Mean value theorem for double integral Let D be an elementary region and f : D → IR be continuous, then for some pt.

(x0 , y0 ) in D we have f (x, y)dA = f (x , y ) • A(D) ∫∫ 0 0 D where A(D) denotes the area of D.

Pf. Since f is continuous on D, it has a maximum value M and a minimum value m. Thus m ≤ f (x, y) ≤ M for all (x, y) ∈ D ∴mA(D) = ∫∫mdA ≤ ∫∫ f (x, y)dA ≤ ∫∫ MdA = M • A(D) DD D 1 i.e. m ≤ ∫∫ f (x, y)dA ≤ M A(D) D Since a continuous function on D takes on every value between its maximum and

minimum values (Intermediate value theorem)., there must be a pt. (x0 , y0 ) ∈ D with 1 f (x , y ) = f (x, y)dA 0 0 ∫∫ A(D) D

Triple Integral Given a continuous function f : C → IR , where C is some rectangular box in IR 3 , we can define the integral of f over C as a limit of sums just as we did for a function of two variables. Briefly, we partition the three sides of C into n equal parts and form n−1 n−1 n−1 the sum S n = ∑∑∑ f (Cijk )∆V i=0 j=0 k=0

th Where cijk ∈ Cijk , the ijk rectangular box in the partition of c and ∆V is the

volume of Cijk .

Def. Let f : C → IR be continuous. If lim S n exists and the limit is independent n→∞

of the pts cijk , we call the limit of S n the triple integral of f over C and denote it by

∫∫∫ f (x, y, z)dV C

Theorem Let f : C → IR be a bounded real-valued function on the rectangular box C and suppose that f is continuous except for a finite set of pts, then f is integrable over C.

Fubini’s theorem Let f be a continuous function with domain C = [a,b]×[c, d]×[u,v]. Then

b d v ∫∫∫ f (x, y, z)dV = ∫∫∫ f (x, y, z)dzdydx C a c u

v b d = ∫∫∫ f (x, y, z)dydxdz u a c

c v b = .... = ∫∫∫ f (x, y, z)dxdzdy d u a

A region W is of type I if

W = {(x, y, z) : a ≤ x ≤ b,φ2 (x) ≤ y ≤ φ1 (x),γ 2 (x, y) ≤ z ≤ γ 1 (x, y)}

where φi ,γ i ,i = 1,2 are continuous functions.

or W = {(x, y, z) : c ≤ y ≤ d,ψ 2 (y) ≤ x ≤ψ 1 (y),γ 2 (x, y) ≤ z ≤ γ 1 (x, y)} where ψ i ,γ i ,i = 1,2 are continuous functions.

Definition For more general bounded sets W ⊆ IR3 and f :W → IR a given

continuous function. Choose a box B that contains W and define f (x,y,z),(x,y,z)∈W f *(x, y, z) = {0,(x,y,z)∈B|W then f * is integrable over B and we can define ∫∫∫ f (x, y, z)dV = ∫∫∫ f *(x, y, z)dV WB Suppose W is of type I. Then either

b φ1 (x) γ1 (x, y) ∫∫∫ f (x, y, z)dV = ∫ ∫ ∫ f (x, y, z)dzdydz W a φ2 (x) γ 2 (x, y)

γ1 (x, y) = ∫∫ ∫ f (x, y, z)dz)dydx D γ 2 (x, y)

e.g. Let W be the region bounded by the planes x=0, y=0, z=2 and the surface z = x 2 + y 2 , x ≥ 0, y ≥ 0. Compute ∫∫∫ xdxdydz W 2 2 Sol. Let γ 1 (x, y) = 2,γ 2 (x, y) = x + y

2 φ1 (x) = 2 − x ,φ2 (x) = 0

a=0 and b = 2 . Then W is a region of type I.

1 2 (2−x2 ) 2 2 ∫∫∫ xdxdydz = ∫∫[ ( ∫xdz)dy]dx W 0 0 x2 + y2

1 2 (2−x2 ) 2 = ∫∫[ (x(2 − x 2 − y 2 )dydx 0 0

3 2 3 (2 − x 2 ) 2 = ∫ x[(2 − x 2 ) 2 − ]dx 0 3

2 2 3 = ∫ x(2 − x 2 ) 2 dx 0 3

1 2 3 = − ∫ (2 − x 2 ) 2 d(2 − x 2 ) 3 0

5 2 2 − 2(2 − x ) 2 = 15 0

5 2 2 8 2 = 2( ) = 15 15

2 2 2 2 x y z e.g. Evaluate I = (x 2 + y 2 + z )dV , where W = {(x, y, z) : + + ≤ 1} ∫∫∫ 2 2 2 W a b c

Sol : Let D be the region obtained by the intersection of W with a plane // to x-y z0

plane that passes through (0,0, z0 ) . Then x 2 y 2 D = {(x, y) : + ≤ 1} z0 z 2 z 2 a 2 (1− 0 ) b 2 (1− 0 ) c 2 c 2 I = x 2 dV + y 2 dV + z 2 dV = I + I + I ∫∫∫ ∫∫∫ ∫∫∫ 1 2 3 W w W

c c z 2 4 I = ( z 2 dxdy)dz = z 2πab(1− )dz = πabc3 3 ∫∫∫∫ c 2 15 −−c Dz c z 2 (Q dxdy = Area of Dz = πab(1− 2 )) ∫∫ c Dz

4 4 Similarly, I = πa 3bc, I = πab3c 1 15 2 15

4 b ∴ I = πabc(a 2 + b 2 + c 2 ) = ( y 2 dxdz)dy 15 ∫∫∫ −bDy

Maps from IR 2 to IR 2 π e.g.1 Let D* = [a,b]×[0, ] where 0

Change of variable theorem Def. Let T : D* ⊆ IR2 → IR2 be a C’ transformation given by x=x(u, v) and y-y(u, v). ∂x ∂x ∂(x, y) The Jacobian of T is = ∂u ∂v = det[DT(u,v)] ∂(u,v) ∂y ∂y ∂u ∂v e.g. The function from IR 2 → IR 2 that transforms polar co-ordinate into Cartesian co-ordinate is given by x = r cosθ , y = r sinθ ∂(x, y) cosθ − r sinθ and its Jacobian is = = r ∂(u,v) sinθ r cosθ

Theorem Let D and D* be elementary regions in IR 2 and let T : D* → D be C’, suppose that T is onto. Then for any integrable function f : D → IR , we have ∂(x, y) ∫∫ f (x, y)dxdy = ∫∫ f (x(u,v), y(u,v) dudv D D* ∂(u,v) Pf.

 ∂x ∂x   ∂x      v ∂u ∂v ∆u ∂u DT(∇ui ) =    = ∆u   ∂y ∂y  0  ∂y        ∂u ∂v   ∂u   ∂x ∂x   ∂x      v ∂u ∂v  0  ∂v DT(∇vj) =    = ∆v   ∂y ∂y  ∆v  ∂y        ∂u ∂v   ∂v 

Area of T(R*) is approximately equal to the absolute value of ∂x ∂x ∂x ∂x ∆u ∆v ∂(x, y) ∂u ∂v = ∂u ∂v ∆u∆v = ∆u∆v = ∂y ∂y ∂y ∂y ∆u ∆v ∂(u,v) ∂u ∂v ∂u ∂v

e.g Let P be the //gram bounded by y=2x, y=2x-2, y=x and y=x +1. Evaluate ∫∫(x, y)dxdy by making the change of variables : x=u-v, y=2u-v D Solution :

T(u, v) = (u-v, 2u-v)

∂(x, y) 1 −1 ∴ = det  = 1 ∂(u,v) 2 −1 Hence ∫∫ xydxdy = ∫∫(u − v)(2u − v)dudv P P*

0 1 = ∫∫(2u 2 − 3uv + v 2 )dudv −2 0

0 2 3u 2v = [ u 3 − + v 2u]1 dv ∫ 0 −2 3 2

2 3 v3 = [ v − v 2 + ]0 3 4 3 −2 2 8 = − [ (−2) − 3 − ] = 7 3 3

2 2 e.g. Evaluate ∫∫ln(x + y )dxdy , where D D = {(x, y) : a 2 ≤ x 2 + y 2 ≤ b 2 , x ≥ 0, y ≥ 0} (0 < a < b) Solution Let x = r cosθ , y = r sinθ i.e. T (r,θ ) = (x, y) π Then T : D* → D is 1-1 onto, where D* = {(r,θ ) : a ≤ r ≤ b,0 ≤ θ ≤ } 2 2 2 2 ∂(x, y) ∴ ∫∫ln(x + y )dxdy = ∫∫lnr drdθ D D* ∂(r,θ ) π b 2 b 2 π = ∫∫rlnr drdθ = ∫ 2rlnrdr a 0 2 a π 1 = [b 2 nb − a 2 na − (b 2 − a 2 )] 2 l l 2 x 2 x 2 (Using integration by parts : x nxdx = nx − ) ∫ l 2 l 4 e.g. Evaluate ∫∫ x 2 + y 2 dxdy , where R = [0,1]×[0,1] R π Solution Let D * = {(r,ϑ) : 0 ≤ θ ≤ and0 ≤ r ≤ secθ}, 1 4 π π D * = {(r,ϑ) : ≤ θ ≤ and0 ≤ r ≤ cscθ} 2 4 2 Define T (r,θ ) = (r cosθ, r sinθ ) , then

T(D1*) = T1 , T(D2 *) = T2 , R = T1 ∪T2 = T(D*)

We have to find the volume of the region under z = x 2 + y 2 and over R.

From symmetry of z = x 2 + y 2 on R, we can see that

∫∫ x 2 + y 2 dxdy = 2∫∫ x 2 + y 2 dxdy R π = 2∫∫( r 2 )rdrdθ D1*

π 4 secθ = 2∫∫( r 2 dr)dθ 0 0

π 2 4 = ∫sec3 θdθ 3 0

π π 4 secθ tanθ π 1 4 sec3 θdθ = [ ] 4 secθdθ + secθdθ ∫ 0 ∫ 0 2 2 0 2 1 π 2 1 = + [ n secθ + tanθ ] 4 = + n(1+ 2) 2 2 l 0 2 2 l

2 2 1 ∴ ∫∫ x + y dxdy = [ 2 + ln(1+ 2)] R 3 Def. Let T :W ⊆ IR3 → IR3 be a C 1 function defined by x=x(u, v, w), y=y(u, v, w), z=z(u, v ,w) . Then the Jacobian of T is ∂x ∂x ∂x ∂u ∂v ∂w ∂(x, y, z) ∂y ∂y ∂y = ∂(u,v, w) ∂u ∂v ∂w ∂z ∂z ∂z ∂u ∂v ∂w Change of variable formula ∂(x, y, z) ∫∫∫ f (x, y, z)dxdydz = ∫∫∫ f (x(u,v, w), y(u,v, w), z(u,v, w) dudvdw DD* ∂(u,v, w) where D* is an elementary region in uvw-space correspond to D in xyz-space, under a co-ordinate change T : (u,v, w) → (x(u,v, w), y(u,v, w), z(u,v, w))

(i) Jacobian for the map defining the change to cylindrical co-ordinates x = r cosθ, y = r sinθ, z = z

cosθ − r sinθ 0 ∂(x, y, z) ∴ = sinθ r cosθ 0 = r ∂(r,θ, z) 0 0 1

∴∫∫∫ f (x, y, z)dxdydz = ∫∫∫ f (r cosθ,r sinθ )rdrdθdz DD* (ii) Consider the spherical co-ordinate system given by x = ρ sinφ cosθ, y = ρ sinφ sinθ, z = ρ cosφ

sinφ cosθ − ρ sinφ sinθ ρ cosφ cosθ ∂(x, y, z) = sinφ sinθ ρ sinφ cosθ ρ cosφ cosθ ∂(ρ,θ,φ) cosφ 0 − ρ sinφ

− ρ sinφ sinθ ρ cosφ cosθ sinφ cosθ − ρ sinφ sinθ = cosφ − ρ sinφ ρ sinφ cosθ ρ cosφ sinθ sinφ sinθ ρ sinφ cosθ (expand along the last row) = − ρ 2 cos 2 φ sinφ sin 2 θ − ρ 2 cos 2 φ sinφ cos 2 θ − ρ 2 sin 3 φ cos 2 θ − ρ 2 sin 3 φ sin 2 θ = − ρ 2 cos 2 φ sinφ − ρ 2 sin 3 φ = − ρ 2 sinφ ∴∫∫∫ f (x, y, z)dxdydz D = ∫∫∫ f (ρ sinφ cosθ, ρ sinφ sinθ, ρ cosφ)ρ 2 sinφdρdθdφ D* 3 e.g. Evaluate ∴ ∫∫∫exp(x 2 + y 2 + z 2 ) 2 dV , where D = {(x, y, z) : x 2 + y 2 + x 2 ≤ 1} D

Sol. Use spherical co-ordinates, then D* = {(ρ,θ,φ) : 0 ≤ ρ ≤ 1,0 ≤ θ ≤ 2π ,0 ≤ φ ≤ π}

3 3 ∴ ∫∫∫exp(x 2 + y 2 + z 2 ) 2 dv = ∫∫∫ρ 2e ρ sinφdρdθdφ D D*

1 ππ2 3 = ∫∫∫ρ 2e ρ sinφdθdφdρ 00 0

1 π 3 = 2π ∫∫e ρ ρ 2 sinφdφdρ 0 0

1 3 = − 2π ρ 2e ρ [cosφ]π dρ ∫ 0 0

1 3 4π = − 4π ρ 2e ρ dρ = [e ρ3 ]1 ∫ 0 0 3

4π = (e −1) 3 e.g. Let D = {(x, y, z) : x 2 + y 2 + z 2 ≤ R 2 } be the ball of radius R and center (0,0,0) ∈ IR3

ππ2 R then Vol(D) = ∫∫∫dxdydz = ∫∫∫ρ 2 sinφdρdθdφ D 0 00

R 3 ππ2 = ∫∫sinφdθdφ 3 0 0

2πR 3 π = ∫sinφdφ 3 0

2πR 3 = [−cosφ]π 3 0 4πR3 = (vol. of solid sphere) 3

Integrals over paths Def. The path integral or the integral of f(x, y, z) along the path σ : I = [a,b] → IR3

b is defined by ∫∫fds ≡ f (x(t), y(t), z(t)) σ '(t) dt σ a

2 2 2 where σ '(t) = x'(t) + y'(t) +z'(t) , ds = elt.of arc length e.g. Let σ :[0,2π ] → IR3 be the helix given by t → (cos t,sin t,t) and let f (x, y, z) = x 2 + y 2 + z 2 , then d d dt σ '(t) = [ cost]2 + [ sin t]2 +( ) 2 dt dt dt

= sin 2 t + cos2t +1 = 2

f (σ (t) = cos 2 t + sin 2 t + t 2 = 1+ t 2

2π t 3 ∴ f (x, y, z)ds = (1+ t 2 ) 2dt = 2[t + ]2π ∫ ∫ 0 σ 0 3

2 2π = (3 + 4π 2 ) 3 If we think of the helix as a wire and f (x, y, z) = x 2 + y 2 + z 2 as the mass density, 2 2π then the total mass of the wire is (3 + 4π 2 ) . 3 Limit definition Subdivide the interval I = [a,b] by a partition

a = t0 < t1 < ... < tn = b

This leads to a decomposition of σ into paths σ i defined on [ti ,ti+1 ] for

0 ≤ i ≤ n −1. Denote the arc length of σ i by ∆Si .

ti+1 i.e. ∆Si = σ '(t) dt ∫t i

When n is large, the arc length ∆Si is small and f(x, y, z) is approximately constant for points on σ i . Then n−1

∴ f (x, y, z)ds = lim f (x, y, z)∆Si ∫ n→∞ ∑ σ i=0

n−1 t = lim i +1 f (x(t), y(t), z(t) γ '(t) dt n→∞ ∑∫t i=0 i

b = ∫ f (x, y, z) γ '(t) dt a when σ (t) = (x(t), y(t)) is a plane curve and f (x, y) ≥ 0 . We can construct a “fence” with base the image of I=[a, b] under σ and with altitude f(x, y) at (x, y). Then

b ∫ f (x, y)ds = ∫ f (x(t), y(t) x'(t) + y'(t)dt σ a

represents the area of a side of this fence.

e.g. Evaluate I = ∫ (x + y)ds σ where σ is the line segments joining (0,0), (1,0) and (1,1) Sol. I = ( ∫ + ∫ + ∫ )(x + y)ds OA AB BO y=0 on OAand ds = dx,

1 1 ∴ ∫ (x + y)ds = ∫ xdx = OA 0 2

x=0 on AB and ds = dy,

1 3 ∴ ∫ (x + y)ds = ∫ (1+ y)dy = AB 0 2 y=x on BO and ds = − 2dx

1 ∴ ∫ (x + y)ds = ∫ − 2x 2dx = − 2 BO 0

∴ I = 2 − 2 “The adventures of Tow Sawyer” Mark Twain at 2 e.g. Find the total mass of a wire σ (t) = (a,at, ) 0 ≤ t ≤ 1, a > 0 with linear 2 2z density ρ(x, y, z) = a

Sol. ds = a 1+ t 2 dt

1 a ∴m = ∫ ρ(x, y, z)ds = ∫ at 1+ t 2 dt = (2 2 −1) σ 0 3

e.g. Tom Sawyer’s aunt has ask him to whitewash both sides of the old fence shown. Tow estimated that for each 25 sq. ft of whitewashing he lets someone do for him, the willing victim will pay 5 cents. How much can Tom hope to earn, assuming his aunt will provide whitewash free of charge?

Solution σ '(t) = (−90cos 2 t sin t,90sin 2 t cost)

∴ σ '(t) = 90sint cost

Area of one side of half of the fence

π y 2 30sin 3 t = ∫ (1+ )ds = ∫ (1+ )90sin t costdt σ 3 0 3

π 2 = 90∫ (sin t +10sin 4 t)costdt 0 sin 2 t π = 90[ + 2sin 5 t] 2 2 0 1 = 90( + 2) = 225 2 Hence, area of one side of fence is 450 sq.ft. Since both sides are to be whitewashed, the total area is 900sq.ft. Thus the amount 900 5 that Tom could realize for the job = $( )( ) = $1.8 25 100