Math 23b Practice Final Summer 2011

Transcription

1 Math 2b Practice Final Summer (1 points) Sketch or describe the region of integration for 1 x y and interchange the order to dy dx dz. f(x, y, z) dz dy dx Solution. 1 1 x z z f(x, y, z) dy dx dz 2. (1 points) Are the following vector fields conservative? If they are, find their potential functions. (a) F(x, y) xy, xy (b) F(x, y) 2x i + 2y(x2 +1) j y 2 +1 (y 2 +1) 2 (c) F(x, y, z) (2xyz + sin x)i + x 2 zj + x 2 yk Solution. Vector field (a) is not conservative, since: N x y x M y Vector field (b) may seem to be conservative at first glance, but the partial derivatives above actually differ by a sign. Vector field (c) is conservative. You can check that curl F, or you can try to find the potential function: 2xyz + sin x dx x 2 yz cos x + 1 (y, z) x 2 z dy x 2 yz + 2 (x, z) x 2 y dz x 2 yz + (x, y)

2 Therefore we may take as our potential function: f(x, y, z) x 2 yz cos x. (1 points) Evaluate the integral cos(x + 2y) sin(2x y) da R where R is the triangle in the x, y-plane bounded by the lines x + 2y 4, y 2x, and the x-axis. Solution. We use the substitution u x+2y, v 2x y. Note the change of variables formula: (x, y) dx dy du dv (u, v) This means that to find the proper Jacobian, we must first invert our transformation, and solve for x and y. This leads to the equations x 1(u + 2v) and y 1 (2u v). 5 5 Therefore our Jacobian is: ( 1 ) 2 J det Next, we deal with the new bounds for our integral. Our old bounds consisted of the lines x + 2y 4, y 2x, and y. The first two convert easily into u 4 and v. For y, we use our (inverted) substitution that we just found: y 1 (2u v) 5 2u v v 2u If we graph the lines u 4, v, and v 2u in the u, v-plane, we find a triangle with vertices at (, ), (4, ), and (4, 8). We put all these substitutions together, and evaluate: 4 2u ( ) 1 cos(u) sin(v) dv du 1 4 cos u( cos v) u du cos u(cos 2u 1) du cos u cos 2u cos u du Page 2

3 From here you can use the substitution cos 2u cos 2 u sin 2 u to solve the integral. 4. (25 points) Evaluate the following line integrals. You may use any theorems or definitions that you like to help you. (a) If is the line from (, ) to (, 9): (x + y) d Solution. We parameterize as γ(t) t, 9t with t [, 1]. Then γ (t), 9, γ (t) 9 1, and our integral becomes: 1 (t + 9t)( 1) dt t + t dt 9 [ t2 + 2 t/2 9 [ ] ] 1 (b) If is the unit circle: x dy y dx x 2 + y 2 Solution. Parameterize the unit circle as γ(t) cos t, sin t, t [, 2π]. Then γ (t) sin t, cos t. We substitute into our integral: x dy y dx 2π (cos t)(cos t) (sin t)( sin t) x 2 + y 2 cos 2 t + sin 2 t 2π 1 1 2π Page

4 (c) If is parameterized by c(t) sin t, cos(2t), cos t, t [, π]: 2 (2y 2 x cos(x 2 )e z2 ) dx + (2y sin(x 2 )e z2 ) dy + (2y 2 z sin(x 2 )e z2 + 1) dz Solution. The vector field in question is conservative. We find that a potential function is: f(x, y, z) y 2 sin(x 2 )e z2 By the Fundamental Theorem, we evaluate f at the two endpoints of, which are (, 1, 1) and (1, 1, ). F d f(1, 1, ) f(, 1, 1) sin(1) () sin(1) (d) If is the triangle with vertices (, ), (, 1), and (1, 2): xy dx + x 2 dy Solution. Here we can use Green s Theorem. We find that: N x M y 2x x x Next we turn to the bounds. The three lines that bound our region are x, y 2x, and y x + 1. Graphing these, it is clear that we should integrate our double integral with respect to y first. Our integral becomes: 1 x+1 2x x dy dx 1 1 x((x + 1) (2x)) dx x x 2 dx 1 2 x2 1 x Page 4

5 (e) If is parameterized by c(t) t, t, e t, t [, 1] (e x sin y)i + (e x cos y)j + z 2 k d Solution. Again, we find that this vector field is conservative, with potential function: f(x, y, z) e x sin y + 1 z Therefore, we can evaluate f at the endpoints of, which are (,, 1) and (1, 1, e): F d f(1, 1, e) f(,, 1) e sin(1) 1 e + 1 Page 5

6 5. (15 points) Find the area of the surface defined by z xy and x 2 + y 2 2. Solution. We may parameterize this surface as Φ(u, v) u, v, uv, but because of the bounds for u and v we will have switch to polar coordinates at some point. For now, we calculate: Φ u 1,, v Φ v, 1, u Φ u Φ v v, u, 1 Φ u Φ v v 2 + u Thus our surface area is given by: Φ u Φ v du dv D D v2 + u du dv Now, to set bounds for this integral, we switch to polar coordinates: D v2 + u du dv 2π This last integral is easy with a u-substitution. 2 ( r 2 + 1)(r) dr dθ Page 6

7 6. (15 points) onsider the region W bounded above by the sphere x 2 + y 2 + z 2 25, and below by the cone z 2 x 2 + y 2. (a) Find the volume of W. Solution. We may choose to do this in either spherical or cylindrical coordinates; it is not yet clear which will be easier. If we want to use spherical, we will need to identify the φ-angle that the cone makes with the z-axis. We take a cross section of the cone by setting x (or y) equal to zero, and find that we are left with the line z y, which has slope. Therefore the angle made with the z-axis is φ arctan π/6. Since this angle is simple, we continue on with spherical coordinates. The integral for volume is: W dv 2π 2π 2π π 6 π 6 π 6 2π 5 ρ 2 sin φ dρ dφ dθ 5 1 ρ sin φ dφ dθ 125 sin φ dφ dθ 125 π 6 cos φ dθ 125 2π ( ) 2 1 dθ ( ) ( ) (2π) 2 25π 125π (b) Integrate the vector field F(x, y, z) x, y, z over each of the two surfaces making up the boundary of W. Solution. We first deal with the spherical top. Using the work we did in part (a), we parameterize this surface as: Φ(u, v) 5 cos u sin v, 5 sin u sin v, 5 cos v, u [, 2π], v [, π/6] Page 7

8 You can calculate that: Φ u Φ v (5 sin v) 5 cos u sin v, 5 sin u sin v, 5 cos v So our surface integral becomes: π 6 2π F ds x, y, z (5 sin v) x, y, z du dv S π 6 2π 75 75π π 6 2π π 6 15 sin v(x 2 + y 2 + z 2 ) du dv sin v du dv sin v dv 75π [ cos v] π 6 [ ] 75π π 75π 2 Next, we turn to the cone, which we parameterize in cylindrical coordinates as: Φ(u, v) u cos v, u sin v, u, u [, 5/2], v [, 2π] We found the upper bound for u by intersecting the equations of the sphere and cone given above. We then calculate: Φ u cos v, sin v, Φ v u sin v, u cos v, Φ u Φ v u cos v, u sin v, u We now compute our surface integral: 5/2 2π F ds u cos v, u sin v, u u cos v, u sin v, u dv du S 5/2 2π u 2 cos 2 v u 2 sin 2 v + u 2 dv du Therefore the entire flux out of the surface is: 75π 75π 2 Page 8

9 (c) State Gauss Theorem, and apply your above work to verify it for this region and integrand. Solution. You may look up the statement of Gauss Theorem either in the textbook or in the notes. According to the theorem, the surface integral is equal to: F ds divf dv S W ( + + ) dv W 9 dv W Since this is the same integral we calculated in part (a), we conclude that the right hand side of this equality is: ( 25π 125π ) 9 75π 75π Therefore the theorem is verified. Page 9

10 7. (15 points) State Stokes Theorem, and verify it by first integrating the vector field F(x, y, z) xyzi + yj + zk over, the triangle with vertices (4,, ), (,, ), and (,, 6), oriented by the ordering of the points, and then doing the corresponding surface integral. Solution. You may look up the statement of Stokes Theorem either in the textbook or in the notes. We first calculate the curl of our vector field: i j k curlf x y z xyz y z ()i ( xy)j + (xz)k, xy, xz Now, we need to find the equation of the plane passing through the given points so we can parameterize it. We do this by first calculating two vectors that lie in the plane, for example 4,, and,, 6. Next, we cross them to find a normal to the plane: 4,,,, 6 18, 24, 12 Using this normal and one of our points, we conclude that the equation of the plane (in standard form) is x + 4y + 2z 12. So, we can parameterize this surface as: Φ(u, v) u, v, 6 2 u 2v which has: Φ u Φ v 2, 2, 1 Note that this is not the same normal that we found above. Also note that this normal is oriented in a way consistent with the orientation of the curve, which is what we want. To identify the bounds for u, v, we notice that these independent variables correspond to the x, y-coordinates. Therefore we can restrict our attention to the x, y plane. Our solid projects onto the x, y plane as the triangle with vertices at (4, ),(, ), and (, ), which corresponds to the bounding lines x, y, and y 4 x +. Page 1

11 We put all this information together in our surface integral: F d curlf ds S 4 4 u u+ 4 u+ 4 u+ ( 4 u +, xy, xz 2, 2, 1 dv du 2uv + u (6 2 ) u 2v dv du 18u 9 8 u du 9u u4 9(16) 9(8) 72 2uv + 6u 2 u2 2uv dv du 6u 2 u2 dv du ) (6u 2 ) u2 du Page 11