Math 350 Solutions for Final Exam Page 1. Problem 1. (10 points) (a) Compute the line integral. F ds C. z dx + y dy + x dz C

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1 Math 35 Solutions for Final Exam Page Problem. ( points) (a) ompute the line integral F ds for the path c(t) = (t 2, t 3, t) with t and the vector field F (x, y, z) = xi + zj + xk. (b) ompute the line integral z dx + y dy + x dz for the path c(t) = ( e t2, ln(t + ), cos(t) ) ) with t. Solution. (a) We have c (t) = (2t, 3t 2, ), so F ds = F (t 2, t 3, t) c (t) dt = = (t 2, t, t 2 ) (2t, 3t 2, ) dt 2t 3 + 3t 3 + t 2 dt = 5 4 t4 + t= 3 t3 t= = = 9 2 (b) This is the integral of the vector field F (x, y, z) = zi + yj + xk. This vector field satisfies the conditions to be a gradient field, and it s easy enough to find that F = f for the function f(x, y, z) = xz + 2 y2. The fundamental theorem of calculus for line integrals says that the value of the integral is given by the difference of the values of f at the endpoints of the curve. So z dx + y dy + x dz = F ds = ( f) ds = f ( c() ) f ( c() ) = f ( e, ln(2), cos() ) f(,, ) Math 35 Final Exam 2/2/4, 9:3 :3am

2 Math 35 Solutions for Final Exam Page 2 = e cos() + 2 (ln 2)2 Problem 2. (5 points) Let be the region = { (x, y) : x 2 and y and x 2 + y 2 9 }. (a) Sketch the region. (b) Write the integral f(x, y) dx dy as a sum of one or more iterated integrals in xy-coordinates. (c) Write the integral f(x, y) dx dy as a sum of one or more iterated integrals in polar coordinates. Solution. (b) For x, the region is x 2 y 9 x 2, while for x 2, the region is y 9 x 2. So 9 x x 2 f(x, y) dx dy = f(x, y) dx dy + f(x, y) dx dy x 2 (c) The vertical line x = 2 intersects the circle x 2 +y 2 = 9 at the point whose angle θ is cos (2/3). So for θ cos ( 2/3), the values of r go from r = to the line x = 2. Since x = r cos θ, that means that r goes from to 2/ cos θ. Then, for cos (2/3) θ π/2, the value of r goes from to 3. Hence cos (2/3) 2/ cos θ f(x, y) dx dy = f(r cos θ, r sin θ) r dr dθ + π/2 cos (2/3) 3 f(r cos θ, r sin θ) r dr dθ. Problem 3. ( points) Find all of the critical points of the function f(x, y) = 3 x3 + 3 y3 2 x2 5 2 y2 + 6y + and classify the critical points as local maxima, local minima, and saddle points. Math 35 Final Exam 2/2/4, 9:3 :3am

3 Math 35 Solutions for Final Exam Page 3 Solution. We have f x (x, y) = x 2 x = x(x ), f y (x, y) = y 2 5y + 6 = (y 2)(y 3). So there are four critical points: For each one we need to compute (, 2), (, 3), (, 2), (, 3). = f xx f yy f 2 xy = (2x )(2y 5) 2 = (2x )(2y 5). Then a point is a local minimum if > and f xx >, it is a local maximum if > and f xx <, and it is a saddle point if <. Note that f xx = 2x. We make a little table: Point (, 2) (, 3) (, 2) (, 3) Value of Value of f xx Type of point Max Saddle Saddle Min Problem 4. ( points) Let f(x, y) be defined by 2x 3 3y 3 if (x, y) (, ), f(x, y) = x 2 + y 2 if (x, y) = (, ). (a) alculate f f (, ) and (, ) directly from the definition. x y (b) Let a and b be non-zero constants, and define a function g(t) = f(at, bt). alculate dg dt (). (c) Let h(t) = (at, bt), so the function g(t) in (b) is g(t) = f(h(t)). The chain rule would say that dg dt () = f(, ) h () = f f (, )a + (, )b. x y oes this agree with your answers from parts (a) and (b)? explain what is going wrong. Solution. (a) We compute If not, f f(h, ) f(, ) (, ) = lim definition of partial derivative, x h h 2h 3 /h 2 = lim definition of f, h h Math 35 Final Exam 2/2/4, 9:3 :3am

4 Math 35 Solutions for Final Exam Page 4 Similarly, = 2. f f(, k) f(, ) (, ) = lim y k h 3k 3 /k 2 = lim k k = 3. (b) For t we have definition of partial derivative, definition of f, g(t) = f(at, bt) = 2(at)3 3(bt) 3 (at) 2 + (bt) 2 = 2a3 3b 3 a 2 + b 2 t. This formula is also true for t =, since g() = f(, ) =. Hence g () = 2a3 3b 3 a 2 + b 2 (In fact, this is g (t) for every value of t.) (c) From (b) we have But using (a) we have g () = 2a3 3b 3 a 2 + b 2. f f (, ) a + (, ) b = 2a 3b. x y These are not the same in general. Indeed, their difference is 2a 3 3b 3 (2a 3b) = 2ab2 + 3a 2 b ab( 2b + 3a) =, a 2 + b 2 a 2 + b 2 a 2 + b 2 so they are the same only if a =, b =, or 3a = 2b. The reason that this does not contradict the chain rule is because the chain rule only applies if the partial derivatives are continuous. In this example, the partial derivatives of f, although they do exist at (, ), are not continuous. Problem 5. (5 points) For each of the following vector fields F, check whether F is conservative. If it is conservative, find a potential function. If it is not conservative, explain why not. (a) F = zi + ( x 2 + z2) j + (x + yz)k. 2 (b) F = (2xy + x)i + 2 (x2 + sin 2 3y)j. Note: espite the new majorities in the House and the Senate, there is not yet a law saying that all (American) vector fields are conservative! Math 35 Final Exam 2/2/4, 9:3 :3am

5 Math 35 Solutions for Final Exam Page 5 (c) Let a be a non-zero constant vector, let r = xi + yj + zk, and let F = a r. Solution. By definition, a vector field F is conservative if it is the gradient of a function F = f. (a) A vector field F = P i + Qj + Rk defined everywhere on a solid region (or even defined everywhere except for a finite set of points) is conservative if and only if its curl is zero, or equivalently, if In this case we have P y = Q x and P z = R x and Q z = R y. P y = and Q x = 2x, P z = and R x =, Q z = z and R y = z. The first line shows that F is not conservative. Alternatively, one computes curl(f ) = 2xi is nonzero. (b) Similarly, a vector field F = P i + Qj in the plane that is defined everywhere in a region is conservative if and only if Q x = P y. In this case Q x (x, y) = 2x = P y (x, y), so F is conservative. We can find an f(x, y) by inspection, or more systematically by integration. Thus if F = f, then Integrating with respect to x gives f x (x, y) = P (x, y) = 2xy + 2 x. f(x, y) = x 2 y + 4 x2 + g(y) for some function g(y) depending only on y. Then we use x 2 + g (y) = f y (x, y) = Q(x, y) = x 2 + sin 2 (3y) to find that g (y) = sin 2 (3y). So now we just need to integrate cos(6y) g(y) = sin 2 (3y) dy = dy = y 2 2 sin(6y). 2 Using this in our formula for f(x, y) gives the desired function, f(x, y) = x 2 y + 4 x2 + y 2 sin(6y) 2 Of course, one can always add a constant. Math 35 Final Exam 2/2/4, 9:3 :3am

6 Math 35 Solutions for Final Exam Page 6 (c) Let a = (a, b, c). Then F = a r = det i j k a b c = (bz cy)i (az cx)j + (ay bx)k. x y z As in (a), we need to check if the curl vanishes. For this vector field, we have F = det i j k / x / y / z = 2ai + 2bj + 2ck. bz cy az + cx ay bx So the curl of this vector field F is constant, and indeed is given by F = 2a. Since this is non-zero, F is not conservative. Problem 6. ( points) Let be the unit circle = { (x, y) : x 2 + y 2 = } oriented in a counter-clockwise direction. Let f(t) and g(t) be functions of one variable with continuous derivatives. Evaluate ( ) ( f(x) + g(y) dx + xg (y) + 3x 7 ) dy. Solution. The easiest way to do this problem is to let be the unit disk, so =, and use Green s theorem. Thus ( ) ( f(x) + g(y) dx + xg (y) + 3x 7 ) dy ( ) ( = f(x) + g(y) dx + xg (y) + 3x 7 ) dy = xg x( (y) + 3x 7 ) ( ) f(x) + g(y) dx dy y using Green s theorem, ( = g (y) + 3 ) g (y) dx dy = 3 dx dy = 3Area() = 3π Math 35 Final Exam 2/2/4, 9:3 :3am

7 Math 35 Solutions for Final Exam Page 7 Problem 7. ( points) Let S be a surface in R 3, and let S be the boundary of S. Let F be a vector field on S with continuous partial derivatives. Suppose that you are given the following information about S and F : (i) S lies in the plane y = 3 (ii) Area(S) = 7 (iii) Length( S) = 25 (iv) (v) div(f ) = x 2 + y 2 z curl(f ) = 3xi yj 2zk Using this information, evaluate the absolute value of the line integral F ds. S Solution. Here we will use Stokes theorem. Note that since S lies in the plane y = 3, the unit normal vector n at every point of S is the vector n = j (or j if we want to point the other direction). We compute F ds = curl(f ) ds Stokes theorem, S S = curl(f ) j ds since n = j, S = y ds from the given formula for curl(f ), S = 3 ds since y = 3 for every point of S, S = 3 ds S = 3Area(S) = 5 since we are told that S has area 7. If we used the other normal, we d get 5, but in any case, the absolute value of the integral is 5. Problem 8. ( points) Let f(x, y) = x 4 + y For any a >, let R a be the rectangle alculate R a = [ a, a] [ a, a]. lim f(x, y) dx dy. a a 2 R a Math 35 Final Exam 2/2/4, 9:3 :3am

8 Math 35 Solutions for Final Exam Page 8 Be sure to explain how you got your answer. Solution. When a is very small, the value of the integral R a very close to f(, ) multiplied by the area of R a. So lim f da = lim a a 2 R a a a f(, ) Area(R a) 2 = lim a a 7 4a 2 2 = 4 7. f da is If you want to be more formal, you can quote the mean value theorem for integrals, which says that R a f da = f(x a, y a ) Area(R a ) for some point (x a, y a ) in R a. Hence lim f da = lim a a 2 a a f(x a, y 2 a ) 4a 2 = 4 lim f(x a, y a ) = 4f(, ), a R a where the last equality comes from the fact that f is continuous and the fact that as a, the square R a shrinks down to the point (, ). Problem 9. ( points) Let S R be the sphere of radius R centered at the origin, taken with outward pointing normal. Let F be the vector field F (x, y, z) = x 3 i + z 3 j + y 3 k. Use the ivergence Theorem to compute F ds. S R Solution. Let Ω R be the solid ball of radius R centered at the origin, so S R is its boundary. Then F ds = F ds S R Ω R = div(f ) dv by the ivergence Theoerm, Ω R = 3x 2 dv since div(f ) = 3x 2. Ω R Math 35 Final Exam 2/2/4, 9:3 :3am

9 Math 35 Solutions for Final Exam Page 9 learly the way to compute this integral is using spherical coordinates. So π 2π R 3x 2 dv = 3(ρ cos θ sin ϕ) 2 ρ 2 sin ϕ dρ dθ dϕ Ω R = π 2π R So we have three integrals to do. π 2π R 3ρ 4 (cos 2 θ)(sin 3 ϕ) dρ dθ dϕ. ρ 4 dρ = 5 R5. 2π cos 2 + cos(2θ) θ dθ = dθ = θ sin(2θ) 2π 4 = π. π sin 3 ϕ dϕ = ( cos 2 ϕ) sin ϕ dϕ = cos ϕ π 3 cos3 ϕ = 4 3. This gives the value 3x 2 dv = 3 Ω R 5 R5 π 4 3 = 4πR5 5 Here s a cleverer way to do the integral using an idea that was described in one of the problem sets. By symmetry, we have 3x 2 dv = 3y 2 dv = 3z 2 dv. Ω R Ω R Ω R But adding them gives an integral that s easy to compute using spherical coordinates, π 2π R 3x 2 + 3y 2 + 3z 2 dv = 3ρ 2 ρ 2 sin ϕ dρ dθ dϕ Ω R = π 2π R = 3 5 ρ5 R since x 2 + y 2 + z 2 = ρ 2, θ 3ρ 4 sin ϕ dρ dθ dϕ 2π = 3 R5 5 2π 2 ( cos ϕ) = 2πR5. 5 Math 35 Final Exam 2/2/4, 9:3 :3am π

10 Math 35 Solutions for Final Exam Page Hence Ω R 3x 2 dv = 3 ΩR 3x 2 + 3y 2 + 3z 2 dv = 4πR5 5 Math 35 Final Exam 2/2/4, 9:3 :3am