Multivariable alculus Jaron Kent-Dobias May 17, 2011 1 Lines in Space By space, we mean R 3. First, conventions. Always draw right-handed axes. You can define a L line precisely in 3-space with 2 points, p and q. If you take the vector from p to q a, you can find the slope of the line. One point and the slope also determines the line. p (p 1, p 2, p 3 ) a (a 1, a 2, a 3 ) Notice that above, three numbers define both a point and a vector. There seems to be some sort of point/vector duality here. All we care about the vector is its magnitude and direction. The point is unique. But really, there is no difference, as every point can define a vector and every vector can define a point. A vector can be linearly translated however. We can write the equation for this line as p + ta L. The direction of the line is defined by ta and is displaced by the point p. In fact, L {p + ta t R}. What we have is a set. We want to find some sort of equation from this. Rephrasing our above set, we have (x, y, z) L (x, y, z) p + ta. We can unpack this vector equation into a set of scalar equations. x p 1 + ta 1 y p 2 + ta 2 z p 3 + ta 3 These are parametric equations of the line with parameter t. We may desire equations that are not parametric, though. Note that we have three linear equations! We simply solve for t. t x p 1 a 1 y p 2 a 2 z p 3 a 3, (a i 0) These are the symmetric equations of L. Note: Two planes in R 3 which are not parallel meat at a line. These symmetric equations of a line are, in fact, two equations of two planes. The above string of equalities is in fact two equalities. A linear equation in x, y, and z describes a plane like a equation in x and y describes a line. 1
2 Planes in R 3 onsider a plane P in R 3. The plane can be uniquely determined by three points p 0, q, and r. These three points can be used to make two vectors and a displacement point, p 0, p 0 q, and pr. Taking the cross product of our vectors results in an orthogonal vector which defines the plane, n. P is affine (it doesn t necessarily contain the origin), a linear translate of a vector subspace. The one vector that is orthogonal is a normal vector to the plane. With the normal vector, all one needs is a point and that vector to define the plane. We want to be able to determine if a point s lies on the plane. s P n (p p 0 ) 0. This is an equation for the plane called the vector equation of the plane. We can now unpack this This tells us that n (a, b, c) p (x, y, z) p 0 (x 0, y 0, z 0 ) a(x x 0 ) + b(y y 0 ) + c(z z 0 ) 0. This is called the scalar equation of a plane. By rearranging, we can get an equation of the form ax + by + cz d where d n p 0 ax 0 + by 0 + cz 0. This is called the linear equation of a plane. Note that the normal vector is easy to read from the linear equation of a plane. Example 1. If we have 7y 103z 79, we immediately recognise this as a plane and that its normal vector is 2.1 Symmetric equations for a line We have already found equations of the form n (0, 7, 103) x p 1 a 1 y p 2 a 2 We can rewrite these to and x p 1 a 1 y p 2 a 2. a 2 x a 1 y a 2 p 1 a 1 p 2 a 3 x a 1 z a 3 p 1 a 1 p 3. 2
3 Functions of several variables Definition 1. The artesian Product of two sets A, B is A B {(a, b) a A, b B}. Definition 2. A map or function F from A to B is a subset of A B such that for every a A, there is a unique b B, where (a, b) F A B. Notation: (a, b) F F (a) b Definition 3. A is the domain of F, B is the codomain, range, or image of F is F (a) {(a) B a A} Example 2. In this course, we will focus on functions where A R n, B R m. We ve dealt with matrices and linear maps in linear algebra. 3.1 Graphs of functions Definition 4. Let f : R n R. The graph of f is Γ { (a, f(a)) R n+1 a R n} Example 3. f : R R Γ {(x, y) y f(x)} The maximum n we can actually draw is 2. For n 3, Γ R too big. Example 4. f : R 2 R. 1. Find at least one point on Γ. f(x, y) x2 a 2 + y2 b 2 2. Slice Γ with planes perpendicular to x, y axes, called sections. 3. Slice Γ with planes perpendicular to the z axis, called level sets. } Γ {(x, y, z) R 3 z x2 a 2 + y2 b 2 If we choose some z c, we get an ellipse; this curve defines the level set of z at c. If z c x2 a 2 + y2 b 2 ; x d d2 a 2 + y2 b 2 z, then we have defined a parabola in y and z. This graph is an elliptic paraboloid. 3
4 Directional Derivatives onsider f : R 2 R. What do f x (a, b) and f y (a, b) mean geometrically? The partial derivative gives the slope of the tangent line to the curve at (a, b) defined by slicing our function with a plane normal to the y or x axis, respectively. What is the rate of change of f : R n R in the direction v? Definition 5 (Directional Derivative). Let f : R n R, u R n be a unit vector. The directional derivative of f(x) in the direction u is f(x + hu) f(x) D u f(x) lim. h 0 h Theorem 1. D u f(x) f(x) u. [Theorem 1] is a philosophical statement about the nature of the gradient -Dagan Karp Proof. Fix a R n. Define g : R R n by g(t) a + tu. f(a + hu) f(a) D u f(a) lim h 0 h f g(h) f g(0) lim h 0 h D(f g)(0) D (f(g(0)) Dg(0) D u f(a) f(a) u. This is what you did in the days of your adolescence, when you first learned the chain rule. -Dagan Karp Example 5. Let f(x, y) x 2 y 2 + 70 be volume at (x, y) in decibels, where x, y are feet from a given origin. Note: f(x, y) (2x, 2y). We can interpret this by evaluating at points. At (1, 2), f(1, 2) (2, 4) db/ft. How about the direction (1, 1) at the point (1, 2)? f(1, 2) Theorem 2. Let f : R n R, u R n, u 1. 1 2 (1, 1) 2 db/ft. 1. D u f is largest when f and u are in the same direction. 2. The magnitude of f is the greatest rate of increase of f. 3. D u f 0 when f u or f 0. 4. f(x) 0 for all x R n iff all directional derivatives are zero (flat everywhere). 4
Example 6. f(x, y) x 2 y 2 + 70. This object is a hyperbolic paraboloid. The height of this saddle tell us the volume at that location. What is the fastest decrease at (1, 2)? f(1, 2) ( 2, 4). How fast is this? 2 5 db/ft. If we walk 3 ft/sec, then how fast is the volume decreasing? 5 Paths 3 ft/sec 2 5 db/ft 6 5 db/sec. Definition 6 (Path). A path in R n is a map c : I R n where I R is an interval. Definition 7 (Velocity). Definition 8 (Acceleration). Definition 9 (Speed). Example 7. v(t) c (t) Dc(t) a(t) v (t). v(t) c : R R 3 The helix is defined by the path c 1 (t) (a cos t, a sin t, bt). The speed is a 2 + b 2 ; note that this is constant, or does not depend on t. The circle is defined by c 2 (t) (a cos t, a sin t). It has constant c 2 (t) a. Definition 10 (Tangent Line). The tangent line to c(t) at c 0 c(t 0 ) is the line through c 0 and parallel to any tangent vector. Note that c(t) must be a differentiable path. That s kind of a gender normative thing; What do you guys think?. Would anyone be offended if I said What do you girls think? Well, what do you girls think? ome on girls. -Dagan Karp Proposition 1. Let c(t) be a differentiable path. if v 0 v(t 0 ) is nonzero, then the tangent line to c(t) at c 0 is r(s) c 0 + sv 0 Remark: This works for all points on a circle. r(t) c 0 + (t t 0 )v 0. Proposition 2. Let x, y : I R n be paths in R n. d dy 1. (x y) x dt dt + y dx dt. d dx 2. If n 3, (x y) dt dt y + x dy dt. Proposition 3. If c(t) has constant magnitude, then c(t) and v(t) are orthogonal. Example 8. So, c 2 (t) (a cos t, a sin t) v 2 (t) ( a sin t, a cos t) c(t) v(t) 0. 5
5.1 Arclength onsider c : [a, b] R 2. The length of this path is the sum of many little pieces of length, or where Because we can express dl as L b a dl, dl (dx) 2 + (dy) 2. c(t) (x(t), y(t)), (dx ) 2 dl + dt Now we can directly calculate L in relation to t: b (dx ) 2 L + dt ta b ta We can treat the above as a definition of length. c (t) dt. ( ) 2 dy dt. dt ( ) 2 dy dt dt Example 9. c 2 (t) (a cos t, a sin t) t [0, 2π] L 2π t0 2π t0 L 2πa. a 2 sin 2 t + a 2 cos 2 t dt a dt Definition 11 (Vector Field). A vector field on R n is a map F : R n R n. Example 10. The above is a constant vector field on R 2. F(x, y) (2, 3) Example 11. F(x, y) ( y, x). Note that (x, y) F(x, y) 0, so any vector in the vector field is orthogonal to its position vector. 6
Example 12. F : R 3 0 R 3 F(x) c x 3 x. Let u x x. Then F(x) c x 2 u Definition 12 (Gradient Field). The vector field F is a gradient field if f : R n R such that f is called the scalar potential function. F(x) f(x). Example 13. Let f : R 3 0 R Note: f(x) F(x). f(x) c x. f x c x x2 + y 2 + z 2 cx ( ) 3 x2 + y 2 + z 2 The same is true for y and z. Definition 13 (Flow Line). A differentiable path c(t) is a flow line of F(x) if c (t) F (c(t)). Example 14. F(x, y) ( y, x) Let c(t) (a cos t, a sin t). c (t) ( a sin t, a cos t) F (c(t)) F(a cos t, a sin t) ( a sin t, a cos t) Therefore, c(t) is a flow line of F. Example 15. F(x, y) (0, x). Find flow lines. Let c(t) (x(t), y(t)).. Solve ODE. x (t) 0 y (t) x(t). So x(t) a and y(t) at + b. c(t) (a, at + b). Example 16. F(x, y) (x, 0). x (t) x(t) y (t) 0. dx dt x x(t) aet and y b. c(t) ae t, b). 7
6 Divergence and url Definition 14 (Del). The del operator on R n is (,,..., x 1 x 2 Definition 15 (Divergence). The divergence of a vector field F on R n is given by x n ). F F 1 x 1 + F 2 x 2 + + F n x n, where F (F 1, F 2,..., F n ). Note that the divergence is a scalar valued function. It represents the expansion or contraction of F near a. Think tiny box around a R n. If div(f) F at a is positive, then there is an expansion at a. If negative, there is a contraction. If 0, then a is incompressible. Example 17. F(x, y, z) (x, y, z). Example 18. F(x, y) ( y, x). div(f) F 1 + 1 + 1 3. F y x + x y 0 + 0 Definition 16 (url). Let F (F 1, F 2, F 3 ) be a vector field on R 3. The curl of F is 0. curl(f) F î ĵ k x y z F 1 F 2 F 3 ( F3 î y F ) ( 2 z ĵ F3 x F ) ( 1 + z k F2 x F ) 1. y This is a vector valued function. It is a vector pointing in the direction of the axis of greatest spin at a point. Its magnitude is proportional to the rate of spin. Example 19. F(x, y, z) ( y, x, 0). curl(f) F î ĵ k x y z y x 0 (0, 0, 2). 8
Example 20. F(x, y, z) (x, y, z). Example 21. (x, y, z) (y 2, 0, 0). î ĵ k F x y z x y z (0, 0, 0). î ĵ k F x y z y 2 0 0 (0, 0, 2y). Definition 17 (Scalar url). Let F be a planar vector field, The scalar curl of F is F(x, y) (M, N). N x M y. Notice that this looks like a component of the ordinary curl. It is a scalar that, when positive, indicates a counterclockwise rotation, and when negative clockwise. Example 22. F(x, y) (y, x). There is a decent Polish beer in the sink. -Dagan Karp N x M y x x y y 1 1 2 Example 23 (Gravity). F(x) x x 3 on R 3. Note that F 0 F 0. Theorem 3. Let f be class 2 function on R 3. Then ( f) 0. Proof. f (f x, f y, f z ) î ĵ k ( f) x y z f x f y f z (f zy f yz, f xz f zx, f yx f xy ) (0, 0, 0) 9
Are we all equally groovy? -Dagan Karp Remark: onservative vector fields are irrotational. Theorem 4. Let F (F 1, F 2, F 3 ) be class 2. Then ( F) 0. 7 Taylor Series/Taylor s Theorem For f : R R, use f(a), f (a), f (a), etc. to approximate f(x) near x a. P n (x) n k0 f (k) (a) (x a) k k! f(x) P n (x) + R n (x), or the nth order Taylor Polynomial plus a error or remainder term. Theorem 5. The first few Taylor Polynomials are lim x a R n (x) (a x) n 0. P 0 (x) f(a) P 1 (x) f(a) + f (a)(x a) P 2 (x) f(a) + f (a)(x a) + f (x) (x a) 2 2 Remark: require that P n (a) f(a), P n (a) f (a), P n (a) f (a),... P n (n) (a) f (n) (a). P n (x), f(x) agree to order n. Let f : R n R be. Our goal is to find polynomials P n (x) such that f(x) P n (x) + R n (x) where lim x a R n (x) x a 0. Let n 2. f(x, y) : R 2 R. We want P 0 (x), P 1 (x), P 2 (x) to agree with f(x) order 0, 1, 2. P 0 (x) f(a) P 1 (x) f(a) + f(a) (x a) P 2 (x) f(a) + f(a) (x a) + 1 (x a)h(f)[x a], 2! where [ ] fxx (a) f H(f) xy (a) f yx (a) f yy (a) 10
Example 24. f(x, y) e x cos y + xy, approx near (0, 0). f x (x) e x cos x + y f y (x) e x sin y + x f x (0, 0) 1 f y (0, 0) 0. f xx (x) e x cos y f xy (x) e x sin y + 1 f yy (x) e x cos y. f xx (0, 0) 1 f xy (0, 0) 1 f yy (0, 0) 1. P 2 (x, y) f(0, 0) + f x (0, 0)x + f y (0, 0)y + f xx(0, 0) 2 1 + x + x2 2 y2 2 + xy. 8 Extrema: Local Max/Min x 2 + f yy(0, 0) y 2 + f xy (0, 0)xy 2 The critical points of f(x) are where f 0. f(x) f(a) + f(a) (x a) + (x a)h(f)[x a] + R 2 (x) Theorem 6. Let f(x) 2 and D det H (f(a)) where a is a critical point of f. 1. If D > 0 and f xx (a) < 0, then a is a local max. 2. If D > 0 and f xx (a) > 0, then a is a local min. 3. If D < 0, a is a saddle point. 4. If D 0, then no information is given. Example 25. f(x) x 2 y + 3xy 2 + xy. lassify the extrema. 0 f 4 cases: (0, 0), ( 1, 0), ( 0, 1 3), ( 1 3, 1 9). ( 2xy + 3y 2 + y, x 2 + 6xy + x ) (y(2x + 3y + 1), x(x + 6y + 1)). D(0, 0) 0 1 < 0 D( 1, 0) 0( 6) ( 1) 2 < 0 ( D 0, 1 ) 0 ( 1) 2 < 0 3 ( D 1 ) 3, 1 29 ( 9 ( 2) 3) 1 2 > 0. Use the theorem to classify the points: 1st, 2nd, and 3rd are saddle points, 4th is a min. 11
8.1 onstrained Optimization Example 26. Find the point on the plane x+2y +z 6 which is closes to the origin. We want to minimise the expression f(x, y, z) x 2 + y 2 + z 2 Let g(x, y, z) x + 2y + z. We want g 6. We can apply a technique called Lagrange multipliers. At local extrema, f g or f λ g. 2x λ1 2y λ2 2z λ1 x + 2y + z 6 λ 2, (1, 2, 1) is our point. It is a global min by geometry. 9 Double Integration Recall for f : R R Recall: f(x, y) : R 2 R xb xa R f(x) dx lim f(xi ) x. x 0 f da Example 27. R semicircle, f(x, y) x. R f da 9.1 Polar oordinates y2 y0 x 4 y 2 lim f(xi, y i ) x y x, y 0 x 4 y 2 x dx dy x2 y 4 x 2 x 2 To convert from artesian coordinates x, y to polar coordinates r, θ, r x + y ( θ tan 1 y ) x To convert from polar coordinates to artesian coordinates, Example 28. R f da x r cos θ y r sin θ da r dr dθ θπ r2 θ0 r0 y0 r cos θr dr dθ 0 Example 29. f(x, y) y and R is defined by x 3, y x, y 0. θ π 4 r3 sec θ f da r 2 sin θ dr dθ 9 2 R θ0 r0 x dy dx 12
10 Triple Integration For f : R 3 R and W R 3, How to compute? 1. hop W into little cubes. W f dv lim Riemann Sum x, y, z 0 2. Stack the cubes in one coordinate direction to determine the inner limits of integration. 3. Project along that axis to obtain a two-dimensional shadow. 4. Use the shadow to determine the remaining limits of integration as done previously. Example 30. Find the volume in the first octant bounded by x 2 +z 2 4 and y 3. The region is contained by a cylinder perpindicular to the x and z axes. z 4 x 2 x2 y2 z 4 x 2 f dv 1 dz da dz dy dx W W R z0 V a dv R y3 y0 dy da x0 y0 z0 z2 x 4 z 2 y3 Example 31. Volume in first octant bounded by 6x + 3y + 2z 6. z0 x0 y0 dy dx dz. 10.1 ylindrical and Spherical oordinates ylindrical coordinates are like polar coordinates, but with a artesian z dimension. artesian to cylindrical, we have x r cos θ The inversion is Example 32. y r sin θ z z r x 2 + y 2 ( θ tan 1 y ) x z z f dv W dv r dr dz dθ. δ(x, y, z) x 2 + y 2 + z 2 of region W, defined by a wedge of height 2, radius 3 and angle π 6. M δ dv W θ π 2 r3 z2 θ π 3 r0 z0 (r 2 + z 2 )r dz dr dθ To convert from 13
Example 33. Volume of sphere of radios R. V dv For spherical coordinates, we have W θ2π rr z R 2 r 2 θ0 r0 2π R θ0 2π θ0 Example 34. Volume of sphere V r0 2R 3 dθ 3 2R3 3 (2π) z R 2 r 2 r dz dr dθ zr R 2 r 2 dr dθ x r cos θ ρ sin φ cos θ y r sin θ ρ sin φ sin θ z ρ cos ( φ θ tan 1 y ) x ρ 2 x 2 + y 2 + z 2 ( φ tan 1 r ) ( tan 1 x2 + y 2 z z dv ρ 2 sin φ dρ dφ dθ dv W θ2π φπ ρr R3 3 R3 3 θ0 φ0 ρ0 θ2π φ2π 4πR3 3 θ0 θ2π θ0 φ0 dθ. ) ρ 2 sin φ dρ dθ dθ sin φ dπ dθ Example 35. The mass of a cone with density δ(x, y, z) x 2 + y 2 + z 2, height 2 and with angle π 4. M δ dv W θ2π θ0 φ π 4 φ0 ρ2/ cos φ ρ2 ρ 3 sin φ dρ dφ dθ 14
Example 36. The mass of the same cone with density δ(x, y, z) ( x 2 + y 2) z. M δ dv W θ2π φ2 z2 θ0 φ0 zr r 2 zr dz dr dθ Other oordinates: Theorem 7. Let T : R 3 R 3 be a change of coordinates taking (α, β, γ) to (x, y, z). Then dv det(j) dα dβ dγ where J DT 11 Line Integrals an we integrate over interesting regions in R 2 or R 3 or R n? Line integrals are integrals over one-dimensional regions of R n. Let be a curve in R n. r(t) parameterise, i.e. {r(t) t [a, b]}. Let F be a vector field on R n such that f : R n R is scalar. 1. f ds 2. 3. f dx F dr Definition 18. s is arclength. Let P be a partition of [a, b] with m parts. f ds is the integral of f over with respect to arclength. P {a p 0 < p 1 < < p m b} x i x i x i 1 r(p i ) (x i, y i ) y i y i y i 1 Let (x i, y i ) be a point on such that x i 1 < x i < x i. The length of the partition P is P Max { x i, y i i 1,..., m} f ds lim m P 0 i1 f(x i, y i ) ( x i ) 2 + ( y i ) 2. 15
Arclength s(t). where r(t) (x(t), y(t)) Example 37. onsider F on R 3 given by τt s(t) r (τ) dτ. τt 0 ds dt d ( t ) r (τ) dτ dt τt 0 ds r (t) dt tb f ds f (r(t)) r (t) dt f dx f dy F dr ta tb ta tb ta tb ta f (r(t)) x (t) dt f (r(t)) y (t) dt F (r(t)) r (t) dt. F(x, y, z) (z, xy, x + z). Integrate F along c given by three line segments travelling one unit in the x direction, one unit in the y direction, and one unit in the z direction. tb F dr F (r(t)) r (t) dt ta F dr + c 1 F dr + c 2 F dr c 3 t1 t0 0 + 1 2 + 3 2 2. (0, 0, t) (1, 0, 0) dt + c 1 {(t, 0, 0) t [0, 1]} c 2 {(1, t, 0) t [0, 1]} c 3 {(1, 1, t) t [0, 1]} t1 t0 What about another path from (0, 0, 0) to (1, 1, 1)? D (0, t, 1) (0, 1, 0) dt + D {(t, t, t) t [0, 1]} F dr t1 t0 t1 t0 1 3 + 3 2 11 6. (t, t 2, 2t) (1, 1, 1) dt ( t 2 + 3t ) dt t1 t0 (t, 1, 1 + t) (0, 0, 1) dt 16
Definition 19 (losed Path). A closed path in R n is a path r(t) : [a, b] R n such that r(a) r(b). F dr F dr. c Theorem 8 (Green s Theorem). Let c be a closed curve bounding a region R in R 2. If R is always on the left of c, and F (M, N) is class 1 and defined on all of R, then F dr (N x M y ) da. L orollary 1. Let c, R be as above. Then the area of R is 1 2 ( y, x) dr. c Theorem 9 (Fundamental Theorem of Line Integrals). Let F f be a conservative vector field. Let c be a curve from A to B. Then F dr f(b) f(a). If c is closed and F is conservative, then c c c F dr 0. Remark: For conservative fields, line integrals are path independent. Proof. c F dr tb ta tb ta tb ta F (r(t)) r (t) dt f (r(t)) r (t) dt d (f (r(t))) dt dt f (r(b)) f (r(a)) f(b) f(a). orollary 2. If F is conservative, then c F dr 0 for any smooth closed curve c. Theorem 10. F is conservative if and only if F dr 0 for every smooth closed curve c. c Proof. We need to find a scalar potential f. Let s fix f(0, 0) 0. For (x, y) R 2, let f(x, y) F dr c where c goes from (0, 0) to (x, y). F (M, N). We must show that f (f x, f y ) (M, N). Theorem 11. Let F (M, N), and let F be class 2. If F is conservative, then N x M y. Proof. F f, be definition. So, f (f x, f y ) (M, N). M y f xy, N x f yx. F 2 f xy f yx. 17
Theorem 12. Let F (M, N) be a vector field on a simply connected region R. If N x M y, then F is conservative. Theorem 13. On a simply connected region, F is conservative if and only if curl(f) 0. 12 Flux Recall: F dr computes the flow of F along. What is the flow of F across a surface in R 3? onsider a constant F and a flat surface S. In this case, Flux F n(area of S). What if F is not constant or S is not flat? Flux of F across S S F n ds. Note that the normal vector is dependent on one s location on the surface. If my net is parallel, I don t catch any fishies -Dagan Karp 12.1 Surface Integrals There are two flavours of surface integrals. 1. Scalar surface integrals. Let X be a surface in R 3. Then the scalar surface integral of the scalar function f over X is given by f ds (a) If δ is mass density, then δ ds is the mass of X. (b) 1 ds is surface area. X 2. Vector surface integrals, or flux integrals. X X X F ds. (a) If F is the velocity of a fluid (m/s), then flux is the volume of fluid passing through the surface per unit time ( m 3 /s ). Remark: Usually orient n up or out. Note that a consistent choice of n always exists for orientable surfaces. To compute: 1. Parameterise the surface. Use two parameters, usually s and t. (Write x, y, z in terms of s, t.) 2. Determine ds or ds in terms of s, t. 3. Turn into double integral over s, t. Note: write X as X(s, t). Parameterise by T (s, t), so that X(s, t) {T (s, t) t, s R}. 18
onsider a surface defined by We can parameterise a surface using T (s, t) (x(s, t), y(s, t), z(s, t)). T s (x s, y s, z s ) T t (x t, y t, z t ) Both of the above vectors are tangent to the surface. Thus, ds (T s T t ) ds dt ds T s T t ds dt. Example 38. Find the flux of F(x, y, z) (x, y, z) through the bowl z x 2 + y 2 over the unit disk. ( F ds x, y, x 2 + y 2) ( 2x, 2y, 1) dx dy X R θ2π r1 θ0 π 2. r0 Example 39. Surface area of a sphere of radius A. X A ds φπ θ2π φ0 4πA 2. r 3 dr dθ θ0 1 A 2 sin φ dφ dθ Example 40. Flux of F (0, 0, 3) over sphere radius 2 in the first octant. X F ds 13 Stokes Theorem φ π 2 3π φ0 θ π 2 θ0 (0, 0, 3) (,, cos φ)4 sin φ dθ dθ. Recall Green s Theorem. Given some region R and some path such that R is to the left of, F dr curl(f) da. Stokes Theorem extends this into 3-space. R Theorem 14 (Stokes Theorem). If S is a surface with boundary curve S consisting of closed oriented paths (oriented so S is on the left when viewed from the top ), then for F vector field on R 3, F dr curl(f) ds. S S 19
Why is this true? If we have a surface that is chopped up into many little pieces, the circulation around those little pieces cancels everywhere within the surface, leaving only the boundary. We attempt to compute the circulation around a little piece of surface so that we have sum them and find the total circulation. F dr curl(f) n ds, where ds is the area of our little piece. Take the limit of sum over all boxes, obtaining curl(f) n ds. S Note: The circulation of F over S is the sum of the curl of F over S. Example 41. Let F ( y, 2x, z). Find F dr, where is the surface defined by a sphere of radius two centered at the origin and bounded by the first octant. Directly, we would need to use 2 parameterisations. Using Stokes Theorem, we have î ĵ k F x y z y 2x z (0, 0, 3). So, F dr (0, 0, 3) ds S φ π 2 θ π 2 3π φ0 θ0 (0, 0, 3) (,, cos φ)4 cos φ dθ dφ Example 42. F ( x 2, 0, 0 ), same path. F dr S 0. curl(f) ds Example 43. F ( yz 2, xz 2, z ). Let 1 be a circle of radius 3 in in the xy-plane, and let 2 be a circle of radius 3 at z 2. We want F dr. 1 2 We make a cylindrical surface bounded by these two paths. Note, that we need to use the inward normal for this region to be to the left of the curves 1 and 2. ( F dr 2xz, 2yz, 2z 2 ) ( R cos θ, R sin θ, 0) ds 72π 1 2 S 20
14 Gauss Theorem/Divergence Theorem Notation: A closed surface integral of a vector field F over a closed surface X is F ds. X Theorem 15 (Gauss Theorem/Divergence Theorem). Let V be a closed and bounded region in R 3 with boundary surface V. Then F ds F dv. V V Example 44. Let X be a sphere of radius 3 and F (2x, 0, 5z). Find the flux of F across X. Flux F ds X Let V be V { (x, y, z) R 3 x 2 + y 2 + z 2 9 } V X. By Gauss Theorem, F ds F dv 7 dv 252π. X V V Example 45. Express x 2 z dv as a surface integral, where V has boundary V. We need to find F such that F x 2 z. By Gauss Theorem, V F ( 0, x 2 yz, 0 ). F x 2 z. x 2 ( z dv 0, x 2 yz, 0 ) ds. V V 21