1 Vector Algebra
We will now define addition, subtraction and scalar multiplication of vectors. Let v, w ∈ Rn then,
v ± w = hv1 ± w1, v2 ± w2, ··· , vn ± wni
λv = λhv1, v2, ..., vni = hλv1, λv2, ..., λvni where v = (v1, ..., vn) and w = (w1, ..., wn) and λ ∈ R.
There is some underlying geometry to the addition of vectors. By our equivalence definition, we can proceed without loss of generality assuming v = ha, bi, w = hc, di and their base-points are at the origin.
2 Due to the geometry above, we say that the sum of two vectors obey the parallelogram law. We will now record some basic properties of vector algebra which will given as exercises. From now on I will choose not to bold vectors i.e v = v.
Fact: (Basic Properties of Vector Algebra) For all vector v, u, w and for all scalars λ,
• v + w = w + v
• u + (v + w) = (u + v) + w
• λ(v + w) = λv + λw
Exercises
(1) Sketch the vectors for v + tw where t = 0, ±1, ±2, ±3. Is there a geometric object in which the heads are describing?
(2) Let v be a nonzero vector and λ be a nonzero scalar. Show that kλvk = |λ|kvk. Why do we need the absolute value?
(3) A unit vector v is a vector such that kvk = 1. Given any vector nonzero vector v in the plane, can you think of a scalar λ such that kλvk = 1? Hint: Use the question above!
(4) Let i = h1, 0i and j = h0, 1i. Show that any vector v in the plane can be written as a linear combination of i, j i.e there exists s, t ∈ R s.t ai + bj = v.
(5) Verify the facts for the basic properties of vector algebra.
3 Vectors in Three Dimensions
A point in R3 = R × R × R is of the form (a, b, c) where it is clear from the decomposition that a, b, c ∈ R. A vector in R3 is of the form ha, b, ci.
Fact: Given P = (a1, b1, c1) and Q = (a2, b2, c2) then d(P,Q) i.e kP − Qk is given by,
p 2 2 2 kP − Qk = (a2 − a1) + (b2 − b1) + (c2 − c1)
Parametric Equations of a Line
Given P0 = (p0, p1, p2) and v = hv0, v1, v2i we have that a parametrization for the line through P0 in the direction of v is given by,
r(t) = OP0 + tv = hp0, p1, p2i + thv0, v1, v2i
1 Exercises
(1) Use vector geometry to show that a line can be parametrized in the way which is given in the beginning of this section. Verify this for both plane and space lines. (2) Show that parameterizations for space lines aren’t unique. Hint: Think of an explicit example!
(3) Let r1(t), r2(s) be two parametrizations for lines in 3-space. Come up with a definition for the intersection of two lines. With this definition verify whether r1(t) = h1, 0, 1i + th3, 3, 5i and r2(s) = h3, 6, 1i + sh4, −2, 7i intersect. (4) Let P,Q ∈ R3. Show that g(t) = (1 − t)P + tQ where −∞ < t < ∞ is a parametrization for the line through P,Q. With this definition show that the midpoint of the segment from P to Q is g(.5).
2 Dot Product and the Angle Between Two Vectors
Let v = ha1, b1, c1i and w = ha2, b2, c2i then the dot product of v, w is defined by,
v · w = a1a2 + b1b2 + c1c2
Theorem: Properties of the Dot Product
(1) 0 · v = v · 0 = 0 (2) v · w = w · v (3) (λv) · w = v · (λw) = λ(v · w) (4) u · (v + w) = u · v + u· and (v + w) · u = v · u + w · u (5) v · v = kvk2
Theorem: (Dot Product and the Angle) Let 0 ≤ θ ≤ π be the angle between two nonzero vectors v, w then we have that,
v · w = kvkkwk cos(θ)
Theorem (Projection) Assume v is a nonzero vector. The projection of u along v is the vector,
u · v u = v || v · v
Lemma (Decomposition) If v is a nonzero vector, then every vector u can be written as the sum of the projection u|| and a vector u⊥ that is orthogonal to v.
1 Exercises
(1) Prove the angle between vectors formula. Hint: If you have a triangle with sides being the vectors v, w, what is the third side? (2) If two nonzero vectors u, v are perpendicular then what is u · v ? (3) Use the above question to prove the projection formula? (4) In the case of projecting u onto a vector v, show that u may be decomposed as suggested in the lemma.
2 The Cross Product
To being this section we will just recall some facts from basic linear algebra. We will start with the determinant of a 2x2 and 3x3 matrix.
Let A be a 2x2 matrix with real entries then the determinant of A is given by,
a b det(A) = = ad − bc c d
The determinant of a 3x3 matrix is expressed in terms of 2x2 determinant called minors.
a1 b1 c1 b2 c2 a2 c2 a2 b2 det(A) = a2 b2 c2 = a1 − b1 + c1 b3 c3 a3 c3 a3 b3 a3 b3 c3
Definition (The Cross Product) The cross product of vector v = ha1, b1, c1i and w = ha2, b2, c2i is the vector,
i j k b1 c1 a1 c1 a1 b1 v × w = a1 b1 c1 = i − j + k b2 c2 a2 c2 a2 b2 a2 b2 c2
Theorem (Geometric Description of the Cross Product) The cross product v × w is the unique vector with the following three properties;
(1) v × w is orthogonal to v, w (2) kv × wk = kvkkwk sin θ where θ is the angle between v, w and 0 ≤ θ ≤ π (3) {v, w, v × w} forms a right-handed system
Theorem (Basic Properties of the Cross Product) Below are some properties of the cross product which you will verify in the exercises.
(1) w × v = −v × w (2) v × v = 0 (3) (λv) × w = v × (λw) = λ(v × w) (4) (u + v) × w = u × w + v × w (5) u × (v + w) = u × v + u × w
1 Cross Products, Area and Volume
We will now give some more geometric properties of the cross product and the determinant function. Below are images of a plane spanned by two vectors and a parallelpiped spanned by three vectors.
Theorem (Area and Volume via Cross Products and Determinants) Let u, v, w be nonzero vectors in R3. Then we have the following relations (show in exercises).
(1) The parallelogram P spanned by v, w has area A = kv × wk u
(2) The parallelepiped P spanned by u, v, w has volume V = |u · (v × w) = det v
w
Exercises
(1) Verify all the basic properties of the cross product (2) Compute the area of a general parallelogram in the plane and volume of a parallelopiped in 3-space. (2) Prove the theorem above in reference to area of parallelograms and parallelpipeds spanned by nonzero vectors. (3) Prove the scalar triple product and show that v × w is orthogonal to both v, w where each are nonzero.
2 Planes in Three-Space
Before exploring planes in 3-space, let us revise our knowledge of line equations in the plane using vectors. Let p0 be a point in the plane and ~p0 be the corresponding position vector. If we choose v to be the direction of the line passing through p0 we have the vector parametrization r(t) = ~p0 + tv. Here if we let v = hv1, v2i then for (x, y) on the line parametrized by r(t) we have h(x, y) − (x0, y0)i ·h−v2, v1i = 0. | {z } ~u
This follows from the fact ~u is in the direction of v which is orthogonal to h−v2, v1i. Simplifying the dot product above we recover −v2x + v2x0 + v1y − v1y0 = 0 ⇒ ax + by = c where a = −v2, b = v1 and c = −v2x0 − v1y. This standard representation of a line in the plane that we are used to. We will now show that this formulation carries on into 3-space i.e the equation of a plane is given by ax + by + cz = d.
Consider a plane P passing through a point (x0, y0, z0). Let ~n = ha, b, ci be a nonzero vector such that for 3 any (x, y, z) ∈ R we have h(x, y, z) − (x0, y0, z0)i · ~n = 0. Geometrically, the set of (x, y, z) such that the above equation is true gives a plane passing through the point (x0, y0, z0). If we simplify the notation above we have that ax + by + cz = ax0 + by0 + cz0. | {z } = d
Given three points P, Q, R which are not collinear, there is a unique plane which passes through all of them. The uniqueness of this plane will come from the fact that it contained the lines which pass through P,R and P,Q i.e having common intersection P . Choosing P, Q, R as described above then the plane is given by h(x, y, z) − P i · (PQ~ × PR~ ). Notice that to compute this equation we needed a point on the plane, therefore we could have chosen P,Q or R.
1 Exercises
(1) Try to come up with a definition which says when two planes are parallel i.e they don’t intersect. Hint: Think of normal vectors! (2) Verify that the planes which use P,Q or R as points in the above image, describe the same plane. (3) What if in the above definition we use the vectors RP,~ QP~ or RQ,~ RP , does the plane equation change?
(4) Given a plane ax + by + cz = d and a vector parametrization of a line r(t) = ~p0 + tv, what would be a “good” definition for the intersection of the line and plane? (5) Given to planes with L (line) as their intersection. What is the angle between the two planes?
2 A Survey of Quadratic Surfaces
Before we begin our study of quadratic surfaces we first remind ourselves of the general conic sections encountered in middle/highschool. These conics are obtained by what we will define as traces of a two-sheeted surfaces. This will be made more precise later.
We will derive these during class discussion but the equations of the following conics in general position are;
x2 y2 (1) Ellipse: + = 1 a2 b2
x2 y2 (2) Hyperbola: − = 1 a2 b2 1 (3) Parabola: x2 4c
(4) Circle: (x − a)2 + (y − b)2 = r2
1 Definition (Parabola): A parabola is the set of all points (x, y) in a plane that are equidistant from a fixed line called the directrix and a fixed point called the focus (not on the line). The midpoint between the focus and the directrix is called the vertex, and the line passing through the focus and the vertex is called the axis of the parabola.
In view of this definition, if we let y = −p and p 6= 0 and choose (x, y) ∈ R2 then we want d((0, p), (x, y)) = d((x, y), (x, −p)) = y + p and so we have x2 = ypy and so we recover equation (3).
Definition (Ellipse): An ellipse is the set of all points (x, y) in a plane, the sum of whose distances from two distinct fixed points, called foci, is constant. The line through the foci intersects the ellipse at two points, called the vertices. The chord joining the vertices is called the major axis, and its midpoint is called the center of the ellipse. The chord perpendicular to the major axis at the center is called the minor axis of the ellipse.
The verification of this formula requires a bit a algebraic manipulation. To start, just take (x, y) to be a vertex and (−c, 0) and (c, 0) to be your foci. Depending on whether your major and minor axis are vertical or horizontal you get the standard forms,
x2 y2 x2 y2 + = 1 or + = 1 a2 b2 b2 a2
2 Definition (Hyperbola): A hyperbola is the set of all points (x, y) in a plane, the difference of whose distances from two distinct fixed points, called foci, is constant. The graph of a hyperbola has two disconnected parts, called branches. The line through the two foci intersects the hyperbola at two points, called vertices. The line segment connecting the vertices is called the transverse axis, and the midpoint of the transverse axis is called the center of the hyperbola.
Again here the derivative is just a lot of algebraic manipulation and for convenience we take (x, y) to be a vertex. The standard form of the equation of a hyperbola with the center at the origin (a, b 6= 0) is,
x2 y2 y2 x2 − = 1 or − = 1 a2 b2 a2 b2
Definition (Circle): A circle is the set of points (x, y) such that the distance from (x, y) and the origin is some constant r > 0.
By the above definition we have d((x, y), (0, 0)) = r ⇒ px2 + y2 = r and so x2 + y2 = r2. This is probably the one definition which wasn’t too surprising.
3 Below we give the graphs for the quadratic surfaces and their equations. The equations will become more clear if we look at how planes in R3 intersect each surface. This will give us a good idea of the curves that lie on our surfaces as in the example for the two-sheeted figure.
4 Cylindrical and Spherical Coordinates
We will now introduce two generalizations of polar coordinates to R3: cylindrical and spherical coordinates. One motivation switching coordinates systems is that later an appropriate coordinate change may solve an integral more easily than one in standard cartesian as in second-year calculus.
Cylindrical Coordinates: Recall that a point in the plane is completely characterized by its distance from the origin and the angle it makes with the positive x-axis. Therefore, if we associate to each point P ∈ R3 the coordinate (r, θ, z) then P describes a point on a cylinder in this coordinate system.
We now show via table how to go back and forth between cylindrical and rectangular systems.
1 Spherical Coordinates: Just as a point P can be described as a point on a cylinder, we can also describe P as a point on a sphere. The picture below shows that to each point P is rectangular coordinates, we can associate to it (ρ, θ, φ) where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π.
The above facts follows quickly from elementary geometry. We again, via table demonstrate how to transition between spherical and rectangular coordinate systems.
2 Exercises
(1) Find an equation of the form z = f(r, θ) for the surfaces; x2 + y2 + z2 = 9 and x + y + z = 1. (2) Geometrically describe the following shapes which arise if in spherical coordinates we let ρ = R, θ = θ0 and φ = φ0. (3) In spherical coordinates find an equation of the form ρ = f(θ, φ) for the following surfaces; x2 + y2 + z2 = 9 and z = x2 − y2. (4) Verify the identities which are given for the transition between rectangular and spherical coordinates.
3 Vector-Valued Functions
The goal of this section is to give you a feel for how to parametrize space curves using vectors. Just as you parametrized planes curves, we will now associate to a particle in space, a vector. A vector-valued function is a function of the form ~r(t) = hx(t), y(t), z(t)i.
The underlying curve C traced by ~r(t) is the set of all points (x(t), y(t), z(t)) where t ∈ dom(~r(t)). A curve in R3 is called a space curve.
Convince yourself using vector geometry that if we wanted to parametrize a circle of radius 2 which is centered at (2, 6, 8) and parallel to the xy-plane then one solution is ~g(t) = h2, 6, 8i + h2 cos t, 2 sin t, 0i.
1 Exercises
(1) Use vector geometry to parametrize a line through the point (1, 1, 1) in the direction of (−2, 1, 1). (2) Give a parametrization for a circle of radius 3, centered at (2, 6, 8) and is parallel to the xz-plane. Explicitly show where 4 points on the circle go using vector geometry. x2 y2 (3) Find a parametrization of the ellipse + is the xy-plane, translated to have center (9, −4, 0). 4 9 1 (4) Parametrize the intersection of the plane y = and the sphere x2 + y2 + z2 = 1. 2 (5) Find a parametrization of the intersection for a sphere of radius 1 and then plane z = 2. What is this geometric shape? (6) In section 12.6 we talked about quadratic surfaces. Parametrize the intersection of planes x = a, y = b, z = c for each of these surfaces. You can pick any values a, b, c, just make sure they intersect non-trivially.
2 Calculus of Vector-Valued Functions
Let r(t) be a vector-valued function and for simplicity let r(t) = hf(t), g(t)i. Suppose that r(t) is differentiable at t0 then,
0 r(t0 + h) − r(t0) hf(t0 + h) − f(t0), g(t0 + h) − g(t0)i 0 0 r (t0) = lim = lim = hf (t0), g (t0)i h→0 h h→0 h
Theorem (Vector-Valued Derivative Are Computed Componentwise) A vector-valued function r(t) = hx(t), y(t), z(t)i is differentiable if and only if each component is differentiable. In this case,
d r0(t) = r(t) = hx0(t), y0(t), z0(t)i dt
Theorem (Differentiation Rules) Assume that r(t), r1(t) and r2(t) are differentiable. Then, 0 0 0 (1) Sum Rule: (r1(t) + r2(t)) = r1(t) + r2(t) (2) Constant Multiple Rule: For any constant c ∈ R, (cr(t))0 = cr0(t). (3) Product Rule: For any differentiable scale-valued function f(t); d (f(t)r(t)) = f(t)r0(t) + f(t)r(t) dt (4) Chain Rule: For any differentiable scalar-valued function g(t); d r(g(t)) = g0(t)r0(g(t)) dt
Theorem (Product Rule for Dot and Cross Products) Assume that r1(t) and r2(t) are differentiable. Then; d (r (t) · r (t)) = r (t) · r0 (t) + r0 (t) · r (t) dt 1 2 1 2 1 2
d (r (t) × r (t)) = [r (t) × r0 (t)] + [r0 (t) × r (t)] dt 1 2 1 2 1 2
1 0 The above shows that the vector r (t0) points in the direction of the tangent path at r(t0). We can 0 parametrize this path as L(t) = r(t0) + tr (t0). Our last remark is about vector-valued integration, which you can probably guess, is also done component-wise. This follows from the fact that, we will say a vector-valued function r(t) is integrable, if and only if each component is and so;
Z Z Z Z r(t) dt = x(t) dt, y(t) dt, z(t) dt [a,b] [a,b] [a,b] [a,b]
Exercises
0 0 0 (1) Show for a differentiable vector valued function r(t) = hx1(t), ..., xn(t)i then r (t) = hx1(t), ..., xn(t)i.
(2) Prove the differentiation rule for r(t) = hr1(t), r2(t)i. (3) Prove the product rule for dot products and cross product.
0 (4) Why does the difference quotient for r (t0) point in the direction of the tangent path? (5) Parametrize a circle, an ellipse and a parabola. Compute the tangent vector at any point you like and draw the tangent line. Lastly, give the equation for the tangent line at that point.
2 Arc Length and Speed
Let r(t) = hx(t), y(t)i be a differentiable vector-valued function on [a, b]. Choose a partition P = {a = t0 < t1 < t2 < ··· < tN = b} of [a, b]. Consider the line segment which extends from r(tj−1) = hx(tj−1), y(tj−1)i to r(tj) = hx(tj−1), y(tj)i. The distance of this segment is given by;
q 2 2 Lj = [x(tj) − x(tj−1)] + [y(tj) − y(tj−1)]
We will set ∆t = tj − tj−1 = (b − a)/N. By the mean value theorem we have;
0 ∗ 0 ∗ x(tj) − x(tj−1) = x (tj )∆t y(tj) − y(tj−1) = y (tj )∆t
∗ for some tj ∈ [tj−1, tj]. Hence, we can rewrite the distance above as;
q 0 ∗ 2 0 ∗ 2 Lj = [x (tj )] + [y (tj )] ∆t
PN q 0 ∗ 2 0 ∗ 2 Thus the total length of the arc is L ≈ j=1 [x (tj )] + [y (tj )] ∆t and so;
N q Z b X 0 ∗ 2 0 ∗ 2 p 0 2 0 2 lim [x (tj )] + [y (tj )] ∆t = [x (t)] + [g (t)] dt N→∞ j=1 a
Theorem (Length of a Path) Assume that r(t) is differentiable and that r0(t) is continuous on [a, b]. Then the length s of the path r(t) for [a, b] is equal to;
Z Z s = kr0(t)k dt = px0(t)2 + y0(t)2 + z0(t)2 dt [a,b] [a,b]
Arc Length Function: Speed by definition, is the rate of change of distance traveled with respect to time t. To calculate speed, we define the arc length function;
Z t s(t) = kr0(u)k du a
ds By the Fundamental Theorem of Calculus; Speed at time t = = kr0(t)k. dt
1 Arc Length Parametrization
Let r(t) be a differentiable vector-valued function defined on [a, b]. If r0(u) 6= 0 for any u ∈ [a, b] then we say r(t) is a regular curve. If kr0(t)k = 1 for all t ∈ [a, b] then we say r(t) is a unit speed parametrization or curve with arc-length parameter. The reason for the latter description is because we find the parameter t using the arc-length function.
R 0 0 Let r(t) be a regular curve and form the arc length integral s(t) = [0,t] kr (u)k du. Since kr (t)k= 6 0 and s(t) is an increasing function then it has an inverse t = g(s). Recall that if g = f −1 then g0 = [f 0 ◦ g]−1. Since we have s0(t) = kr0(t)k then;
1 1 g0(s) = = s0(g(s)) kr0(g(s))k
1 If we let r (s) = r(g(s)) then kr0 (s)k = kr0(g(s))g0(s)k = kr0(g(s))k · = 1 and so r (s) is a 1 1 kr0(g(s))k 1 reparametrization of r and the parameter t = g(s) gives the curve unit speed.
2 Curvature
Imagine traveling along a smooth curve traced out by a vector parametrization r(t). We know that r0(t) points in the direction of the curve at time t. Therefore T (t) = r0(t)/kr0(t)k does so as well. We refer to T as the unit tangent vector at time t. Observe that the change in T corresponds to a bend in the curve i.e kdT/dtk is a perfect measure of how the curve bends. However, the curvature then becomes dependent on how fast you travel along the curve. If we adopt a unit speed parametrization then curvature becomes independent of time and we get the following definition.
Definition: Let r(s) be a arc-length parametrization and T the unit tangent vector. The curvature at r(s) is the quantity;
dT κ(s) = ds
Since computing unit speed parameterizations are nearly impossible in regular application, we would like R 0 to have curvature defined for any time parameter t. Recall that s = s(t) = [a,t] kr (u)k du and when computing the inverse g of s(t) we have g(s) = g(s(t)) = t i.e we can express t as a function of s. Thus, by Chain rule we have the following;
dT dT dt dT ds dT dT T (t(s)) = T (t) ⇒ = · ⇒ · = = ν(t) ds dt ds ds dt dt ds
Since κ = kdT/dsk, itt now follows that kT 0(t)k = ν(t)κ(t)
Formula for Curvature If r(t) is a regular parametrization, then the curvature at r(t) is;
kr0(t) × r00(t)k κ(t) = kr0(t)k3
Curvature of a Graph in the Plane The curvature at the point (x, f(x)) on the graph y = f(x) is equal to;
|f 00(x)| κ(x) = 3 (1 + f 0(x)2) 2
1 Unit Normal Vector
Recall that T 0 and T are orthogonal. The unit vector in the direction of T 0, assuming it is nonzero, is called the unit normal vector and is denoted N;
T 0(t) Unit normal vector = N(t) = kT 0(t)k By convenience, we let N be the unit normal vector which is in the direction for which the curve is bending.
Exercises
(0) Show that T 0 and T are orthogonal (1) Verify the curvature formula for a space curve. (2) Verify the curvature formula for a plane curve. (3) Draw a few space and plane curves then draw the corresponding unit tangent and normal vectors on them for at least three points.
2 Motion in Three-Space
The velocity vector v(t) = r0(t) and points in the direction of motion. The acceleration vector is the second derivative r00(t) which we will denote as a(t).
Notation: v(t) = r0(t), ν(t) = kv(t)k, a(t) = r00(t).
Newton’s Second Law states force = mass × acceleration or F = ma. A more general statement is the vector law F = ma, where F is the net force vector acting on the object and a is the acceleration vector. If we have varying force, we write F(r(t)) for the force acting on a particle with position vector r(t) at time t. We then have a reformulation of NSL;
F(r(t)) = ma(t) ⇒ F(r(t)) = mr00(t)
We will now attempt to understand the acceleration vector a(t). Recall that the unit tangent and normal vectors are given by;
v(t) T 0(t) T (t) = N(t) = kv(t)k kT 0(t)k
From our notation we have T (t)kv(t)k = T (t)ν(t) = v(t) and so by the Product Rule,
v0(t) = r00(t) = a(t) = T (t)ν0(t) + ν(t)T 0(t)
Recall from the definition of curvature in 13.4 we have T 0(t) = ν(t)κ(t)N(t) and so we have;
0 2 a = aT T + aN N, aT = ν (t) aN = κ(t)ν(t)
The coefficient aT (t) is called the tangential component and aN (t) the normal component of acceleration. We will now give some more useful formulas for the tangential and normal components;
a · v a = a · T = a = a · N = pkak2 − |a |2 T kvk N T a · v a · v a T = v a N = a − a T = a − v T v · v N T v · v
1 Function of Two or More Variables
First year calculus was all about functions of the form f : D ⊂ R → R. i.e functions of the form f(x) = x2, x3 + 2x, 3x + 1 and so forth. We will now study functions of two or more variables say, f(x, y) = 2x + y, 2x2 + y, x2 + y2 and etc. In our compact notation above, we can write this as;
n f : D ⊂ R → R i.e f(x) = f(x1, x2, ..., xn)
In single-variable calculus we used graphs to visualize important features of a function. Graphs in the multivariable case will play a similar role. Recall that the graph of a function or graph(f) is defined by;
graph(f) = {(x, f(x): x ∈ D, f(x) ∈ R}
In the above if we assume f : D ⊂ Rn → R then graph(f) ⊂ Rn × R. In this section we will explore various techniques which will help us understand the graphs of functions of two or more variables.
Definition: A vertical trace of a graph is the intersection of a vertical plane with the graph.
Definition: A horizontal trace of a graph is the intersection of a horizontal plane with the graph.
1 Definition: Give a function f : D ⊂ Rn → R, a level curve is the set of all x ∈ D with f(x) = c. Definition: A contour map is a plot in the plane for the level curves f(x, y) = c.
For functions of more the two variables, say f(x, y, z) then graph(f) ⊂ R4. It is not possible to visualize its image, but it is possible to draw the level surfaces i.e the set of (x, y, z) s.t f(x, y, z) = c.
Exercises
(1) Draw at least 3 horizontal and vertical traces for the quadratic surfaces mentioned in section 12.6. (2) What are all the possible horizontal traces you can get for a donut? (3) Using level curves, build the equations for at least 4 quadratic surfaces. (4) Draw contour plots for 5 quadratic surfaces.
2 CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES
1. Definitions n m n 1.1. Limit. Let f : R → R some function, x0 = (x1, . . . , xn) ∈ R and y0 = m (y1, . . . , ym) ∈ R . Then
lim f(x) = y0 x→x0 if and only for x ”close to” x0, f(x) is ”close to” y0. In other words, given > 0, there exists δ depending on and x0 (so we write δ = δ(, x0)) such that
n m dR (x, x0) < δ ⇒ dR (f(x), y0) < where the distance on the left is taken in Rn, whereas the one on the right-hand side is taken in the target space Rm. We will drop the subscript Rn from d, and we will often write kx − x0k instead of d(x, x0). We will also drop the bold-face notation. n m Note: a function f : R → R is clearly given by a row vector f = (f1, . . . , fm) where fi’s are the components of f. In fact, fi = pi ◦ f (see below). Then limx→x0 f(x) = y0 if and only if limx→x0 fi(x) = yi, i = 1, . . . , m.
n m n 1.2. Continuity. A function f : R → R is continuous at x0 ∈ R iff
lim f(x) = f(x0) x→x0 We say that f is continuous (everywhere) if it is continuous at every point of the domain Rn.
Note: if f = (f1, . . . , fm), where f1, . . . , fm are the components of f, then f is n continuous iff f1, . . . , fm are continuous, as functions : R → R.
2. Tools Operations with limits: addition, subtraction, multiplication, division (when pos- sible), etc. n m Components. If f : R → R , f = (f1, . . . , fm), then limx→x0 f(x) = y0 =
(y1, . . . , ym) iff limx→x0 fi(x) = yi, for every i = 1, . . . , m. n Projections. The functions pi : R → R, pi(x1, . . . , xn) = xi are continuous, and this can be easily checked with the (, δ)- definition of continuity. Composition of functions. If f : Rn → Rm is continuous and g : Rm → RN is continuous, then g ◦ f : Rn → RN is continuous, where g ◦ f(x) := g(f(x)).
f g Rn −−−−→ Rm −−−−→ RN 1 2 CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES
3. Examples 1. f : R3 → R, f(x, y, z) = x sin(z). Then f = p1 · (sin ◦p3) = continuous. 2. f : R3 → R2, f(x, y, z) = (xy, x + y + z2). • f1(x, y) = xy = p1 · p2 = cont. 2 2 • f2(x, y) = x + y + z = p1 + p2 + (p3) = cont.
Therefore f = (f1, f2) is continuous. 2 (√ x , (x, y) 6= (0, 0) 2 2 3. f : R2 → R, f(x, y) = x +y Proof that f is continuous. 0, (x, y) = (0, 0) On R2 − (0, 0), f is continuous since it is a ratio of continuous functions (with 2 p1 non-vanishing denominator), namely f = √ 2 2 = continuous. ·◦(p1+p2) NB: To make sure we understand why is the denominator continuous, we can write the denominator as the composition of functions
2 2 √ p1+p2 · R2 −−−−→ [0, ∞) −−−−→ R We are thus left to show that f is continuous at (0, 0), in other words to prove that x2 lim p = 0 (x,y)→(0,0) x2 + y2 Squeeze: x2 x2 0 ≤ ≤ √ = |x| px2 + y2 x2 However, as (x, y) → (0, 0) necessarily x → 0 and hence |x| → 0, therefore by the 2 Pinching Lemma √ x → 0 as (x, y) → 0. x2+y2 Different way of saying it ((, δ)-proof). Note that p x2 d((x, y), (0, 0)) ≤ ⇔ x2 + y2 ≤ ⇒ |x| ≤ ⇒ ≤ |x| ≤ px2 + y2 ( xy 2 2 , (x, y) 6= (0, 0) 4. f : R2 → R, f(x, y) = x +y . 0, (x, y) = (0, 0) Proof that f is not continuous at (0, 0). Consider the followin sequences of points in R2: 1 x = ( , 0), k ≥ 1 k k 1 1 y = ( , ), k ≥ 1 k k k Then xk → (0, 0) and yk → (0, 0) as k → ∞, and yet
f(xk) = 0 → 0, k → ∞ 1 1 f(y ) = → , k → ∞ k 2 2 which shows that the limit lim f(x) x→(0,0) does not exist. In particular, the function f is not continuous at (0, 0). CONTINUITY OF MULTIVARIABLE FUNCTIONS. EXAMPLES 3
5. Let ( x sin(1/y), y 6= 0 g(x, y) = 0, y = 0 Determine the points (x, y) ∈ R2 where g is continuous. 6. Tricky example. Let f : R2 → R defined by ( 2 xy , (x, y) 6= (0, 0) f(x, y) = x2+y4 0, (x, y) = (0, 0)
Show that limk→∞ f(xk) = 0, for any sequence xk → (0, 0) such that xk, k ≥ 1 is on a fixed line passing through (0, 0). Is f continuous at (0, 0)? Partial Derivatives
The partial derivatives are the rates of change with respect to each variable separately. Before generalizing, let’s first consider a function f(x, y) of two variables. Now let us consider graph(f);
Consider now the intersection of the graph with a plane y = b i.e a vertical trace;
If you were to look at the projection of this curve in the x, z plane, you can now as yourself, what is the rate of change of f at (a, b). Here b is fixed and so this rate of change, becomes the standard derivative for a function of one variable;
d f(a + t, b) − f(a, b) f(a, b) = lim dt t→0 t In view of the image above, this is precisely the slope of the tangent line in the direction of the x-coordinate. We will choose to denote this as fx(a, b). Similarly we have;
d f(a, b + t) − f(a, b) fy(a, b) = f(a, b) = lim dt t→0 t
1 We will do more applications of the following fact later, but observe that if ∆x and ∆y are extremely small, then we can estimate fx(a, b) and fy(a, b) as follows;
f(a + ∆x, b) − f(a, b) ≈ fx(a, b)∆x
f(a, b + ∆y) − f(a, b) ≈ fy(a, b)∆y
Just as with functions of one-variable, we have higher-order partial derivatives. For notational purposes we will also introduce notation which is used by Leibniz; ∂f ∂f f (a, b) = (a, b) f (a, b) = (a, b) x ∂x y ∂y
Now, higher order partials are also defined with difference quotients, but assuming that everything is nice and limits exist we have;
∂ ∂f ∂ ∂f ∂ ∂f ∂ ∂f f = f = f = f = xx ∂x ∂x yy ∂y ∂y xy ∂y ∂x yx ∂x ∂y
The latter two are called mixed partials. The immediate questions to as with mixed partials is, does the order matter? Luckily we have a very powerful result to answer that question.
Clairaut’s Theorem: Equality of Mixed Partials If fxy and fyx are both continuous functions on a disk D, then fxy(a, b) = fyx(a, b) for all (a, b) ∈ D.
Exercises
(1) Compute the equation of the tangent lines whose slopes are fx(a, b) and fy(a, b).
(2) Using the estimation for fx and fy, calculation (.98)(.42) to the nearest 3 decimal points.
2 Differentiability and Tangent Planes
In single variable calculus, we considered the linear approximation of a graph for a function y = f(x). Similarly, we would like to get a linearization for a function f(x, y). This linearization will be called the tangent plane. Let (a, b) ∈ dom(f) and consider its representation (a, b, 0) in three-dimensions. From the image above the tangent plane must contain the lines;
L1(x) = f(a, b) + fx(a, b)(x − a) and L2(y) = f(a, b) + fy(a, b)(y − a)
which are the lines in the direction of the coordinate axis. The equation of a plane which passes through (a, b, f(a, b)) is given by;
a0 b0 a0(x − a) + b0(y − b) + c0(z − f(a, b)) = 0 ⇐⇒ z − f(a, b) = − (x − a) − (y − b) c0 c0
Let A = −a0/c0 and B = −b0/c0 then we have z − f(a, b) = A(x − a) + B(y − b). Now if we fix y = b then we get the line z − f(a, b) = A(x − a) which implies A = fx(a, b) and similarly B = fy(a, b) is we let x = a. Therefore, we recover our plane equation;
z = L(x, y) = fx(a, b)(x − a) + fy(a, b)(y − b) + f(a, b)
1 Linear Approximation and Differentials
If (x, y) is near (a, b) f(x, y) ≈ L(x, y) where L(x, y) = fx(a, b)(x − a) + fy(a, b)(y − b) + f(a, b). To recover something new from this, set x = a + h and y = b + k then;
f(a + h, b + k) ≈ f(a, b) + fx(a, b)h + fy(a, b)k
We can also write the linear approximation in terms of the change in f;
∆f = f(x, y) − f(a, b), ∆x = x − a, ∆y = y − b ⇒ ∆f ≈ fx(a, b)∆x + fy(a, b)∆y
Finally, the linear approximation is often expressed in terms of differentials;
∂f ∂f df = f (x, y)dx + f (x, y)dy = dx + dy x y ∂x ∂y
Here df represents the change in height of the tangent plane for given changes dx and dy in x, y, whereas ∆f is the change of the function itself. The linear approximation tellls us that ∆f ≈ df.
2 Exericses
(1) Use linear approximation to estimate (3.99)3(1.01)4(1.98)−1 (2) Attempt to reconstruct the tangent plane to make sure the concept is clear. (3) If you rewrite the equation of the tangent plane by moving the z to the right hand side, can you give the normal vector? This will be useful for the next section. (4) Not all functions have tangent planes for every point in their domain. Read the assumptions matter section to get a brief understanding why.
3 The Gradient and Directional Derivatives
The gradient of a function f(x, y) at a point P = (a, b) is the vector;
∇f(p) = hfx(a, b), fy(a, b)i
In three variables, if P = (a, b, c) then we have;
∇f(p) = hfx(a, b, c), fy(a, b, c), fz(a, b, c)i
Also in times we will just write ∇f if the point of evaluation is understood i.e ∇f = hfx, fy, fzi.
d Theorem (Chain Rule for Paths) If f and c(t) are differentiable, then f(c(t)) = ∇f(c(t)) · c0(t). dt
Proof: By definition
d f(x(t + h), y(t + h)) − f(x(t), y(t)) f(c(t)) = lim dt h→0 h
Set ∆f = f(x(t + h), y(t + h)) − f(x(t), y(t)), ∆x = x(t + h) − x(t), ∆y = y(t + h) − y(t). The using linearity we have;
∆f = fx(x(t), y(t))∆x + fy(x(t), y(t))∆y + e(x(t + h), y(t + h))
Now set h = ∆t and divide by ∆t;
∆f f (x(t), y(t))∆x + f (x(t), y(t))∆y e(x(t + h), y(t + h)) = x y + ∆t ∆t ∆t
The last term tends to zero (shown in book) and we have;
d ∆f ∆x ∆y 0 f(c(t)) = lim = fx(x(t), y(t)) lim + fy(x(t), y(t)) lim = ∇f(c(t)) · c (t) dt ∆t→0 ∆t ∆→0 ∆t ∆t→0 ∆t
1 Consider a line through the point P = (a, b) in the direction of a unit vector ~u = hh, ki then c(t) = (a + th, b + tk). The derivative of f(c(t)) at t = 0 is called the direction derivative of f with respect to ~u at P , and is denoted D~uf(P ); d f(a + th, b + tk) − f(a, b) D~uf(a, b) = f(c(t))|t=0 = lim dt t→0 t By the Chain Rule for Paths, if we let c(t) = (a + th, b + tk) and ~v = hh, ki then;
0 D~vf(a, b) = ∇f(a, b) · c (0) = ∇f(a, b) · ~v
It also follows that if λ ∈ R then we have;
Dλ~vf(P ) = λD~vf(P )
Theorem (Computing Directional Derivatives) If ~v 6= ~0, then ~u = ~v/k~vk is the unit vector in the direction of ~v, and the directional derivative is given by;
1 D f(P ) = ∇f(P ) · ~v ~u k~vk
We will now get some important properties of the gradient. Let ∇f(P ) 6= ~0 and ~u be a unit vector. Then recall that by the dot product;
D~uf(P ) = ∇f(P ) · ~u = k∇f(P )k cos θ
In the above θ is the angle between ∇f(P ) and ~u. Therefore, the gradient vector points in the direction of the maximum rate of increase, and this maximum rate is k∇f(P )k.
Another key property is that gradient vectors are normal to level curves. Suppose P lies on the level curve f(x, y) = k. We parametrize this level curve by a path c(t) such that c(0) = P and c0(0) 6= ~0. Then f(c(t)) = k for all t and so by Chain Rule;
d d ∇f(P ) · c0(0) = f(c(t))| = k = 0 dt t=0 dt
This shows that the gradient vector is orthogonal to the tangent line at P and so we have our claim. The equation of the tangent plane to the level surface F (x, y, z) = k at P = (a, b, c) is given by;
∇F (P ) · hx − a, y − b, z − ci = 0
2 The Chain Rule
We will now extend the Chain Rule for Path that we derived in the previous section to more general functions. Let f(x, y, z) be a function and x = x(s, t), y = y(s, t), z = z(s, t). Then we have;
∂f d = f(c(s)) = ∇f(c(s)) · hx , y , z i ∂s ds s s s ∂f d = f(c(t)) = ∇f(c(t)) · hx , y , z i ∂t dt t t t
In single-variable calculus, we used implicit differentiation to compute dy/dx when y is defined implicitly as a function of x through an equation f(x, y) = 0. We will now apply to functions of several variables. Suppose that x is defined implicitly by an equation;
F (x, y, z) = 0
Then z = z(x, y) is a function of x, y. Treating F (x, y, z) as a composite function with x, y as independent variables then by Chain Rule;
d F (c(x)) = ∇F (c(x)) · c0(x) ⇒ F + F z = 0 dx x z x
If Fz 6= 0, we solve for zx and get;
Fx Fy zx = − zy = − Fz Fz
1 Optimization in Several Variables: Part I
Introduction: For a function of one variable f : R → R we have that if a ∈ R has the property that f(x) ≤ f(a) or f(a) ≤ f(x) for all x with |x − a| < then f 0(a) = 0. We aim to come up with criteria for functions of two (or more variables) in which given some B = B(P, r) in the domain of our function, we have f(P ) ≥ f(x, y) of f(P ) ≤ f(x, y) for all (x, y) ∈ B and P = (a, b). We know make this more precise with the following definition.
Definition (Local Extreme Values): A function f(x, y) has a local extremum at P = (a, b) if there exists an open ball B(P, r) such that;
(1) Local maximum: f(x, y) ≤ f(a, b) for all (x, y) ∈ B(P, r) (2) Local minimum: f(x, y) ≥ f(a, b) for all (x, y) ∈ B(P, r)
In the case of a function of one variable, if a ∈ R is a local extremum then the tangent line is horizontal i.e the linear approximation for the function has derivative zero. We can expect the same for the tangent plane which is the linear approximation for a function f(x, y) at a point (a, b). Since our approximation is given by z = f(a, b) + fx(a, b)(x − a) + fy(a, b)(y − b) then we get a horizontal tangent plane when fx(a, b) = fy(a, b) = 0. However, intuition isn’t enough so we have to justify this thought.
1 The image above illustrates the phenomenon that we expect to happen. We will now associate common terminology to such points.
Definition (Critical Point) : A point P = (a, b) in the domain of f(x, y) is called a critical point if;
(1) fx(a, b) = 0 or fx(a, b) does not exist, and
(2) fy(a, b) = 0 or fy(a, b) does not exist.
Remark: We’ve made sense of fx(a, b) = fy(a, b) = 0 but why the sudden, ”or does not exists”? Think of functions of one variable like f(x) = |x| or g(x) = x(2/3). Then have minimum values at x = 0 but there derivatives aren’t defined there. Therefore, to allow instances of such points, we amend this extra condition in the definition.
2 Theorem: If f(x, y) has a local minimum or maximum at a point P = (a, b) then (a, b) is a critical point of f(x, y).
Proof: Without loss of generality, lets just assume that f(x, y) has a local minimum at P = (a, b). Define g : R → R by g(x) = f(x, b). Then it follows that g attains its minimum in some neighborhood of a ∈ R 0 since at g(a) we know f(a, b) is a local minimum. Therefore, we have g (a) = 0 = fx(a, b) or does not exist. Just to make this more clear observe that,
0 g(a + h) − g(a) f(a + h, b) − f(a, b) g (a) = lim = lim = fx(a, b) h→0 h h→0 h
0 Similarly if we define h(x) = f(a, x) we argue by symmetry and obtain that h (b) = 0 = fy(a, b) for does not exist.
Just as in the one variable case (image on left), we can encounter functions of several variable (image on right) which have points in their domain which are a neither a local min or local max!
However, just as in the one variable case, there is a Second Derivative Test for determining the type of a critical point (a, b) of a function f(x, y) in two variables. The result we use relies on the sign of the 2 discriminant D = D(a, b) which is defined as D = D(a, b) = fxx(a, b)fyy(a, b) − fxy(a, b). We will not give a proof of this facts, but rather enjoy the consequences.
Theorem: Let P = (a, b) be a critical point of f(x, y). Assume fxx, fyy, fxy are continuous near P then;
(1) If D > 0 and fxx(a, b) > 0, then f(a, b) is a local minimum.
(2) If D > 0 and fxx(a, b) < 0, then f(a, b) is a local maximum. (3) If D < 0, then f has a saddle point at (a, b). (4) If D = 0 then the test in inconclusive.
3 Optimization in Several Variable: Part II
We now switch gears and ask when can be certain a function f(x, y) obtains global extrema i.e values in which f(a, b) ≤ f(x, y) or f(a, b) ≥ f(x, y) for all (x, y) in the domain of our given function. First we need to develop some terminology and then we will arrive at a sufficient condition.
Definition (Bounded Domain): We sat that a domain D is bounded if there is a number M > 0 such that D is contained in a ball of radius M centered at the origin.
Definition (Interior point): An interior point of D is a point such that it is contained in some open disk D(P, r) which is also contained in D.
Definiton (Boundary point) : A boundary point of D is a point such that every disk centered at this point contains points in D and points outside of D.
Definition (Closed and Open): A domain D is open if every point of D is an interior point. Conversely D is said to be closed if it contains all of its boundary points.
4 We now have the language to introduce the following theorem and as one could probably guess, the sufficient condition for which a function attains its min and max is on closed and bounded regions.
Theorem: Let f(x, y) be a continuous function on a closed, bounded domain D in R2 then,
(1) f(x, y) takes on both a minimum and a maximum value on D. (2) The extreme values occur either at the critical points in the interior of D or at points on the boundary of D.
Remark: Due to the lengthiness of the result, we will also just enjoy the consequences of the result. Observe that this is just a higher dimensional result of the Extreme Value Theorem encountered in first-year calculus. The result hinges on the fact the continuous image of a closed and bounded region, is closed an bounded. Therefore if f : R2 → R and D ⊂ R2 is closed and bounded then so is f(D) ⊂ R. Since ever closed and bounded set in R contains its supremum and infimum, we have the result.
5 Lagrange Multipliers: Optimizing with a Constraint
Theorem (Lagrange Multipliers) Assume that f(x, y) and g(x, y) are differentiable. If f(x, y) has a local minimum or a local maximum on the constraint curve g(x, y) = 0 at P = (a, b), and if ∇g(P ) 6= ~0, then there exists a scalar λ such that;
∇f(P ) = λ g(P )
Proof: Let c(t) be a parametrization of the constraint curve g(x, y) = 0 near P , chosen so that c(0) = P and c0(0) 6= ~0. Then f(c(0)) = f(P ), and by assumption, f(c(t)) has local min or max at t = 0. Thus, t = 0 is a critical point of f(c(t)) and;
d f(c(t))| = ∇f(P ) · c0(0) = 0 dt t=0
This shows that ∇f(P ) is orthogonal to the tangent vector c0(0) to the curve g(x, y) = 0.
The gradient ∇g(P ) is also orthogonal to c0(0), therefore ∇f(P ) and ∇g(P ) are parallel.
1 Integration in Two Variables
Consider a function f : D ⊂ R2 → R. Suppose first that D is a rectangular region R i.e D = [a, b] × [c, d]. Just as in the case for functions of one variable, we wish to calculate the volume of the solid region between graph(f) and z = 0, over the domain D.
We will now run the program, which consists of partitioning the region D, choosing sample points in each sub-square to get height of sub-rectangular box, and then take the limit as the partition get extremely small.
(1) Subdivide [a, b] and [c, d] by choosing partitions; a = x0 < x1 < ··· < XN , c = y0 < y1 < ··· < yM = d.
(2) Create an N × M grid of sub-rectangles Rij
(3) Choose a sample point Pij in each Rij
Since each Rij = [xi−1, xi] × [yj−1, yj] then its area is given by;
∆Aij = ∆xi ∆yj where ∆xi = xi − xi−1 and ∆yj = yj − yj−1.
1 Next we form the Riemann sum with the function values f(Pij) which you can think of as heights;
N M N M X X X X SN,M = f(Pij) ∆Aij = SN,M = f(Pij) ∆xi∆yj i=1 j=1 i=1 j=1
From the image above, we see that each summand f(Pij) ∆xi∆yj gives a volume for each sub-rectangular box and so the sum is an approximation via these boxes. In the final step we take P = {{xi}, {yj}} to be notation for the partition and kPk = max{∆xi, ∆yj}. Therefore, as kPk → 0, the boxes approximate the solid region under the graph more closely i.e;
N M ZZ X X Volume = f(x, y) dA = lim f(Pij) ∆Aij kPk→0 D i=1 j=1
Theorem (Linearity of the Double Integral) Assume that f(x, y) and g(x, y) are integrable over a rectangle R then;
ZZ ZZ ZZ C(f(x, y) + g(x, y)) dA = C f(x, y) dA + C g(x, y) dA
R R R
If f(x, y) = C is a constant function then we have;
N M ZZ ZZ X X C dA = C 1 dA = C · lim ∆Aij = C · Area(R) kPk→0 R R i=1 j=1
2 Our main tool for evaluating double integrals is the Fundamental Theorem of Calculus (FTC). To use FTC, we can express the double integral as an interated integral, which is an expression of the form;
Z b Z d f(x, y) dy dx a c
Theorem (Fubini’s Theorem) The double integral of a continuous function f(x, y) over a rectangle R = [a, b] × [c, d] is equal to the interated integral;
ZZ Z b Z d Z d Z b f(x, y) dA = f(x, y) dydx = f(x, y) dxdy x=a y=c y=c x=a R
3 Double Integrals over More General Regions
Suppose we wish now to integrate over regions D whose boundaries are simple closed curves which are also smooth. To do this, imagine a rectangular box R = [a, b] × [c, d] which bounds D.
˜ ˜ ˜ Given a function f(x, y) on D, we define a new function f(x, y) on R s.t f = f|D and f(R\ D) = 0. Thus, the double integral of f over D is defined as the integral of f˜ over R;
ZZ ZZ f(x, y) dA = f˜(x, y) dA
D R
The existence of this integral is in question since the jump from f(x, y) to zero may allow discontinuities but the following theorem gives the existence.
1 Theorem If f(x, y) is continuous on a closed domain D whose boundary is a closed, simple, piecewise RR smooth curve, then D f(x, y) dA exists.
Given this new function f˜ which extends f to a rectangular region, we have;
ZZ N m X X ˜ X f(x, y) dA ≈ f(Pij) ∆xi∆yj = f(Pij) ∆xi∆yj D i=1 j=1 Pij ∈D
Moreover, if we let f(x, y) = C be a constant function then we have;
ZZ ZZ C · Area(D) = C dA = C 1 dA
D D
We now direct our attention to integration over regions D in the plane which are between two graphs. We say D is vertically simple if it is of the form below;
We say D is horizontally simple if it is of the form;
2 Theorem If D is vertically simple with description a ≤ x ≤ b and g1(x) ≤ y ≤ g2(x) then;
ZZ Z b Z g2(x) f(x, y) dA = f(x, y) dydx a g1(x) D
If D is a horizontally simple region with description c ≤ y ≤ d and g1(y) ≤ x ≤ g2(y) then;
ZZ Z d Z g2(y) f(x, y) dA = f(x, y) dxdy c g1(y) D
Proof: Suppose without loss that D is vertically simple. Choose R = [a, b] × [c, d] containing D. Then we have;
ZZ Z b Z d (∗) f(x, y) dA = f˜(x, y) dydx a c D
˜ ˜ Since f(x, y) is zero outside of D, then for fixed x ∈ [a, b], f(x, y) = 0 unless g1(x) ≤ y ≤ g2(x). Therefore,
Z d Z g2(x) f˜(x, y) dy = f(x, y) dy c g1(x)
Substituting this in for (*) then we have;
ZZ Z b Z g2(x) f(x, y) dA = f(x, y) dydx a g1(x) D
3 Triple Integrals
Let f : B ⊂ R3 → R where B is a cube in three-space. Our goal is to generalize the double integral, i.e compute the volume under a four-dimensional object. The program is the same, (1) partition B into smaller cubes, (2) express the volume as a triple sum, and (3) take the limit as the partition goes to zero i.e increasing the number of cubes to get a better approximation.
First write B = [a, b] × [c, d] × [p, q]. Next we will subdivide the cube into sub-cubes;
Bijk = [xi−1, xi] × [yj−1, yj] × [zk−1, zk]
by choosing the following partitions of the three intervals;
a = x0 < x1 < ··· < xN = b
c = y0 < y1 < ··· < yM = d
p = z0 < z1 < ··· < zL = q
The volume of Bijk is ∆Vijk = ∆xi∆yj∆zk where ∆xi = xi − xi−1, ∆yj = yj = yj−1 and ∆zk = zk − zk−1. As before, we now pick a sample point Pijk lying in each Bijk and form the Riemann sum:
N M L X X X SN,M,L = f(Pijk) ∆Vijk i=1 j=1 k=1
1 If we choose to denote the partition above as P = {{xi}, {yj}, {zk}} and let kPk = max{∆xi, ∆yj, ∆zk} and if SN,M,L → L as kPk → 0 then we say f in integrable over B. The limit value is denoted;
ZZZ f(x, y, z) dV = lim SN,M,L B kPk→0
Theorem (Fubini’s Theorem for Triple Integrals): The triple integral of a continuous function f(x, y, z) over a cube B = [a, b] × [c, d] × [p, q] is equal to the iterated integral;
ZZZ Z b Z d Z q f(x, y, z) dV = f(x, y, z) dz dy dx B x=a y=c z=p
Furthermore, the iterated integral may be evaluated in any order.
2 Now instead of a cube, we wish to integrate over a solid region W that is simple (image given below).
In the image we have W = {(x, y, z):(x, y) ∈ D} and z1(x, y) ≤ z ≤ z2(x, y). The domain D is the projection of W onto the xy-plane. As in the case of the double integral, we define the triple integral of f(x, y, z) over W by;
ZZZ ZZZ f(x, y, z) dV = f˜(x, y, z) dV W B
where B is a cube containing W and;
f(x, y, z) , if (x, y, z) ∈ W f˜(x, y, z) = 0 , if (x, y, z) 6∈ W
Theorem (Triple integrals over simple regions) The triple integral of a continuous function f over the region W = (x, y) ∈ D, z1(x, y) ≤ z ≤ z2(x, y) is equal to the interated integral;
! ZZZ ZZ Z z2(x,y) f(x, y, z) dV = f(x, y, z) dz dA W D z=z1(x,y)
Furthermore, the volume V of a region W is defined as the triple integral of the constant function f(x, y, z) = 1 i.e;
ZZZ V = 1 dV W
3 RRR Example: (Solid Region with a Rectangular Box) Evaluate W z dV , where W is the region between the planes z = x + y and z = 3x + 5y lying over the rectangle D = [0, 3] × [0, 2].
4 Integration in Polar, Cylindrical, and Spherical Coordinates
Polar coordinates are convenient when the domain of integration is an angular sector or a polar rectangle; R : θ1 ≤ θ ≤ θ2, r1 ≤ r ≤ r2.
Here we suppose that r1 ≥ 1 and that all radial coordinates are nonnegative. Recall that rectangular and polar coordinates are related by x = r cos θ and y = r sin θ. Thus, we can write a function f(x, y) in polar coordinates as f(r cos θ, r sin θ). The Change of Variables Formula for a polar rectangle R is;
ZZ Z θ2 Z r2 f(x, y) dA = f(r cos θ, r sin θ) · r drdθ θ1 r1 R
To derive the equation above, we estimate the area ∆A of the small polar rectangle shown in the above image. If ∆r, ∆θ are very small, then this polar rectangle is very nearly an ordinary rectangle of sides ∆r and r∆θ, and therefore ∆A ≈ r∆r∆θ.
1 Now, decompose R into an N × M grid of small polar rectangles Rij as above, and choose a sample point Pij in Rij. If Rij is small and f(x, y) is continuous then;
ZZ f(x, y) dxdy ≈ f(Pij) · Area(Rij) ≈ f(Pij) rij ∆r∆θ
Rij
In practice we will choose the partition to be regular i.e ∆θ = (θ2 − θ1)/M and ∆r = (r2 − r1)/M. The integral over R is the sum;
ZZ X ZZ X X f(x, y) dxdy = f(x, y) dxdy ≈ f(Pij) · Area(Rij) ≈ f(rij cos θij, rij sin θij) rij ∆r∆θ i,j i,j i,j R Rij
The above gives a Riemann sum for the double integral of r · f(r cos θ, r sin θ) over the region r1 ≤ r ≤ r2 and θ1 ≤ θ ≤ θ2.
Theorem (Double Integral in Polar Coordinates) For a continuous function f on the domain D : θ1 ≤ θ ≤ θ2, r1(θ) ≤ r ≤ r2(θ);
ZZ Z θ2 Z r2(θ) f(x, y) dA = f(r cos θ, r sin θ) r drdθ θ1 r1(θ) D
2 Cylindrical coordinates are useful when the domain has axial symmetry-that is, symmetry with respect to an axis. Recall that we have the relations;
x = r cos θ y = r sin θ z = z
Theorem (Triple Integrals in Cylindrical Coordinates) Suppose W can be described as the region between two surfaces; z1(r, θ) ≤ z ≤ z2(r, θ), lying over a domain D in the xy-plane with polar description; D : θ1 ≤ θ ≤ θ2, r1(θ) ≤ r ≤ r2(θ). A triple integral over W can be written as an iterated integral;
! ! ZZZ ZZ Z z2(r,θ) Z θ2 Z r2(θ) Z z2(r,θ) f(x, y, z) dV = f(x, y, z) dz dA = f(r cos θ, r sin θ, z) r dz drdθ z1(r,θ) θ1 r1(θ) z1(r,θ) W D
3 We will now come up with a Change of Variable Formula in spherical coordinates. Recall that the relation between rectangular and spherical coordinates is given by;
x = ρ cos θ sin φ y = ρ sin θ sin φ z = ρ cos φ We claim that dV = ρ1 sin φ dρdφdθ where this represents the volume of a small spherical wedge.
Define a small spherical wedge W by the inequalities;
W : θ1 ≤ θ ≤ θ2, φ1 ≤ φ ≤ φ2, ρ1 ≤ ρ ≤ ρ2
In the above image; when the increments ∆θ = θ2 − θ1, ∆φ = φ2 − φ1 and ∆ρ = ρ2 − ρ1 are small, the spherical wedge is nearly a box with sides ∆ρ, ρ1∆φ and ρ1 sin φ1∆θ and volume;
2 Volume(W) ≈ ρ1 sin φ1 ∆ρ∆φ∆θ
4 3 Following the usual steps, we decompose W into N spherical sub-wedges Wi with increments;
θ − θ φ − φ ρ − ρ ∆θ = 2 1 ∆φ = 2 1 ∆ρ = 2 1 N N N
and choose a sample point P1 = (ρi, θi, φi) in each Wi. Assuming f is continuous, the following approximations holds for large N;
ZZ 2 f(x, y, z) dV ≈ f(Pi) · Volume(Wi) ≈ f(Pi) ρi sin φi ∆ρ∆θ∆φ
Wi
Taking the sum over i, we obtain;
ZZ X 2 f(x, y, z) dV ≈ f(Pi)ρi sin φi ∆φi∆θ∆φ W i
The sum on the right is a Riemann sum for the function f(ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ)ρ2 sin φ and so we have the following theorem.
Theorem (Triple Integrals in Spherical Coordinates) For a given region W defined by;
θ1 ≤ θ ≤ θ2, φ1 ≤ φ ≤ φ2, ρ1(θ, φ) ≤ ρ ≤ ρ2(θ, φ) the triple integral RRR f(x, y, z) dV is equal to; W
Z θ2 Z φ2 Z ρ2(θ,φ) f(ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ)ρ2 sin φ dρdφdθ θ1 φ1 ρ1(θ,φ)
5 Change of Variables
A function G : X → Y from a set X to another set Y is often called a map or mapping. In this section, we consider maps G : D → R2 defined on a domain D ⊂ R2. We will write G(u, v) = (x(u, v), y(u, v)) and let A = (u1, u2),B = (v1, v2), c ∈ R. We say G is a linear map if;
G(A + cB) = G(A) + cG(B) ⇒ G(u1 + cv1, u2 + cv2) = G(u1, u2) + cG(v1, v2)
We will now study how area changes under a mapping. The Jacobian determinant or simply Jacobian of a map;
G(u, v) = (x(u, v), y(u, v)) is the determinant given by;
xu xv Jac(G) = = xuyv − xvyu yu yv
Theorem (Jacobian of a Linear Map) The Jacobian of a linear map G(u, v) = (Au + Cv, Bu + Dv) is constant with value;
AC Jac(G) = = AD − BC BD
Under G, the area of a region D is multiplied by |Jac(G)|, that is; Area(G(D)) = |Jac(G)| · Area(D).
One other useful approximation is the fact that if given a domain D which is small and P is a sample point in D then;
Area(G(D)) ≈ |Jac(G)(P )| · Area(D)
1 We now arrive at the more general Change of Variables Formula where G is a C1 mapping i.e continuous.
1 Theorem (Change of Variables Formula) Let G : D0 → D be a C mapping that is one-to-one on the interior of D0. If f(x, y) is continuous then;
ZZ ZZ f(x, y) dxdy = f(x(u, v), y(u, v))|JacG| dudv
D D0
The Change of Variables Formula has the same form in three (or more) variables as in two variables. Let G : W0 → W be a mapping from a 3-dimensional region W0 in (u, v, w)-space to a region W in (x, y, z)-space say,
x = x(u, v, w) y = y(u, v, w) z = z(u, v, w)
The Jacobian Jac(G) is the 3 × 3 determinant;
xu xv xw
Jac(G) = yu yv yw
zu zv zw
The Change of Variables Formula states;
dx dy dz = |Jac| du dv dw
1 More precisely, if G is C and injective on the interior of W0, and if f is continuous then;
ZZ ZZ f(x, y, z) dx dy dz = f(x(u, v, w), y(u, v, w), z(u, v, w)) · |Jac(G)| du dv dw
W W0
2 Vector Fields
A vector field F assigns to each point P a vector F(P ) that represents the velocity of some physical event at that point. A three-dimensional vector field can be represented by it’s component functions;
F(x, y, z) = hF1(x, y, z),F2(x, y, z),F3(x, y, z)i
A vector field F which is the gradient of some differentiable function V ie. F = ∇V is called a conservative vector field and V is said to be a potential function for F. We now give a necessary condition for a vector field F to be conservative. This result is referred to as the cross-partial property of a conservative vector field.
Theorem (Cross-Partial Property of a Conservative Vector Field) If the vector field F = hF 1,F 2,F 3i is conservative then;
1 2 2 3 3 1 Fy = Fx Fz = Fy Fx = Fz
Theorem (Uniqueness of Potential Functions) If F is conservative on an open connected domain, then any two potential functions of F differ by a constant.
1 Line Integrals
The line integral just like all integrals, defined through the process of sub-division, summation and passage to the limit. The interpretation of the line integral is not the usual area between the graph of a function and the x-axis. It can be viewed geometrically as the area of a fence with changing height and boundary curve C.
(1) Divide the curve C into N consecutive arcs C1, ..., CN .
(2) Choose a sample point Pi ∈ Ci Pn Pn (3) Form the Riemann sum: i=1 f(Pi) · l(Ci) = i=1 f(Pi) · ∆si R PN (4) If it exists, then the integral of f over C is C f(x, y, z) ds = lim i=1 f(Pi) · ∆si. ∆si→0
In practice, line integrals are computed using parameterizations. Suppose C has a parametrization c :[a, b] → Rn which we will denote as c(t). We will suppose also that c has a continuous derivative.
(0) Let {a = t0 < t1 < ··· < tN−1 < tN = b} be a partition of [a, b]. n (1) Divide C into N consecutive arcs C1, ..., CN where Ci :[ti−1, ti] → R ∗ ∗ (2) Choose sample points Pi = c(ti ) where ti ∈ [ti−1, ti] R 0 (3) Using the arc-length formula we have: l(Ci) = ∆si = kc (t)k dt. [ti−1,ti] 0 0 ∗ R (4) Since c is continuous, we can choose the sample points above so that kc (t )k∆ti = by i [ti−1,ti] Mean-value theorem for integrals. PN PN ∗ 0 ∗ (5) Therefore we have; i=1 f(Pi)∆si = i=1 f(c(ti ))kc (ti )k∆ti R 0 PN ∗ 0 ∗ (6) Now by definition of the integral: [a,b] f(c(t))kc (t)k dt = lim∆t→0 i=1 f(c(ti ))kc (ti )k∆ti.
1 Theorem (Computing a Scalar Line Integral) Let c(t) be a parametrization of a curve C with t ∈ [a, b]. If f(x, y, z) and c0(t) are continuous then;
Z Z b f(x, y, z) ds = f(c(t))kc0(t)k dt C a where ds = kc0(t)k dt.
Consider a walk along a mountain range. Then the total work exerted or energy expended is one example of a quantity represented by a vector line integral. Depending on the path you choose, the amount of work needed will vary (except for a certain vector field). Therefore, to compute a vector line integral without any ambiguity, we must choose a uniform path of travel along a curve. This choice of direction is called orientation.
We will define the line integral of a vector field F over a curve C as the scalar line integral of the tangential component of F.
If we let T = T(P ) denote the unit tangent vector at a point P on C pointing in the positive direction then the tangential component of F at P is the dot product;
F(P ) · T(P ) = kF(P )k cos θ
Definition (Vector Line Integral) The line integral of a vector field F along an oriented curve C is the the integral of the tangential component of F ;
Z Z F · ds = F · T ds C C
2 If we have a parametrization for C which we assume to be piecewise smooth and if we suppose this parametrization c(t) is regular then we have;
c0(t) T = kc0(t)k
In term of the arc length differential ds = kc0(t)k dt we have: (F · T) ds = F(c(t)) · c0(t) dt. We know arrive at the following theorem, whose proof we’ve just rendered.
Theorem (Computing a Vector Line Integral) If c(t) is a regular parametrization of an oriented curve C for t ∈ [a, b] then;
Z Z Z b F · ds = F · T ds = F(c(t)) · c0(t) dt C C a
Suppose we know F = hF1,F2,F3i and write ds = hdx, dy, dzi then: F · ds = F1dx + F2dy + F3dz
If we have a parametrization c(t) = hx = x(t), y = y(t), z = z(t)i then we have;
dx dy dz ds = c0(t) = hx0(t), y0(t), z0(t)i = , , dt dt dt dt so that;
dx dy dz F · ds = F (c(t)) + F (c(t)) + F (c(t)) dt 1 dt 2 dt 3 dt
Therefore we get the following formula;
Z Z b dx dy dz F1dx + F2dy + F3dz = F1(c(t)) + F2(c(t)) + F3(c(t)) dt C a dt dt dt
3 Conservative Vector Fields
1 2 Parametrized Surfaces and Surface Integrals
We know wish to parametrize surfaces in terms of maps G : D ⊂ R2 → S where S represents a surface in R3. We will show how to parametrize surfaces such as the cone, cylinder, sphere, torus (donut) and graphs of functions.
We wish to find a parametrization of the surface S given by x2 + y2 = z2. Observe that any horizontal trace which is given by intersecting the plane z = u with the surface is of the form: x2 + y2 = u2 i.e a circe of radius u. This a parametrization if given by;
G :(u, v) 7→ (u cos v, u sin v, u)
Now we will consider one for a cylinder of radius R which is given by x2 + y2 = R2. We can 2-parameter family to describe the cylinder; {(θ, z) : 0 ≤ θ < 2π, −∞ < z < ∞}.
G :(θ, z) 7→ (R cos θ, R sin θ, z)
1 We know arrive at the sphere. Recall that the cylinder and the sphere’s parametrization were already given when we did a coordinate change to cylindrical and spherical coordinates. The cartesian equation of a sphere with radius R is given by x2 + y2 + z2 = R2.
G :(θ, φ) ∈ [0, 2π) × [0, π] 7→ (R cos θ sin φ, R sin θ sin φ, R cos φ)
Finally, consider the graph of a function z = f(u, v). We can give it the simple parametrization;
G :(u, v) 7→ (u, v, f(u, v))
2 We now move to calculating surface area of a parametrizing surfaces. Suppose G is a map which is injective and suppose that G is continuously differentiable.
The tangent vectors to these grid curves is given by;
Gu = Tu(P ) = hxu, yu, zui
Gv = Tv(P ) = hxv, yv, zvi
The parametrization G is called regular at P if the following cross product is nonzero;
~n(P ) = Tu(P ) × Tv(P )
In this case, Tu and Tv span the tangent plane to S at P and ~n(P ) is a normal vector to the tangent plane. We call ~n(P ) a normal to the surface S.
3 Let G :(u, v) ⊂ D 7→ S be a parametrization of a surface S. Let Rij be a division of D into small rectangles of size ∆u × ∆v. The image G(Rij) = Sij is a “curved” parallelogram. If we allow ∆u, ∆v to be sufficiently small, then the curved parallelogram Sij has approximately the same area as a genuine parallelogram. Thus we have;
~ ~ Area(Sij) ≈ kPQ × PS
where the RHS are the two vectors which span the parallelogram.
Now we use the linear approximation to estimate the vectors PQ~ and PS~ ;
~ PQ = G(uij + ∆u, vij) − G(uij, vij) ≈ Gu(uij, vij)∆u = Tu∆u
~ PQ = G(uij, ∆v + vij) − G(uij, vij) ≈ Gu(uij, vij)∆v = Tu∆v
Thus we have Area(Sij) ≈ kTu∆u × Tv∆vk = kTu × T)vk∆u∆v. Since ~n(uij, vij) = Tu × Tv and Area(Rij) = ∆u∆v we obtain;
Area(Sij) ≈ k~n(uij, vij)k · Area(Rij)
4 The entire surface S is the union of the small patches Sij, so we can apply the approximation on each patch to obtain; X X Area(S) = Area(Sij) ≈ k~n(uij, vij)k∆u∆v i,j i,j
The sum on the RHS is a Riemann sum for the double integral of k~n(u, v)k over the parameter domain D. As ∆u, ∆v tend to zero, these Riemann sums converge to a double integral, which we take as the definition of the surface area; ZZ Area(S) = k~n(u, v)k du dv
D
Now we can define the surface integral of a function f(x, y, z); ZZ f(x, y, z) dS
S
(0) Let G be a parametrization for S and divide the parameter domain D into small patches Sij
(1) Choose a sample point Pij = G(uij, vij) in each small path Sij P (2) For each small patch, form the sum i,j f(Pij) · Area(Sij) (3) The limit as ∆u, ∆v 6 0 (if it exists) is the surface integral;
ZZ X f(x, y, z) dS = lim f(Pij) · Area(Sij) ∆u,∆v→0 S i,j P P (4) Using our approximation we have; i,j f(Pij) · Area(Sij) ≈ i,j f(G(uij, vij))k~n(uij, vij)k ∆u∆v
(5) The RHS is just the Riemann sum for the double integral of f(G(u, v))k~n(u, v)k over the parameter domain D.
Theorem (Surface Integrals and Surface Area) Let G(u, v) be a parametrization of a surface S with parameter domain D. Assume that G is continuously differentiable, injective and regular (except possibly at the boundary of D) then; ZZ ZZ f(x, y, z) dS = f(G(u, v)) k~n(u, v)k du dv | {z } S D dS
Furthermore, if f(x, y, z) = 1, then we obtain the surface area of S;
ZZ Area(S) = k~n(u, v)k du dv
D
5 Surface Integrals of Vector Fields
We now wish to calculate flux or rates of flow through a surface. Since flux through a surface goes from one side of the surface to the other, we need to specify a positive direction of flow. This is done by means of orientation, which it will be sufficient to have a choice of unit normal vector ~en(P ) at each point P ∈ S.
The normal component of a vector field F at a point P on an oriented surface S is the dot product;
F(P ) · ~en(P ) = kF(P )k cos θ
Here we see that the dot product measures how much substance flowed through the surface S at a specific point P by means of giving the normal direction of the vector field at P . The vector surface integral (flux) is defined as the integral of the normal component;
ZZ ZZ F · dS = (F · ~en) dS
S S
1 Let G(u, v) be a regular parametrization for a surface S then;
~n(u, v) ~e = ~e (u, v) = n n k~n(u, v)k
We can now rewrite the definition of the vector surface integral to give;
ZZ ZZ ZZ ~n(u, v) F · dS = (F · ~e )k~n(u, v)k dudv = F(G(u, v)) · k~n(u, v)k dudv n k~n(u, v)k S D D
Theorem (Vector Surface Integral) Let G(u, v) be a oriented parametrization of an oriented surface S with parameter domain D. Assume that G is injective and regular, except possibly at points on the boundary of D then;
ZZ ZZ F · dS = F(G(u, v)) · ~n(u, v) dudv
S D
We now wish to calculate flow rates which is the measure of the volume of water that flows through a surface per unit time. To compute the flow rate we issue the steps in program format;
(0) Let ~v be the velocity vector field (1) At each point P ∈ S,~v(P ) is the velocity vector of the fluid particle located at the point P (3) We claim that the flow rate through a surface S is equal to the surfaces integral of ~v over S. Suppose first that S is a rectangle of area A and ~v is a constant vector field with value ~v0 perpendicular to the rectangle.
(4) The particle travels at speed k~v0k, say in meters per second i.e a given particle flows through S within a one-second time interval it its distance to S is at most k~v0k meters - in other words, if its velocity vector passed through S
2 (5) Therefore the block of fluid passing though S in a one-second interval is a box of volume kv0k · A.
(6) Thus we have the equation Flow rate = (velocity)(area) = k~v0kA
If the fluid flows at an angle θ relative to S, then the block of water is a parallelepiped of volume Ak~v0k cos θ. If ~n is a normal to S of length equal to the area of A (which we can always have!), then we can write the flow rate as the dot product; Flow rate = Ak~v0k cos θ = ~v0 · ~n.
In the general case, the velocity vector field ~v is not constant, and the surface S may be curved. To compute flow rate, we choose a parametrization G(u, v) and we consider a sufficiently small rectangle of size ∆u × ∆v that is mapped by G to a small patch S0 of S. For any sample point G(u0, v0) in S0, the vector ~n(u0, v0)∆u∆v is a normal vector of length approximately the area of S0. The patch is nearly rectangular, so we have;
Flow rate though S0 ≈ ~v(u0, v0) · ~n(u0, v0)∆u∆v
3 The total flow per second is the sum of the flows through the small patches. As usual, the limits of the sums as ∆u, ∆v → 0 is the integral of ~v(u, v) · ~n(u, v), which is the surface integral of ~v over S.
Flow Rate Through a Surface For a fluid with velocity vector field ~v we get the following beautiful equality; ZZ Flow rate across the surface S (volume per unit time) = ~v · dS
S
4 Green’s Theorem
Introduction: For a conservative vector field F, the circulation around every closed path is zero. For vector fields in the plane, Green’s Theorem tell us what happens when F is not conservative.
R R Preliminaries: Recall that another notation for the line integral C F · ds is C F1 dx + F2 dy + F3 dz. In this notation ds = hdx, dy, dzi so that F · ds = F1 dx + F2 dy + F3 dz. In terms of a parametrization c(t) = (x(t), y(t), z(t)) we get ds = hx0(t), y0(t), z0(t)i dt. Hence we get the following formula,
Z Z b 0 0 0 F1 dx + F2 dy + F3 dz = F1(c(t))x (t) + F2(c(t))y (t) + F3(c(t))z (t) dt C a
Also, recall that if C is a smooth oriented curve which is a union of n smooth curves C1, ..., Cn then,
n Z X Z F · ds = F · ds C i=1 Ci
Green’s Theorem: Let D be a domain whose boundary ∂D is a simple closed curve, oriented counterclockwise. Then, I ZZ ∂F2 ∂F1 F1dx + F2dt = − dA ∂D D ∂x ∂y
Proof: We will do a special case and suppose D can be described as the union of two graphs y = g(x) and y = f(x) with g(x) ≤ f(x). We can split Green’s theorem up into two equations,
I ZZ I ZZ ∂F1 ∂F2 (1) F1 dx = − dA (2) F2 dy = dA ∂D D ∂y ∂D D ∂x
1 We will prove the first equation and then the second. Observe that,
I I I F1 dx = F1 dx + F1 dx ∂D C1 C2
To compute these line integral we parametrize the graphs. Here y = g(x) can be parametrized by c1(t) = (t, g(t)) and y = f(x) can be parametrized by c2(t) = (t, f(t)) where a ≤ t ≤ b. Taking into account the clockwise orientation we have,
I Z Z F1 dx = F1 dx − F1 dx ∂D c1 c2
In both parametrization x = t, so dx = dt and so,
I Z b Z b F1 dx = F1(t, g(t)) dt − F1(t, f(t)) dt ∂D a a
∂F1 Now applying the Fundamental Theorem of Calculus to ∂y (t, y) as a function of y with t held constant;
Z y=f(t) ∂F1 (t, y) dy = F1(t, f(t)) − F1(t, g(t)) y=g(t) ∂y
Thus we have that,
I Z b Z f(t) ZZ ∂F1 ∂F1 F1dx = − (t, y)dy dt = − dA ∂D t=a y=g(t) ∂y D ∂y
Then other equation is proven in a similar fashion, expressing ∂D as the union of the graphs x = f1(y) and x = g1(y).
2 Fun Application
While it may seem like overkill and an abuse of required mathematics for this application, it is sort of nice. We will show that Green’s Theorem can be used to find a formula for the are of the domain D enclosed by a simple closed curve C.
Let F(x, y) = h−y, xi then we have D1F2 − D2F1 = 2 and so by Green’s Theorem,
I ZZ −y dx + x dy = 2 dx dy = 2Area(D) C D
Exercises
3 More Challenging Exercises
(1) Find a formula for the areas of a circle of radius R, and an ellipse.
(2) Complete the proof for Green’s Theorem i.e shoe that equation (2) holds.
H H (3) Show that if C is a simple closed curve, then C −y dx = C x dy and both integral are equal to the area enclosed by C.
4 Stoke’s Theorem
Before introducing Stoke’s theorem we must talk about oriented surfaces with boundary and orientation induced on the boundary. Below we give some examples of surfaces with a boundary curve, multiple boundary curves and closed surfaces (surfaces without boundary).
We will now make clear what we mean by induced boundary. Let S be the open “cylinder” given below. Since S is oriented, we have a choice of outward and inward pointing normal vector for all points p ∈ S.
Suppose we choose the outward pointing normal. We think of the outward pointing normal as a means of the up direction then we assign a direction to ∂S by declaring that if we walk along the boundary curves, the surface S lies to our left. This gives the directions on the boundary in part (A). Similarly if we choose the inward pointing normal, the surface is to our right and so our travel must reverse part (A).
Now we define curl(F) = ∇ × F where we have;
∂ ∂ ∂ ∇ = , , ∂x ∂y ∂z
One immediate consequence is that is F = ∇V then curl(F) = ~0. Now suppose S is an oriented surface with parametrization G : D → S, where D is a domain in the plane bounded by smooth, simple closed curves, and G is injective and regular, except possibly on the boundary of D.
1 Theorem (Stokes’ Theorem) For surfaces S as described above we have;
I ZZ F · ds = curl(F) · dS ∂S S
The integral on the left is defined relative to the boundary orientation of ∂S. If S is closed i.e ∂S = ∅, then the surface integral on the right is zero.
Lastly, we have a analogous result for vector fields which are the curl of another vector field. We write this as F = curl(A) and refer to A as the vector potential for F. Stokes’s Theorem tells us that for any two surfaces S1,S2 with the same oriented boundary C we have;
ZZ ZZ I F · dS = F · dS = A · ds C S1 S2
Theorem (Surface Independence for Curl Vector Fields) If F = curl(A then the flux of F through a surface S depends only on the oriented boundary ∂S and not on the surface itself;
ZZ I F · dS = A · ds S ∂S
2 Divergence Theorem
We now conclude this course with the last fundamental theorem called the divergence theorem. There are huge applications of this theorem as well as the others which if we have time, we will explore them. First we will set the frame-work to apply the theorem. Let S be a closed surface that encloses a 3-D region W. In other words S is the boundary of W.
The derivative appearing in the Divergence Theorem is the divergence of a vector field F = hF1,F2,F3i defined by; ∂F ∂F ∂F div(F) = 1 + 2 + 3 ∂x ∂y ∂z
Theorem (Divergence Theorem) Let S be a closed surface that encloses a region W in R3. Assume that S is piecewise smooth and is oriented by the normal vectors pointing to the outside of W. Let F be a vector field whose domain contains W then;
ZZ ZZZ F · dS = div(F) dV S W
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