Math 211, Sections 01,05 Solutions to Practice Problems for the Final Exam

Math 211, Sections 01,05 Fall 2018 Solutions to Practice Problems for the Final Exam Z Z 1. Let F~ (x, y, z) = hy, z 2 , xi. Compute F~ · d~r and F~ · d~r, where C1 is the helix x = 4 cos t, C1 C2 y = 4 sin t, z = t from (4, 0, 0) to (4, 0, 2π), and C2 is the line segment also running from (4, 0, 0) to (4, 0, 2π). Answer. For the first integral, ~r(t) = h4 cos t, 4 sin t, ti for 0 ≤ t ≤ 2π, and so F~ (~r(t)) = h4 sin t, t2 , 4 cos t and ~r 0 (t) = h−4 sin t, 4 cos t, 1i. Thus, Z Z 2π Z Z 2π −8 + 8 cos(2t) + 4 cos t dt + −16 sin2 t + 4t2 cos t + 4 cos t dt = F~ · d~r = C1 0 0 2π 4t2 cos t dt. 0 Integrating by parts (with u = 4t2 , dv = cos t dt, so du = 8t dt, v = sin t) on the second piece gives Z 2π  2π  2 2π 8t sin t dt. = − 8t + 4 sin(2t) + 4 sin t 0 + 4t sin t 0 − 0 A second integration by parts (u = 8t, dv = sin t dt, du = 8 dt, v = − cos t) then gives us Z 2π  2π  2π = −16π + 0 + 8t cos t 0 − 8 cos t dt = −16π + 16π − 0 − 8 sin t 0 = 0. 0 For the second integral, ~r(t) = h4, 0, ti for 0 ≤ t ≤ 2π. So F~ (~r(t)) = h0, t2 , 4i and ~r 0 (t) = h0, 0, 1i. Z Z 2π ~ Thus, F · d~r = 4 dt = 8π. C2 0 2. Evaluate the following line integrals. Z (a). z dx − y dy − x dz, where C is parametrized by ~r(t) = 5~i − sin t~j − cos t~k, for 0 ≤ t ≤ π/4. C Z √ (b). (y + 2xey ) dx + (x + x2 ey ) dy, where C is parametrized by ~r(t) = t~i + ln t~j for 1 ≤ t ≤ 4. ZC (c). y cos x dx + x sin y dy, where C is the boundary of the triangle with vertices (0, 0), (π/2, 0), C and (0, π/2), oriented clockwise. Answer. (a). [Note: the curl of F~ = hz, −y, −xi is h0, 2, 0i 6= ~0, which means F~ is not conservative, so we can’t use the Fundamental Theorem of Line Integrals.] We have ~r 0 (t) = h0, − cos t, sin ti. So the integral is Z π/4 Z π/4 h− cos t, sin t, −5i · h0, − cos t, sin ti dt = − sin t cos t − 5 sin t dt 0 0 √ √ π/4 1 1 1 5 2 10 2 − 21 2 = − sin t + 5 cos t =− · + − (−0 + 5) = . 2 2 2 2 4 0 ∂Q ∂P (b). The 2D curl of F~ = hy + 2xey , x + x2 ey i is − = 1 + 2xey − 1 − 2xey = 0, which means ∂x ∂y F~ is conservative, since the domain of F~ is R2 and hence is simply-connected. To write F~ = ∇f , we would have fx = y + 2xey , and therefore f = xy + x2 ey + g(y), which gives fy = x + x2 ey + g 0 (y). Thus, we can choose g = 0, and we have F~ = ∇f , where f (x, y) = xy + x2 ey . The starting point of C, when t = 1, is h1, 0i, and the ending point, when t = 4, is h2, ln 4i. Thus, by the Fundamental Theorem of Line Integrals, the integral is Z (y + 2xey ) dx + (x + x2 ey ) dy = f (2, ln 4) − f (1, 0) = 2 ln 4 + 4 · 4 − (0 + 1) = 15 + 2 ln 4. C (c). The curve is in the plane and closed, so we can use Green’s Theorem. We compute ∂Q ∂P − = sin y − cos x. Note that the curve is oriented negatively, and the hypoteneuse of the ∂x ∂y triangle is the line y = π/2 − x. Thus, if D denotes the (filled-in) triangle, the integral is Z π/2 Z π/2−x ZZ Z cos x − sin y dy dx sin y − cos x dA = y cos x dx + x sin y dy = − 0 0 D C Z Z π/2 π/2 π/2−x π π ( − x) cos x + cos( − x) − 1 dx dx = y cos x + cos y = 2 2 y=0 0 0 R Integrating by parts (with u = x, dv = cos x dx), x cos x dx = x sin x + cos x + C, so our integral is π/2 π π π π π π sin x − x sin x − cos x − sin( − x) − x = − − 0 − 0 − − (0 + 0 − 1 − 1 − 0) = 2 − . 2 2 2 2 2 2 0 R [Note: This integral is actually not that bad to evaluate directly as C F~ · d~r. The contributions from the two vertical legs of C easily work out to zero, and the contribution from the hypoteneuse quickly turns into an integral very similar to the integral dx above.] 3. Find the work done by the force F~ (x, y, z) = hx2 y, z, 2x − yi moving an object along the straight line segment from (1, 1, 1) to (2, −3, 3). Answer. We parametrize the line segment by ~r(t) = (1 − t)h1, 1, 1i + th2, −3, 3i = h1 + t, 1 − 4t, 1 + 2ti, for 0 ≤ t ≤ 1. Thus, ~r0 (t) = h1, −4, 2i and F~ (~r(t)) = h(1 + t)2 (1 − 4t), 1 + 2t, 1 + 6ti. Z 1 Z 1 2 Hence, the work is (1 + 2t + t )(1 − 4t) − 4(1 + 2t) + 2(1 + 6t) dt = −1 + 2t − 7t2 − 4t3 dt 0 0 1 7 7 10 = −t + t − t3 − t4 = −1 + 1 − − 1 − (0 + 0 − 0 − 0) = − . 3 3 3 0 [The negative answer means the object moved against the force. This is like when an object moves up; the work done by gravity is negative.] 2 ~ be the vector fields 4. Let F~ and G F~ (x, y, z) = (x sin y + z)~i + xy~j + xz~k, ~ G(x, y) = (ex + cos(xy) − xy sin(xy))~i + (2y − x2 sin(xy))~j. (a). Compute curl F~ and div F~ . (b). Is F~ conservative (on R3 )? If so, express it as a gradient; if not, explain why not. ~ conservative (on R2 )? If so, express it as a gradient; if not, explain why not. (c). Is G ~i ~j ~k ~ Answers. (a). curl F = ∂x ∂y ∂z = h0, 1 − z, y − x cos yi, x sin y + z xy xz ~ and div F = sin y + x + x = 2x + sin y. (b). No, F~ cannot be conservative, since curl F~ 6= ~0. ~ is (c). The 2D curl of G ∂Q ∂P − = [−2x sin(xy) − x2 y cos(xy)] − [−x sin(xy) − x sin(xy) − x2 y cos(xy)] = 0, ∂x ∂y ~ must be conservative. Any potential function f must satisfy so (since R2 is simply-connected), G 2 fy = 2y − x sin(xy), and hence f = y 2 + x cos(xy) + g(x). Then fx = cos(xy) − xy sin(xy) + g 0 (x) ~ = ∇f , where f (x, y) = must equal P , which means g 0 (x) = ex . Antidifferentiating, it follows that G ex + y 2 + x cos(xy). 2 2 5. Let C be Z the circle x + y = 4, oriented counterclockwise. (xy 2 − y 3 ) dx + (x3 + x2 y) dy. Compute C Answer. Letting D denote the disk of radius 2 centered at the origin, and noting that ∂Q ∂P − = 3x2 + 2xy − 2xy + 3y 2 = 3(x2 + y 2 ), Green’s Theorem gives ∂x ∂y Z 2π Z 2 ZZ Z 3 2 3r3 dr dθ = 2π · r4 = 24π. 3(x2 + y 2 ) dA = (xy 2 − y 3 ) dx + (x3 + x2 y) dy = 4 0 0 0 D C 6. Determine whether each of the following vector fields is conservative. If it is, find a potential function for it. ~ (a). F~ (x, y) = hx2 + 2x sin y, 5 + x2 cos yi (b). G(x, y, z) = hy + z, 2z, x + yi z 2~ ~ ~ ~ (c). H(x, y, z) = 3x i + j + ln y k y ∂ ∂ Answer. (a) The 2Dcurl of F~ is (5 + x2 cos y) − (x2 + 2x sin y) = 2x cos y − 2x cos y = 0. [Since ∂x ∂y the domain is all of R2 , which is simply connected, F~ must be conservative. But no need to say all that, because I’m about to find an explicit potential function.] To find a potential function f (x, y), note that we would have to have fx = x2 + 2x sin y, so f = x3 + x2 sin y + g(y). Thus, fy = x2 cos y + g 0 (y); but on the other hand, we need fy = 5 + x2 cos y. 3 x3 Thus, g 0 (y) = 5; we can choose g(y) = 5y. That is, f (x, y) = + 5y + x2 sin y. 3 ~i ~j ~k ~ = ∂x (b). curl G = h1 − 2, blah, blahi = 6 ~0. [I didn’t bother to compute the rest once ∂y ∂z y + z 2z x + y ~ is not conservative. it was clear that the curl was nonzero.] Thus, G ~k ~i ~j ~ is {(x, y, z) ∈ = h1/y − 1/y, 0 − 0, 0 − 0i = ~0. [Since the domain of H ∂x ∂y ∂z 2 3x z/y ln y 3 ~ is conservative; but as in part (a), I don’t need to R : y > 0}, which is simply connected, we know H say that, since I’m about to compute an explicit potential function.] ~ we must have fx = 3x2 , so f = x3 + g(x, y). Thus, fy = gy , but To find f (x, y, z) such that ∇f = H, fy is supposed to be z/y. Hence, f = x3 + z ln y + h(z). [Note: the reason it’s ln y instead of ln |y| is because we already know the domain is only the region where y > 0.] Last, fz = ln y + h0 (z), but on the other hand, we need fz = ln y. Therefore, h(z) is constant, and we may as well choose that ~ = ∇f , where f (x, y, z) = x3 + z ln y. constant to be zero. Thus, H ~ = (c). curl H xy 2 x2 y E , . (1 + x2 )2 1 + x2 ~ (a). Show that F Z is conservative, and find an associated potential function. √ (b). Compute F~ · d~r, where C is the curve parametrized by ~r(t) = hcos(πt3 ), 1 + t3 i, for 7. Let F~ (x, y) = 0 ≤ t ≤ 2. D C Answer. (a). [I could check that the 2Dcurl of F~ is zero, but since we’re going to have to find a potential function anyhow, I’m going to go directly to that.] We seek a function f (x, y) such that ∇f = F~ . Since I don’t relish antidifferentiating P with respect to x, note that we must have x2 y 2 fy = x2 y/(1 + x2 ), and therefore f = + g(x). 2(1 + x2 ) xy 2 2xy 2 (1 + x2 ) − 2x3 y 2 0 + g (x) = + g 0 (x). However, we are supposed to have Thus, fx = 2(1 + x2 )2 (1 + x2 )2 fx = xy 2 /(1 + x2 )2 , and therefore we must have g 0 (x) = 0. Thus, g is constant, and we may as well x2 y 2 choose that constant to be zero. Hence, F~ = ∇f , where f (x, y) = . 2(1 + x2 ) (b). Z C runs from ~r(0) = h1, 1i to ~r(2) = h1, 3i. Thus, by the Fundamental Theorem of Line Integrals, 9 1 F~ · d~r = f (1, 3) − f (1, 1) = − = 2. 4 4 C 8. Compute the divergence and the curl of the following vector fields. Decide whether each is the gradient of something, or the curl of something, or both, or neither. (a). F~ (x, y, z) = hx + y, y + z, x + zi. ~ (b). G(x, y, z) = x2~i − xey~j + 2xyz~k. ~ (c). H(x, y, z) = hx − yz, y − xz, xy − 2zi. Answer. [Skipping writing out the 3 × 3 determinants for curl, since I’m assuming you’re mostly using this to check your answers:] (a). div F~ = 1 + 1 + 1 = 3, and curl F~ = h−1, −1, −1i. Thus, F~ is neither grad nor curl of anything. ~ = 2x − xey + 2xy, and curl G ~ = h2xz, −2yz, −ey i. Again, G ~ is neither grad nor curl of (b). div G anything. ~ = 1 + 1 − 2 = 0, and curl H ~ = h2x, −2y, 0i. Thus, although H ~ is not grad of anything (since (c). div H ~ ~ ~ ~ is all of R3 , which has no ~ curl H 6= 0), H is curl of something, since div H = 0 and the domain of H (avocado-like) holes. [But we weren’t asked what it is curl of, so I won’t bother trying to figure that out. We didn’t talk about how to find such a thing, anyhow.] Z p 9. Compute 1 + x3 dx + 2xy dy, where C is the triangle with vertices (0, 0), (1, 0), and (1, 3), C oriented counterclockwise. Answer. Integrating that nasty function (especially around a three-piece contour) is not an inviting √ prospect. So let’s use Green’s Theorem. With F~ = hP, Qi = h 1 + x3 , 2xyi, we have Qx − Py = 2y − 0 = 2y. Based on a sketch of the triangle in the plane (see me if you need to see that), Green’s Z p Z 1 Z 3x Z 1 Z 1 3x 1 2 3 Theorem says 1 + x dx + 2xy dy = 2y dy dx = y dx = 9x2 dx = 3x3 = 3. C 0 0 0 y=0 0 10. Evaluate Z Z the following surface integrals. ~ where S is the part of the paraboloid z = x2 + y 2 above the (a). hx, −1, 2x2 i · dS, S 2 2 region upward. Z Z in the xy-plane bounded by x = 1 − y and x = y − 1, oriented p ~ where S is the portion of the cone z = x2 + y 2 beneath (b). hx, 2x + z, y − zi · dS, S Z the Z plane z = 2, oriented downward. ~ where S is the tetrahedron with vertices (c). hx2 , xy, −2xzi · dS, S (0, 0, 0), (1, 0, 0), (0, 1, 0), and (0, 0, 1), oriented outward. 0 Answers. (a). Parametrize S by hx, y, x2 + y 2 i for (x, y) ∈ D, where D is the region in the xy-plane bounded by x = 1 − y 2 and x = y 2 − 1. [Note: in spite of the formula for z, polar coordinates is a bad idea, given the shape of D.] We have ~rx = h1, 0, 2xi and ~ry = h0, 1, 2yi, and so ~rx × ~ry = h−2x, −2y, 1i, which is indeed pointing upward [look at the z-coordinate]. Meanwhile, the two curves bounding D intersect when y 2 − 1 = 1 − y 2 , i.e., y 2 = 1, i.e., y = ±1; and the curve x = y 2 − 1 is to the left (i.e., smaller x). So Z 1 Z 1−y2 ZZ 2 ~ hx, −1, 2x2 i · h−2x, −2y, 1i dx dy hx, −1, 2x i · dS = −1 y 2 −1 S Z 1 Z 1 Z 1 Z 1−y2 2 2 2 2 4y − 4y 3 dy 2y[(1 − y ) − (y − 1)] dy = −2x + 2y + 2x dx dy = = −1 2 y 2 −1 1 4 = 2y − y −1 −1 −1 = (2 − 1) − (2 − 1) = 0. (b). In cylindrical coordinates, S is given by z = r, for 0 ≤ r = z ≤ 2. So parametrize S by ~r(r, θ) = hr cos θ, r sin θ, ri, for 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π. Then ~rr = hcos θ, sin θ, 1i, and ~rθ = h−r sin θ, r cos θ, 0i. Thus, ~rr × ~rθ = h−r cos θ, −r sin θ, ri, which points upwards, i.e., the wrong way. We should instead use its negative, hr cos θ, r sin θ, −ri. Thus, Z 2π Z 2 ZZ ~= hx, 2x + z, y − zi · dS hr cos θ, 2r cos θ + r, r sin θ − ri · hr cos θ, r sin θ, −ri dr dθ. 0 0 S Z 2π Z 2 = r2 cos2 θ + 2r2 cos θ sin θ + r2 sin θ − r2 sin θ + r2 dr dθ Z0 2 0 Z 2π 1 (1 + cos 2θ) + 2 sin θ cos θ + 1 dθ = r2 dr 2 0 0 h r3 2 ih 3 8 2π i 1 = θ + sin 2θ + sin2 θ = (3π − 0) = 8π. 3 0 2 4 3 0 (c). Since S encloses a solid region, we may use the Divergence Theorem. We have divhx2 , xy, −2xzi = 2x + x − 2x = x. Thus, writing E for the solid tetrahedron, and noting that the slanted plane of E is givenZ Zby ZZ Z x + y + zZ= 1, Z Z 1 1−x 1−x−y ~= hx2 , xy, −2xzi · dS x dV = x dz dy dx S Z Z E 0 0 0 Z Z 1 1 1−x 1 1 2 1−x 1 2 2 = x − x − xy dy dx = (x − x )y − xy dx = x(1 − x)2 − x(1 − x)2 dx 2 2 y=0 0Z 0 0 0 1 1 1 3 1 1 1 1 1 1 1 = x − 2x2 + x dx = x4 − x3 + x2 = − + = . 2 0 8 3 4 0 8 3 4 24 3 11. Let F~ (x, y, z) = hx2 + y 2 , cos(xz), yz + sec(x RR )i. Let S be the surface of the rectangular box ~ [0, 1] × [0, 2] × [1, 3], oriented outward. Compute S F~ · dS. Answer. We Zhave div F~ = 2x + y, and so Zby Z the Divergence Theorem, ZZ 1Z 2Z 3 1 2 ~= F~ · dS 2x + y dz dy dx = 4x + 2y dy dx S Z 0 0 1 0 0 Z 1 1 2 1 = 4xy + y 2 dx = 8x + 4 dx = 4x2 + 4x = 8. 0 y=0 0 0 ~ 12. Let G(x, y, z) = hx, y, z 2Zi.ZLet S be the portion of the sphere x2 + y 2 + z 2 = 4 in the first octant, ~ · dS. ~ oriented upward. Compute G S Answer. [Note: We can’t use the Divergence Theorem because the surface isn’t closed.] In spherical coordinates, S would be given by ρ = 2, for 0 ≤ θ ≤ π/2 and 0 ≤ φ ≤ π/2. Thus, we can parametrize S by ~r(θ, φ) = h2 sin φ cos θ, 2 sin φ sin θ, 2 cos φi Then ~rθ = h−2 sin φ sin θ, 2 sin φ cos θ, 0i, and ~rφ = h2 cos φ cos θ, 2 cos φ sin θ, −2 sin φi. We compute ~rθ × ~rφ = h−4 sin2 φ cos θ, −4 sin2 φ sin θ, −4 sin φ cos φi = −4 sin φhsin φ cos θ, sin φ sin θ, cos φi, which points downward. (Note that the z-component is negative for θ, φ ∈ [0, π/2].) Since that is the wrong way, we can make it right by thinking of φ as the first variable, and θ as the second, which ZZ ~ reverses the sign on ~rφ × ~rθ . Thus, hx, y, z 2 i · dS Z π/2 Z π/2 S
h2 sin φ cos θ, 2 sin φ sin θ, 4 cos2 φi · 4 sin φhsin φ cos θ, sin φ sin θ, cos φi dφ dθ Z0 π/2 Z0 π/2  Z π/2  Z π/2

 2 3 = (4 sin φ) 2 sin φ + 4 cos φ dφ dθ = dθ 8 sin φ 1 − cos2 φ + 2 cos3 φ dφ 0 0 0 0 Z 0   3 u u4  1 1 1  14π [u = cos φ, du = − sin φ dφ] = −4π 1−u2 +2u3 du = 4π u− + = . = 4π 1− + 3 2 0 3 2 3 1 =