In general, the formula is S f ds = D f(φ(u, v)) Φ u Φ v da. To compute surface area, we choose f = 1. We compute

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1 alculus III Test 3 ample Problem Answers/olutions 1. Express the area of the surface Φ(u, v) u cosv, u sinv, 2v, with domain u 1, v 2π, as a double integral in u and v. o not evaluate the integral. In general, the formula is f d f(φ(u, v)) Φ u Φ v da. To compute surface area, we choose f 1. We compute Φ u Φ v cosv, sinv, u sin v, u cosv, 2 2 sinv, 2 cosv, u cos 2 v + u sin 2 v 2 sinv, 2 cosv, u 4 + u 2. Thus we have Area 2. Use Green s theorem to compute 1 2π 4 + u2 dv du. F dr, where F(x, y) xy 2, x 2 y, and consists of the part of the parabola y x 2 between ( 1, 1) and (1, 1), along with the line joining (1, 1) to ( 1, 1). Green s theorem says that F dr ( Q x P ) da, where is the region bounded by. Here we have Q x P 2xy 2xy 4xy. Also, the region is y-simple, and can be described by x 2 y 1 with 1 x 1. Thus we have ) F dr 4 xy dy dx 4 (x x5 dx. 1 x Use Green s theorem to compute the integral is the unit circle. 1 F dr, where F(x, y) xy 2, x and Greens theorem says this is equal to the double integral ( Q x P ) da, where is the unit circle. Using polar coordinates, this gives us ( Q F dr x P ) da (1 2xy) da 2π 1 (1 2r cos θr sin θ)r dr dθ 2π 1 (r 2r 3 sin θ cos θ) dr dθ π.

2 4. Use Green s theorem and a vector field of your choice to show that the area of the ellipse x 2 a + y2 1 is πab. 2 b2 ee Example 3 on page 19 in the textbook for a solution. 5. Use tokes theorem to compute is the part of the plane y + z 2 inside the cylinder x 2 + y 2 1. tokes theorem says that curlfd curlf d, where F(x, y, z) 1, xy 2, xy 2 and F dr b a F(r(t)) r (t) dt. (The second equality is just the formula for integrating vector fields along curves.) The boundary of is a curve running around the cylinder x 2 + y 2 1 and lying on the plane y + z 2. Together, these facts imply that we may parametrize the curve via r(t) cos t, sin t, 2 sin t. Then we have 2π F dr 1, costsin 2 t, costsin 2 t sin t, cost, cost dt 2π sin t dt. 6. Use the parametrization Φ(φ, θ) a cosθ sin φ, a sin θ sin φ, a cosφ to show that the area of the sphere of radius a is 4πa 2. We need to integrate d Φ φ Φ θ. To this end, we have d Φ φ Φ θ a cosθ cosφ, a sin θ cosφ, a sin φ a sin θ sin φ, a cosθ sin φ, Thus we have a 2 cos θ sin 2 φ, a 2 sin θ sin 2 φ, a 2 sin φ cosφ a 2 sin φ. Area() 2π π a 2 sin φ dφ dθ 4πa Verify tokes theorem for the vector field F(x, y, z) xz, xy, 1 2 y2 and is the part of the paraboloid z 1 x 2 y 2 above the xy-plane.

3 We first compute F dr, where is the unit circle in the xy-plane (which is the boundary of ). For this we have r(t) cos t, sin t,, which gives F dr 2π, sin t cost, 1 2 sin2 t sin t, cost, dt 2π cos 2 t sin t dt. For the other side of tokes theorem, we parametrize by Φ(u, v) u, v, 1 u 2 v 2, where u 2 + v 1 1. This gives Φ u Φ v 1,, 2u, 1, 2v 2u, 2v, 1. We also compute that curl F y, x, y. Thus we have curlf d v, u, v 2u, 2v, 1 da (4uv + v) da. Using polar coordinates, this is equal to 2π 1 8. Use tokes theorem to evaluate (4r 3 sin θ cos θ + r 2 sin θ) dr dθ. F dr, where F(x, y, z) y 2, x, z 2 and is the part of the plane x + y + z 1 in the positive octant x, y, z. tokes theorem says that instead of computing this line integral (which goes around the three sides of a triangle), we can instead compute curlf d. For this, we parametrize with Φ(u, v) u, v, 1 u v, and note then that Φ u Φ v 1, 1, 1 and curlf,, 1 2y. Thus we have curlf d,, 1 2v 1, 1, 1 da (1 2v) da, where is the triangle in the uv-plane bounded by the positive axes and the line v 1 u. This integral therefore equals 1 1 u (1 2v) dv du 1 (u u 2 ) du Find the upward flux of the vector field F(x, y, z) 2x,, 1 across the disk x 2 +y 2 4, z.

4 Let s act as if this surface does not happen to be in the xy-plane, and let s actually parametrize the thing as a portion of the graph of z. Thus we have Φ(u, v) u, v,. From this we find that Φ u Φ v,, 1. Also F(Φ(u, v)) 2u,, 1. Thus we have F d 2u,, 1,, 1 da da area(). The domain of the parametrization is the circle of radius 2, which has area 4π. 1. Find the upward flux of the vector field F(x, y, z) y, x, z across the surface Φ(u, v) u cosv, u sinv, 1 u 2, where the domain in the uv-plane is the region with u 1 and v 2π. We compute and F(Φ(u, v)) u sinv, u cosv, 1 u 2, Φ u Φ v cosv, sin v, 2u u sinv, u cosv, 2u 2 cos v, 2u 2 sin v, u. We want the upward flux, so the z-coordinate of our normal vector needs to be positive. The z-coordinate of Φ u Φ v is u, which for us is between and 1, so this is the correct normal direction. o the integral is Fd F(Φ(u, v, )) (Φ u Φ v ) da 1 2π 1 2π u sin v, u cosv, 1 u 2 2u 2 cosv, 2u 2 sin v, u dv du (4u 3 sin v cosv + u u 3 ) dv du 1 (2πu 2πu 3 ) du π Express the following integral as an iterated integral in polar coordinates r and θ: xy 2 da, where is the portion of the plane described by x 2 + y 2 9, x. This portion of the plane has r 3 and π/2 θ π/2. Thus we have π/2 3 xy 2 da r 4 sin 2 θ cosθ dr dθ. π/2 12. arefully sketch the region of integration for the integral the order of integration and evaluate the integral y e x2 dxdy. Then reverse

5 The region lies between the lines x 3y and x 3, at points where the y-coordinates lie between and 1. This is the triangle shown below. Reversing the order of integration means treating this as a y-simple region, for which we get the bounds y x/3, x 3. Thus the new integral is 3 x/3 e x2 dy dx [ye x2] yx/3 y dx e u du 1 x3 1 6 ex2 6 (e9 1). x [ xe x2] dx 13. A thin metal sheet occupies the region in the xy-plane bounded by the parabolas x 2y 2 and x 1 + y 2 and lying above the x-axis. The density of the sheet at the point (x, y) is given by f(x, y) y + 2. Find the mass of the sheet. This region is x-simple, where 2y 2 x 1 + y 2 and y 1. Thus we have 1 1+y 2 1 (y + 2) da (y + 2) dxdy (2 + y 2y 2 y 3 ) dy 19 2y 12 2

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