Practice Problems for Exam 3 (Solutions) 1. Let F(x, y) = xyi+(y 3x)j, and let C be the curve r(t) = ti+(3t t 2 )j for 0 t 2. Compute F dr.

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1 1. Let F(x, y) xyi+(y 3x)j, and let be the curve r(t) ti+(3t t 2 )j for t 2. ompute F dr. Solution. F dr b a 2 2 F(r(t)) r (t) dt t(3t t 2 ), 3t t 2 3t 1, 3 2t dt t 3 dt t Evaluate the line integral x 2. x 3 y ds, where is the portion of the graph y x2 /2 for Solution. Note that this is the line integral of a scalar function with respect to arc length. We can parametrize as r(t) t, 1 2 t2 for t 2. Then so that for f(x, y) x 3 /y: x 3 y ds b a 2 5 r (t) 1, t and r (t) 1 + t 2, f(r(t)) r (t) dt 2 t t2 1 + t2 dt 2t 1 + t 2 (use substitution u 1 + t 2, du 2t dt) 2 u du (t3/2 1). 3 u3/ Let F(x, y) (ax 2 y + y 3 + 1)i + (2x 3 + bxy 2 + 2)j be a vector field, where a and b are constants. (a) Find the values and a and b for which F is conservative. Solution. P ax 2 y + y and Q 2x 3 + bxy so P y ax 2 + 3y 2 and Q x 6x 2 + by 2. Thus, P y Q x if and only if a 6 and b 3. (b) For these values of a and b, find f(x, y) such that F f. Math 257 1

2 Solution. f x 6x 2 y + y f 2x 3 y + xy 3 + x + g(y). Therefore, f y 2x 3 +3xy 2 +g (y). Setting this equal to Q gives 2x 3 +3xy 2 +g (y) 2x 3 +3xy 2 +2 so g (y) 2 and hence g(y) 2y + K where K is a constant. Hence, f(x, y) 2x 3 y + xy 3 + x + 2y + K. (c) Still using the values of a and b from part (a), compute such that x e t cos t, y e t sin t, t π. F dr along the curve Solution. The curve starts at (1, ) and ends at ( e π, ), so by the fundamental theorem for line integrals (page 175), F dr f( e π, ) f(1, ) e π Let F(x, y, z) (cos x + 2y 2 + 5yz)i + (4xy + 5xz)j + (5xy + 3z 2 )k on R 3. Show that F is conservative, and find a potential function for F. Solution. Writing F P i + Qj + Rk, the necessary and sufficient condition for F to be conservative is that curl(f). This means that R y Q z, R x P z, and Q x P y. But R y 5x Q z, R x 5y P z, and Q x 4y + 5z P y. Thus F is conservative. To find a potential function f(x, y, z) such that f F, first set f x P and integrate with respect to x to get f(x, y, z) sin x + 2xy 2 + 5xyz + g(y, z). Now differentiate with respect to y and set the result equal to Q to get f y 4xy + 5xz + g y 4xy + 5xz. Thus g y so g(y, z) h(z) and f(x, y, z) sin x + 2xy 2 + 5xyz + h(z). Now differentiate the result with respect to z and set the result equal to R to get f z 5xy + h (z) 5xy + 3z 2. Thus, h (z) 3z 2 so integrating gives h(z) z 3 + K where K is a constant. Thus, f(x, y, z) sin x + 2xy 2 + 5xyz + z 3 + K. 5. Verify that the vector field F(x, y) (y 2 y cos x)i + (2xy sin x + 1)j is conservative. Then compute F dr, where is the curve x t3 t, y 1 + t 2, 1 t 2. Math 257 2

3 Solution. To verify that F is conservative, compute P y y (y2 y cos x) 2y cos x, Q x (2xy sin x + 1) 2y cos x. x Since the two partial derivatives are equal, F is conservative. We can thus evaluate the line integral by the fundamental theorem for line integrals. We must first find a potential function f with f F. Thus, we must have f x y 2 y cos x and f y 2xy sin x + 1. Integrating f x with respect to x gives f(x, y) xy 2 y sin x + g(y). Differentiate this with respect to y to get f y 2xy sin x + g (y). Setting this equal to the expression for f y found earlier gives g (y) 1 so g(y) y + K. Thus, our potential function is f(x, y) xy 2 y sin x + y + K. The fundamental theorem for line integrals then gives F dr f dr f(r(2)) f(r(1)) f(6, 5) f(, 2) (6(25) 5 sin(6) K) ( K) sin(6). 6. Let be the circle x 2 + y 2 1, oriented counterclockwise. Find (2xy + e x ) dx + (x 2 sin y + 3x) dy. Solution. Since is a simple closed curve, Green s Theorem applies. First compute Q x P y x (x2 sin y + 3x) y (2xy + ex ) (2x + 3) (2x) 3. We can then calculate, using Green s Theorem (note that the inside of is D {x 2 + y 2 1}, the disc of radius 1 centered at the origin): ( Q P dx + Q dy x P ) da 3 da y D 3(Area(D)) 3 π(1) 2 3π. 7. Let be the ellipse x2 a + y2 1. Evaluate the line integral 2 b2 x a i + y 2 b j. 2 D F dr where F(x, y) Math 257 3

4 Solution. Parametrize by x a cos t, y b sin t, t 2π. Then 2π ( ) a cos t b sin t F dr i + j ( a sin ti + b cos tj) dt a 2 b 2 2π ( ) a 2 cos t sin t + b2 sin t cos t dt a 2 b 2 2π 3 dt. 8. Let be the positively oriented closed curve formed by the parabola y x 2 running from (, 1) to (1, 1) and by a straight line running back from (1, 1) to (, 1). Evaluate the line integral y 2 dx + 4xy dy in two ways: (a) directly, and Solution. Let 1 + 2, where 1 is the curve running along the parabola and 2 is the horizontal segment. Parametrize 1 by x(t) t, y(t) t 2, t 1. Then dx dt and dy 2t dt, so the line integral over 1 is 1 y 2 dx + 4xy dy 1 ((t 2 ) 2 + 4tt 2 (2t)) dt 1 9t 4 dt 9 5 t Parametrize 2 by x(t) t, y(t) 1, t 1. Then dx dt and dy, so the line integral over 2 is 2 y 2 dx + 4xy dy 1 dt t 1 2. Putting all this together we get y 2 dx + 4xy dy y 2 dx + 4xy dy + y 2 dx + 4xy dy (b) by using Green s theorem. Solution. y 2 dx + 4xy dy 2y da 1 1 R x 2 2y dy dx. Math 257 4

5 The inner integral is Thus, the outer integral is 1 x 2 2y dy y 2 1 x 2 1 x 4. 1 (1 x 4 ) dx ) (x x Let F(x, y, z) yi+zj 2xk and let be a simple closed curve in the plane x+y+z 1. Show that F dr. Solution. Apply Stokes Theorem. Let S be the part of the plane x + y + z 1 inside. Then Stokes theorem states that F dr (curl F) n ds, S where n is the unit normal vector to the surface S. Since S is the plane x + y + z 1, the normal vector is n i + j + k. The curl of F is 3 i j k curl F x y z i + 2j k. y z 2x Thus, and hence (curl F) n ( i + 2j k) i + j + k 3 F dr (curl F) n ds. S 1. Let Σ be the unit sphere centered at the origin, with the outward normal orientation, and let F(x, y, z) x 3 i + y 3 j + z 3 k. ompute the integral: F n ds. Hint: Use the divergence theorem. Σ Solution. Let B be the unit ball, so that Σ is the boundary of B. The divergence of F is div F 3x 2 + 3y 2 + 3z 2. Then the divergence theorem gives F n ds div F dv 3(x 2 + y 2 + z 2 ) dv. Σ B Math B

6 This last integral can be evaluated in spherical coordinates ρ, φ, θ as 2π π 1 3ρ 2 ρ 2 sin φ dρ dφ dθ 12 5 π. 11. Let f be a scalar field (i.e., function) and F a vector field. Determine if each of the following expressions is meaningful. If it is meaningful, indicate whether it is a scalar field or a vector field. Expression Meaningful Scalar Vector f Yes x div F Yes x curl f No curl( f) Yes x F No div( F) Yes x Math 257 6