x − 2) + (y − 2) ≤ 1 , 3 2 2 4 3 2 2 S2 = (x, y, z) ∈ R :(x − 2) + (y − 2) ≤ z , z = 2 = (x, y, 2) ∈ R :(x − 2) + (y − 2) ≤ 4 .

53 So we get hF , dSi = hF , dSi + hF , dSi + hF , dSi = 0. ‹ + ¨ + ¨ + ¨ + S S S1 S2 As what we want is the ﬂux of the vector ﬁeld on S:

hF , dSi = − hF , dSi − hF , dSi. ¨ + ¨ + ¨ + S S1 S2

The unit outward-pointing normal vector of S2 is N 2 = (0, 0, 1), and the outward-pointing normal vector of S1 is N 1 = (0, 0, −1). The outward-pointing normal vector of S is indeed a downward- pointing vector, by the ﬁgure, like the normal vector N given in the statement of the exercise.

Moreover, hF , dSi = hF , N 2i dS = dS on S2, and hF , dSi = hF , N 1i dS = − dS on S1. Therefore, the flow through S2 is A(S2) = 16π and the flow through S1 is −A(S2) = −π. So we reach our final solution

hF , dSi = π − 16π = −15π. ¨S+ 42. Flux of the vector ﬁeld F (x, y, z) = (1, 0, 2) through the piece of elliptic paraboloid 3 2 2 2 S = (x, y, z) ∈ R : z = x + 4y , z ≤ 3y + 1 oriented by the normal vector with positive vertical component. [ Hint: You can add a non ﬂat cover to the surface.]

Resolution:

Surface S of Exercise 42

Notice that S is a piece of a parabolic cylinder of equation z = x2 + 4y2 that is cut by a parabolic cylinder of equation z = 3y2 + 1, and satisﬁes x2 + 4y2 ≤ 3y2 + 1, or x2 + y2 ≤ 1, so we can write S = ϕ(D) for the cartesian parameterization ϕ(x, y) = (x, y, f(x, y) = x2 + 4y2) on the circle 2 2 > > D = {(x, y): x +y ≤ 1}. The associated normal vector is ϕx∧ϕy = (−fx, −fy, 1) = (−2x, −8y, 1) , which has positive vertical component, so this parameterization provides the orientation for S that we are asked. Therefore

hF , dSi = hF ◦ ϕ, ϕx ∧ ϕyi dx dy ¨S ¨D

= h(1, 0, 2), (−2x, −8y, 1)i dx dy = 2 − 2x dx dy = 2A(D) = 2π, ¨D ¨D

54 where we have used that D 2x dx dy = 0, because under the change (x, y) = (−u, v) the integrand 2x changes sign but the domain˜ D remains the same.

Domain W closed by S and the cover Sc of Exercise 42

Alternatively, we can try to apply Gauss’ theorem on a solid W with a boundary ∂W containing S. 2 2 2 Thus we introduce W = {(x, y, z): x + 4y ≤ z ≤ 3y + 1} whose boundary ∂W = S ∪ Sc contains S 2 2 2 and the “cover” surface Sc = {(x, y, z): z = 3y + 1, z ≥ x + 4y } = ψ(D), for the parameterization ψ(x, y) = (x, y, g(x, y) = 3y2 + 1) on the same circle D = {(x, y): x2 + y2 ≤ 1}. The associated normal > > vector is ψx ∧ψy = (−gx, −gy, 1) = (0, −6y, 1) , which has positive vertical component and therefore points outward W . On the other hand, our previous parameterization ϕ of S points now inward W . So if we call ∂W + = + + S ∪ Sc the boundary of W oriented by the outward-pointing normal vector to W , the orientation of S+ is just opposite to the orientation of S, and thus

hF , dSi = − hF , dSi. ¨S ¨S+ We now apply Gauss’ theorem on W . Since the vector field F (x, y, z) = (1, 0, 2) is solenoid, that is, div F = 0, the flux of F through the whole boundary ∂W = S ∪ Sc will be zero, so the flux of F entering W through S will be equal to the flux of F exiting W through Sc:

0 = div F dx dy dz = hF , dSi = hF , dSi + hF , dSi + + + ˚W ‹∂W ¨S ¨Sc Then

hF , dSi = − hF , dSi = hF , dSi = h(1, 0, 2), (0 − 6y, 1)i dx dy = 2A(D) = 2π. + + ¨S ¨S ¨Sc ¨D

43. Compute the circulation −y2 dx + z dy + x dz where C is the triangle obtained when intersecting ˛C the plane 2x + 2y + z = 6 with the three axis of coordinates, oriented by the unit normal vector N(x, y, z) = (2/3, 2/3, 1/3).

Resolution: The curve C has 3 diﬀerent pieces, so the direct computation of the circulation of the vector ﬁeld F (x, y, z) = (−y2, z, x) along C involves the computation of 3 line integrals. Instead, we are going to apply Stokes’ theorem to compute one single double integral:

−y2 dx + z dy + x dz = hF , d`i = hrot F , dSi, ˛C+ ˛∂S+ ¨S+

55   i j k −1

where rot F = ∂x ∂y ∂z = −1, and S = {(x, y, z): z = f(x, y) = 6 − 2x − 2y, (x, y) ∈ D},

−y2 z x 2y where D = {(x, y): x + y ≤ 3, x ≥ 0, y ≥ 0}. Parameterizing S = ϕ(D) with the cartesian parameterization ϕ(x, y) = (x, y, f(x, y) on D, the associated normal vector ϕ ∧ ϕ = (−f , −f , 1)> = √ x y x y (2, 2, 1)> = 9N is oriented like N. Therefore

3 3−x −y2 dx + z dy + x dz = h(−1, −1, 2y), (2, 2, 1)i dx dy = dx 2y − 4 dy ˛C+ ¨S+ ˆ0 ˆ0 3 3 2 y=3−x 2 = dx y − 4y y=0 = (3 − x) − 4(3 − x) dx ˆ0 ˆ0 3 x3 x=3 = x2 − 2x − 3 dx = − x2 − 3x = 9 − 9 − 9 = −9. ˆ0 3 x=0

3 Consider the piecewise regular surface S = S1 ∪ S2 ⊂ R deﬁned by

3 2 2 S1 = {(x, y, z) ∈ R : x + y = 1, 0 ≤ z ≤ 1},

3 2 2 2 S2 = {(x, y, z) ∈ R : x + y + (z − 1) = 1, z ≥ 1}, oriented by the outward-pointing normal. Given the vector field F (x, y, z) = (x + xz + yz2, y + xyz3, x2z4), compute the flux hrot F , dSi. ¨S 3 44. Consider the piecewise regular surface S = S1 ∪ S2 ⊂ R defined by

3 2 2 S1 = {(x, y, z) ∈ R : x + y = 1, 0 ≤ z ≤ 1},

3 2 2 2 S2 = {(x, y, z) ∈ R : x + y + (z − 1) = 1, z ≥ 1}, oriented by the outward-pointing normal. Given the vector ﬁeld F (x, y, z) = (x + xz + yz2, y + 3 2 4 xyz , x z ), compute the ﬂux Shrot F , dSi. ˜ Resolution: First of all, we plot the surface S. We know that S1 is the surface of a cylinder and that S2 is the surface of a sphere. Considering their domains, the piecewise regular surface S = S1 ∪ S2 looks like this:

Surface S of exercise 44.

56 As we are asked to compute the ﬂux S hrot F , dSi given the vector ﬁeld F (x, y, z) = (x + xz + yz2, y + xyz3, x2z4), we will use Stoke’s˜ theorem

hrot F , dSi = hF , d`i, ¨S+ ˛C+ and so we need to compute the circulation of F along a counter-clockwise oriented curve C+ which 3 2 2 corresponds to the boundary of S. This curve is deﬁned by C = ∂S = {(x, y, z) ∈ R : x + y = 1, z = 0}. The parameterization of C = ∂S is

(x, y, z) = σ(θ) = (cos θ, sin θ, 0), 0 ≤ θ ≤ 2π, and therefore σ0(θ) = (− sin θ, cos θ, 0), F (σ(θ)) = (cos θ, sin θ, 0). We then need to check that our parameterization σ is actually oriented counter-clockwise. To do so we take the point σ(0) = (cos 0, sin 0, 0) = (1, 0, 0), and see how the vector σ0(0) = (− sin 0, cos 0, 0) = (0, 1, 0), is oriented. As we can see in the picture below, C is indeed oriented counter-clockwise (C+).

Surface S of exercise 44 viewed from below with a tangent vector T = σ0(t) of the boundary C = ∂S.

Finally, by Stoke’s theorem:

b 2π hrot F , dSi = hF , dì = hF (σ(θ)), σ0(θ)i dθ = h(cos θ, sin θ, 0), (− sin θ, cos θ, 0)i dθ = ¨S+ ˛C+ â ˆ0 2π = (− sin θ cos θ + sin θ cos θ) dθ = 0. ˆ0 Alternatively, we could have also realized that we did not need to compute the parameterization of C, as the vector field on C is F = (x, y, 0). Consequently, the circulation to be computed is

hF , d`i = x dx + y dy = 0, ˛C+ ˛C+ as x2 + y2 = 1 on C implies x dx + y dy = 0.

57 45. Verify the Stokes theorem with the vector ﬁeld F (x, y, z) = (2z, x, y2) and the piece of the circular paraboloid 3 2 2 S = {(x, y, z) ∈ R : z = 4 − x − y , z ≥ 0} oriented so that the boundary C = ∂S is traveled counter-clockwise.

Resolution: Stokes’ theorem states that

hF , d`i = hrot F , dSi, ˛∂S+ ¨S+ so, in order to verify it we need to compute both sides of the equality and see if they yield the same result. Before computing any integrals, let’s represent the exercise graphically.

Graph for exercise 45.

We see that we have a circular paraboloid going downwards on the z axis which is cut by the plane z = 0. We are going to study the part of the paraboloid over this plane. Also a normal vector is plotted on it, which should be upward-pointing, that later on we will see why it is useful. Let’s start with the left part of the equation:

b hF , dì = hF , dì = hF (σ(θ)), σ0(θ)i dθ, ˛∂S+ ˛C+ â where C = ∂S = σ([a, b]) is parameterized counter-clockwise by σ, according to the statement of the exercise. In this case it is easy to use a cylindrical parameterization of the paraboloid S: x = r cos θ, y = r sin θ, z = 4 − x2 − y2 = 4 − r2, or (x, y, z) = ψ(r, θ) = (r cos θ, r sin θ, 4 − r2). If we compute the intersection of the paraboloid and the plane, we get r = 2 for the boundary C = ∂S, which happens to be a circle of radius 2:

C = σ([0, 2π]), σ(θ) = (2 cos θ, 2 sin θ, 0).

Now we proceed to calculate the scalar product between F evaluated at this parameterization and the derivative of σ(θ).

F (σ(θ)) = (0, 2 cos θ, 4 sin2 θ), σ0(θ) = (−2 sin θ, 2 cos θ, 0), hF (σ(θ)), σ0(θ)i = 4 cos2 θ.

58 Finally we compute the ﬁrst integral. ! 2π 2π 1 + cos (2θ) θ θ=2π sin(2θ)θ=2π hF , d`i = 4 cos2 θ dθ = 4 dθ = 4 + = 4π. ˛C+ ˆ0 ˆ0 2 2 θ=0 4 θ=0

Now we have to compute the other part of the equation of the Stokes theorem and see if it yields the same result. To do that, ﬁrst we compute the curl of F

i j k

rot F = ∂x ∂y ∂z = (2y, 2, 1).

2z x y2

Secondly, we use a cartesian parameterization for S = ϕ(D) and we get the associated normal vector ϕx ∧ ϕy

ϕ(x, y) = (x, y, 4 − x2 − y2), (x, y) ∈ D = {(x, y):(x2 + y2 ≤ 4}

i j k

ϕ = (1, 0, −2x), ϕ = (0, 1, −2y), ϕ ∧ ϕ = 1 0 −2x = (2x, 2y, 1), x y x y 0 1 −2y

and therefore the vector surface element is upward-pointing:

dS = ϕx ∧ ϕy dx dy = (2x, 2y, 1) dx dy.

Before doing any more calculations we should make sure that the parameterization ϕ of our surface S is consistent with the orientation given by C. To do that, we evaluate it at a random point, for example for x = 0 and y = 0 we have dS = (0, 0, 1) dx dy, which is an upward-pointing vector and therefore a vector pointing out of the paraboloid as seen in the ﬁgure, so we know that our normal vector is consistent with C, which is traveled counter-clockwise. Let’s keep on going with the calculations then.

hrot F , dSi = h(2y, 2, 1), (2x, 2y, 1)i dx dy = 4xy + 4y + 1 dx dy.

To make the last integration on the disk D easier, we will change to polar coordinates.

2π 2 hrot F , dSi = 4xy + 4y + 1 dx dy = (4r2 cos θ sin θ + 4r sin θ + 1) r dθ dr. ¨S+ ¨D ˆ0 ˆ0 We can compute this integral as the sum of three diﬀerent ones. But before doing that, notice that by symmetry of the cos and the sin integrated between 0 and 2π, the two ﬁrst integrals will be zero. Therefore, we have a very simple integral to compute

2π 2 hrot F , dSi = r dr dθ = 4π. ¨S+ ˆ0 ˆ0

We have veriﬁed then Stokes’ theorem.

3 2 2 46. Let us consider the piece of a cylinder given by S = {(x, y, z) ∈ R : x + y = 4, 0 ≤ z ≤ 5}. (a)Compute the flux of the vector field F (x, y, z) = (2x, y, 3z) through the surface S oriented by the unit normal vector N(x, y, z) = (x/2, y/2, 0). (b)Compute the circulation of the vector field F (x, y, z) = (2x, y, 3z) along the boundary ∂S oriented x y by the normal vector N = , , 0 . 2 2 (c)Compute the circulation of the vector field F (x, y, z) = (2x, y, 3z) along each connected component

59 x y of the boundary ∂S where S is oriented by the normal vector N = , , 0 . 2 2

Resolution: (a)We are asked to compute the flux through the surface S so we will use the flux definition formula, however we must first parameterize our surface. As we know our surface is a cylinder we will be using cylindrical parameterization with r2 = 4: (x, y, z) = ϕ(θ, z) = (2 cos θ, 2 sin θ, z), (θ, z) ∈ D = {(θ, z): 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 5} . With this parameterization S = ϕ(D) and we can now consider the flux as:

hF , dSi = hF ◦ ϕ, ϕθ ∧ ϕzi dz dθ. ¨S ¨D First of all we need to evaluate our ﬁeld F on the parameterization: F (ϕ(θ, z)) = (4 cos θ, 2 sin θ, 3z).

We can proceed to ﬁnd the vector product ϕθ ∧ ϕz. Let’s begin by ﬁnding the derivatives with respect to θ and z: ϕθ = (−2 sin θ, 2 cos θ, 0), ϕz = (0, 0, 1). If we compute the vector product we get the following result:

ϕθ ∧ ϕz = (2 cos θ, 2 sin θ, 0).

Substituting in the integral and computing the scalar product hF ◦ ϕ, ϕθ ∧ ϕzi we get 2π 5 hF , dSi = 4 2 cos2 θ + sin2 θ dθ dz = 20(2π + π) = 60π. ¨S ˆ0 ˆ0

The last step is to check whether this result that we achieved using the vector ϕθ ∧ ϕz is the same orientation as the vector N given by the statement. Let us consider a point such that θ = π/2 and z = 0. If we substitute in the parameterization we can check that this is the point p = (x = 0, y = 2, z = 0). Now we can compute the vector N and the vector ϕθ∧ and check for linear dependence and orientation. N(p) = (0, 1, 0),

ϕθ ∧ ϕz(p) = (0, 2, 0). When we compute the vectors we can see that they are linearly dependent and they both are outward normal vectors. Therefore we must not change the sign of the result. We can now determine that the flux of the vector field F across the surface S oriented by the normal vector N will be equal to 60π. (b) First of all we must deduce the orientation of the two circles of the boundary ∂S of S with a figure and check the normal vector direction. This is essential as we will need these orientations to strike the problem afterwards.

Cylinder S with orientations of Exercise 46(b).

60 In the ﬁgure we can see the outward-pointing normal vector N, and with the right hand rule we know it induces a counter-clockwise circle (black circle). We must now look at the tangency of this induced circle with the boundaries of S.

Observing the tangency in the top circle C5 at z = 5, we see that it induces on it a clockwise orientation. Observing the tangency in the lower circle C0 at z = 0, we see that it induces on it a counter-clockwise orientation. This will be very useful when we strike the problem using Stoke’s theorem because we must know all these orientations to apply:

hF , dì = hrot F , dSi. ˛∂S+ ¨S+ Now that we know the induced orientation of the boundary ∂S we can face the problem. We have three ways of facing this problem that must lead us to the same result, using Stokes’ theorem, computing the circulation by its definition and finding a potential function. 1. Stokes’ theorem. To be able to apply Stokes’ theorem the following conditions must be met:

We must have a surface S with a boundary ∂S: We do have the surface of a cylinder and the curves are the upper lower covers perimeter. ∂S must be a curve or union of curves: We have two circles, one located on z = 5 and another on z = 0 which are the circular paths on the covers. The vector F must belong to C1 in S and ∂S: F is C1 everywhere. ∂S must be oriented by N of S leaving S on the left side: The statement of the problem is providing us with a normal vector.

As the four conditions are met, we can apply Stokes’ theorem. Our plan will be to ﬁnd rot F and then compute its scalar product with dS = N dS, as the above Stoke’s theorem states that the integral of this scalar product on S will be the same as the circulation of F on the boundary of S. In our case this theorem will be very useful as the rotational is (0, 0, 0). Indeed

rot F = (∂y(3z) − ∂z(y), −∂x(3z) + ∂z(2x), ∂x(y) − ∂y(2x)) = (0, 0, 0), so that hF , d`i = hrot F , dSi = 0. ˛∂S+ ¨S+ 2. Computing the circulation. The direct computation of the circulation must be done in the two components of ∂S = C0 ∪ C5. If we parameterize both circles with a standard counter-clockwise parameterization:

C0 = σ0 ([0, 2π]) , with (x, y, z) = σ0(α) = (2 cos α, 2 sin α, 0),

C5 = σ5 ([0, 2π]) , with (x, y, z) = σ5(α) = (2 cos α, 2 sin α, 5), which only diﬀer on the value of z, then

hF , dì = hF , dì + hF , dì = hF , dì − hF , dì, + ˛∂S ‰C0 C5 ˛σ0 ˛σ5 where we have used a counter-clockwise orientation in C0 and a clockwise orientation in C5. As

0 σj(α) = (−2 sin α, 2 cos α, 0), j = 0, 5,

F (x, y, z) = (2x, y, 3z) and z = j on each circle Cj, j = 0, 5, we have that

0 F (σj(α)) = (4 cos α, 2 sin α, 3j)), hF (σ(α)) , σ (α)i = −4 cos α sin α,

61 so we have all the ingredients to compute the circulation of F along any circle of ∂S 2π 2π 0 2 α=2π hF , dì = hF (σj(α)) , σji dα = − 4 cos α sin α dα = − 2 sin α α=0 = 0, ˛σj ˆ0 ˆ0 for j = 0, 5 so that hF , dì = 0. ˛∂S+ 3. Finding a potential function. As rot F = 0, it is easy to the potential function f(x, y, z) = y2 3z2 x2 + + for F which satisfies ∇f~ = F . Therefore the circulation of F along a curve is equal to 2 2 the difference of the potential function at the endpoints of C. If C is a closed curve, the endpoints of C coincide, and therefore the circulation of F along any closed curve is zero. In particular, hF , dì = ˛∂Cj

0 along the two circles Cj, j = 0, 5, that form the boundary of S, so that, again, hF , dì = 0. ˛∂S+ Summarizing, we can conclude that the circulation of F along the boundary of the cylinder S is zero, having calculated it in three different ways. (c) We have to consider all the components of the boundary. In our case, we study two curves, represented in the figure below.

Cylinder with both curves indicated from exercise 46(c).

In order to calculate both circulations, we will apply Stokes’ theorem

hF , d`i = hrot F , dSi. ˛∂S+ ¨S+ So as to apply Stokes’ theorem, we will need an auxiliary invented surface. We will use those that have as a boundary the previously mentioned curves 3 2 2 C1 = ∂S1 for S1 = (x, y, z) ∈ R : x + y ≤ 4, z = 5 = ϕ1(D), ϕ1(r, θ) = (r cos θ, r sin θ, 5), 3 2 2 C2 = ∂S2 for S2 = (x, y, z) ∈ R : x + y ≤ 4, z = 0 = ϕ2(D), ϕ2(r, θ) = (r cos θ, r sin θ, 0), where the parameterizations ϕ1, ϕ2 of S1, S2 are deﬁned on the common domain D = {(r, θ): 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2} .

We compute div F = ∇ ∧ F for our vectorial ﬁeld:

i j k ∂ ∂ ∂ ∇ ∧ F = = (0, 0, 0). ∂x ∂y ∂z 2x y 3z We now have everything we need to apply the theorem, so we proceed.

62 Circulation through ∂S1 We ﬁrst check if our parameterization ϕ1 for S1 is well-oriented by calculating the associated normal vector ϕ1r ∧ ϕ1θ:

ϕ1r = (cos θ, sin θ, 0), ϕ1θ = (−r sin θ, r cos θ, 0), ϕ1r ∧ ϕ1θ = (0, 0, r).

Since 0 ≤ r ≤ 2, ϕ1r ∧ϕ1θ is upward-pointing, and C1 = ∂S1 should be counter-clockwise oriented for the application of Stokes’ theorem. Since we want C1 to be clockwise oriented, we will have to change the sign of the circulation.

hF , d`i = − hF , d`i = − hrot F , dSi = 0. ˛C −=∂S ˛C +=∂S ¨S+ | {z } 1 1 1 1 =0

Circulation through ∂S2 We repeat the same procedure as before. We ﬁrst check if our parameterization ϕ2 for S2 is well-oriented by calculating the associated normal vector ϕ2r ∧ ϕ2θ:

ϕ2r = (cos θ, sin θ, 0), ϕ2θ = (−r sin θ, r cos θ, 0), ϕ2r ∧ ϕ2θ = (0, 0, r).

Since 0 ≤ r ≤ 2, ϕ2r ∧ϕ2θ is upward-pointing, and C2 = ∂S2 should be counter-clockwise oriented for the application of Stokes’ theorem, as it is.

hF , d`i = hrot F , dSi = 0. ˛C +=∂S ¨S+ | {z } 2 2 =0

Therefore we can conclude the circulation of the vector ﬁeld F along each connected component of the boundary ∂S is zero.

We notice that the parameterizations of S1, S2 were not really necessary. As long as div F = 0, Stokes’theorem immediately gives hF , dì = 0, j = 1, 2. ˛Cj 1 1 1 47. (a) For the surface S = (x, y, z): z = x2 + y2, z ≤ 1, x, y ≥ 0 oriented at the point , , by √ 2 2 2 the vector N = (1, 1, −1)/ 3, compute the flux of the vector field F = (1, 0, 0) through S, applying Stokes’ theorem. [ Hint: Use that a constant vector field F = v admits as a potential vector field the 1 linear vector field G = 2 v ∧ r.] (b) Compute the flux of the vector field F = (1, 0, 0) through S, applying Gauss’ theorem. (c) Compute the flux of the vector field F = (1, 0, 0) through S, just using the definition of flux. Resolution: (a) Let us state Stokes’ theorem for a vector field G

hrot G, dSi = hG, d`i. ¨S+ ˛∂S+

3 where S ⊂ R is a surface oriented by a unit normal vector N, and its boundary ∂S is oriented by the vector N leaving S on the left side.

63 Surface S and its oriented boundary of Exercise 47(a).

As F = (1, 0, 0) is constant, using the hint for v = (1, 0, 0), the vector ﬁeld G given by

i j k 1 1 1 1 G = v ∧ r = (1, 0, 0) ∧ (x, y, z) = 1 0 0 = (0, −z, y) , 2 2 2 2 x y z satisﬁes

i j k 1 ∂ ∂ ∂ rot G = ∇ ∧ G = = (1, 0, 0) = F . 2 ∂x ∂y ∂z 0 −z y In accordance to that, now we have by Stokes’ Theorem: 1 1 hF , dSi = hrot G, dSi = hG, d`i = −z dy + y dz = y dz − z dy. ¨S+ ¨S+ ˛∂S+ 2 ˛∂S+ 2 ˛∂S+

The boundary ∂S has three diﬀerent curves, so that ∂S = C1 ∪ C2 ∪ C3, with n π o C = (x, y, z): z = x2 + y2, z = 1, x, y ≥ 0 = (x, y, z) = σ (θ) = (cos θ, sin θ, 1), 0 ≤ θ ≤ , 1 1 2 2 2 2 C2 = (x, y, z): z = x + y , z ≤ 1, x = 0, y ≥ 0 = (x, y, z) = σ2(y) = (0, y, y ), 0 ≤ y ≤ 1 , 2 2 2 C3 = (x, y, z): z = x + y , z ≤ 1, x ≥ 0, y = 0 = (x, y, z) = σ3(x) = (x, 0, x ), 0 ≤ x ≤ 1 . 1 We now compute the circulations hG, d`i = y dz − z dy using the parameterizations σj ˆCj 2 ˆσj written above for j = 1, 2, 3, and later on we will study if they follow the correct orientation with respect to the normal vector N of S.

1 1 π/2 1 y dz − z dy = − cos θ dθ = − , (because z = 1, y = cos θ, dy = cos θ dθ, dz = 0), 2 ˆσ1 2 ˆ0 2 1 1 π/2 1 y dz − z dy = 2y2 dy − y2 dy = , (because z = y2, y = y, dy = dy, dz = 2y dy), 2 ˆσ2 2 ˆ0 6 1 y dz − z dy = 0, (because z = x2, y = 0, dy = 0, dz = 2x dx). 2 ˆσ3 √ Now we check the correct orientations for Stokes’ theorem. As the normal vector N = (1, 1, −1)/ 3 points down, we should have

64 • the path C3 beginning at (0, 0, 0) and ending at (1, 0, 1) which agrees with σ3,

• the path C1 beginning at (1, 0, 1) and ending at (0, 1, 1) which disagrees with σ1,

• the path C2 beginning at (0, 1, 1) and ending at (0, 0, 0) which disagrees with σ2. Therefore 1 hF , dSi = y dz − z dy ¨S+ 2 ˛∂S+ 1 1 1 = y dz − z dy + y dz − z dy + y dz − z dy 2 ˆ + 2 ˆ + 2 ˆ + C1 C2 C3 1 1 1 = − y dz − z dy − y dz − z dy + y dz − z dy 2 ˆσ1 2 ˆσ2 2 ˆσ3 1 1 1 = − − 0 = . 2 6 3

(b) In order to apply Gauss’ theorem a closed surface is needed, it will be achieved by enclosing S with the intersections of W and the x = 0 plane, the y = 0 plane and the z = 1 plane:

3 2 2 W = {(x, y, z) ∈ R : x + y ≤ z ≤ 1, x ≥ 0, y ≥ 0}.

The ﬁrst ﬂat cover S1, as already mentioned, will close S with the intersection of W and the x = 0 plane. We introduce x = 0 in W :

3 2 S1 = {(x, y, z) ∈ R : y ≤ z ≤ 1, x = 0, y ≥ 0}.

Its outward-pointing normal vector N1 and the normal component FN1 of the vector ﬁeld F are, respectively: N1 = (−1, 0, 0),

FN1 = hF , N1i = h(1, 0, 0), (−1, 0, 0)i = −1.

The second ﬂat cover S2 will close S with the intersection of W and the y = 0 plane:

3 2 S2 = {(x, y, z) ∈ R : x ≤ z ≤ 1, x ≥ 0, y = 0}.

Its outward-pointing normal vector N2 and the normal component FN2 of the vector ﬁeld F are, respectively: N2 = (0, −1, 0),

FN2 = hF , N2i = h(1, 0, 0), (0, −1, 0)i = 0. The third, and last, ﬂat cover will close S with the intersection of W and the z = 1 plane:

3 2 2 S3 = {(x, y, z) ∈ R : x + y ≤ 1, x ≥ 0, y ≥ 0}.

Its outward-pointing normal vector N3 and the normal component FN3 of the vector ﬁeld F are, respectively: N3 = (0, 0, 1),

FN3 = hF , N3i = h(1, 0, 0), (0, 0, 1)i = 0.

65 Flat covers S1,S2,S3 with their respective outward-pointing normal vectors.

Applying Gauss’ Theorem:

hF , dSi = hF , dSi+ hF , dS1i+ hF , dS2i+ hF , dS3i = div F dx dy dz. ‹ + ¨ + ¨ + ¨ + ¨ + ˚ ∂W S S1 S2 S3 W

As div F = 0 and hF , dSi = FN dS: hF , dSi = − hF , dS1i− hF , dS2i− hF , dS3i = − −1 dS1 = A(S1). ¨ + ¨ + ¨ + ¨ + ¨ + S S1 S2 S3 S1 Hence, 1 1 1 2 2 hF , dSi = A(S1) = dy dz = 1 − y dy = . ¨S+ ˆ0 ˆy2 ˆ0 3 We conclude that the ﬂux going through the surface S oriented by the outward-pointing normal N is 2/3. (c) We use the parameterization based on cylindrical coordinates S = ϕ(D) with ϕ(r, θ) = (r cos θ, r sin θ, z = f(r)), f(r) = r2 which satisﬁes

     2  cos θ −r sin θ i j k −2r cos θ 2 ϕr = sin θ , ϕθ =  r cos θ  , ϕr ∧ ϕθ = cos θ sin θ 2r = −2r sin θ ,

2r 0 −r sin θ r cos θ 0 r and we see that the normal vector associated to this parameterization ϕr ∧ ϕθ is always an upward- pointing vector, since its z-component is positive: r > 0.

S of Exercise 47(c) on the (r, z) plane and on the (x, y, z) space with outward-pointing normal vectors.

As x, y ≥ 0 and 0 ≤ θ ≤ π , 2 n π o D = (r, θ): 0 ≤ r ≤ 1, 0 ≤ θ ≤ . 2

66 The ﬂux is given by

hF ◦ ϕ, ϕr ∧ ϕθi dr dθ, ¨D 2 where F ◦ ϕ = (1, 0, 0) and hF ◦ ϕ, ϕr ∧ ϕθi = −2r cos θ. Then

π 1 π 2 2 2 −2 cos θ 2 hF ◦ ϕ, ϕr ∧ ϕθi dr dθ = −2r cos θ dr dθ = dθ = − . ¨D ˆ0 ˆ0 ˆ0 3 3

1 1 1 2 Let’s check now the sign of the ﬂux. Substituting ( 2 , 2 , 2 ) = (r cos θ, r sin θ, r ), we get r cos θ = 2 1 r sin θ = r = 2 , so that on this point the normal vector provided by the parameterization is

√ 1 r3 ϕ ∧ ϕ = (−2r2 cos θ, −2r2 sin θ, r) = (−r, −r, z) = −√ (1, 1, −1) = − N, r θ 2 2 2 so we have to change the sign of the flux. The flux is . 3 Equivalently, we can notice that the given unitary normal vector N is a downward-pointing vector. This already implies the change of sign after the computation of the flux.