Math 20E Midterm II(ver. a)

Transcription

1 Name: olutions tudent ID No.: Discussion ection: Math 20E Midterm IIver. a) Fall 2018 Problem core 1 /24 2 /25 3 /26 4 /25 Total /100

2 1. 24 Points.) Consider the force field F 5y ı + 3y 2 j. Compute the work done in moving a particle along graph of y sin x, from, 1) to 0,0). Parametrize our curve with: where t 0. Its velocity vector is: c ct) t,sin t), t) ı + cos t j. Note that our parametrization is in the correct direction. We compute: F d s c 5 cos t 0 F c t) dt, 5y ı + 3y 2 j) ı + cos t j) dt, 5y + 3y 2 cos t dt, 5 sin t + 3 sin 2 t cos t dt, + sin 3 t ) + 0 1)), 6.

3 2. 25 Points.) Let be the part of the surface x 2 + y 2 z that lies between z 5 and z 2. et up the integral x 2 y 3 + z d, so that it is ready to be integrated complete, with limits of integration). Do not compute the integral. This is a surface of revolution. We parametrize in several steps. First parametrize the curve r z in the rz-plane this is a sidewise parabola) using the variable u with where 5 u 2. z u, r u 2 + 3, Now add the parameter v to represent rotation about the z-axis. Our parametrization is now where u,v) [ 5,2] [0,2π]. z u, r u 2 + 3, θ v, Now use x r cos θ, y r sin θ, z z to obtain the final version of our parametrization: where u,v) [ 5,2] [0,2π]. x u 2 + 3) cos v, y u 2 + 3) sin v, z u, Now compute: T u 2u cos v) ı + 2u sin v) j + k, And thus so The integrand translates to: T v u 2 + 3) sin v ) ı + u 2 + 3) cos v ) j. T u T v u 3 + 3) cos v ) ı + u 2 + 3) sin v ) j + 2uu 2 + 3) k, T u T v u 3 + 3) 2 + 4u 2 u 2 + 3) 2 u 2 + 3) 1 + 4u 2. x 2 y 3 + z u 2 + 3) 2 cos 2 v u 2 + 3) 3 sin 3 v + u u 2 + 3) 5 cos 2 v sin 3 v + u. And finally: 2π 2 x 2 y 3 +z d x 2 y 3 +z) T u T v du dv [ 5,2] [0,2π] 0 5 u 2 + 3) 5 cos 2 v sin 3 v + u ) u 2 +3) 1 + 4u 2 du dv.

4 3. a) 13 Points.) Let be an oriented surface. how that if a vector field F is parallel to the normal vector of at every point on, then F d ± F d. b) 13 Points.) Find a parametrization of the surface x yz 2 and use it to find the tangent plane at 20, 5,2). a) F d F n d, where n is the unit normal to our surface direction compatible with orientation) F n cos θ d, where θ is the angle between F and n, ± F d, because F is parallel to n, so θ 0 plus sign) or θ π negative sign). b) We think of this surface as a graph over the yz-plane. o we use a parametrization based on y and z: x uv 2, y u, z v. Now compute T u v 2 ı + j, T v 2uv ı + k. And so the normal at each point is given by: T u T v ı v 2 j 2uv k. The point 20, 5,2) corresponds to u 5, v 2, so the normal at our point of interest is: T u T v ı 4 j + 20 k. o the equation of our plane is: x + 20) 4y + 5) + 20z 2) 0.

5 4. 25 Points.) Let be the closed surface that consists of: the part of the paraboloid z x 2 + y that lies below z 36, and the part of the plane z 36 where x 2 + y Let Fx, y, z) x ı + y j z) k. Calculate Let be oriented by the outward-pointing normal.) F d We compute two flux integrals: the first over the paraboloid z x 2 + y 2 100, 100 z 36, and the second over the disk x 2 + y 2 64, z 36. We can parametrize the paraboloid in many ways; let s think of it as a graph over the xy-plane: x u, y v, z u 2 + v 2 100, where u 2 + v Now we compute T u ı + 2u k, so T v j + 2v k, T u T v 2u ı 2v j + k. Note that our T u T v always points upwards. This means our parametrization is in the wrong orientation, so we need to insert a minus sign in our integral: F d x ı + y j z) k ) 2u ı 2v j + k ) du dv paraboloid u ı + v j u 2 + v 2 100)) k ) 2u ı 2v j + k ) du dv 2u 2 2v u 2 + v 2 100)) du dv 2u 2 2v u 2 + 2v du dv 197 du dv 197 π 64. Note that we expected this integral to be positive because the k component of F is always negative.) Now for the disk: we parametrize using u,v) u,v, 36), with u 2 + v T u ı and T v j, so T u T v k. Note that this is the correct orientation. o our integral is: F d disk x ı + y j z) k ) k du dv 3 + 2z du dv ) du dv 69 du dv 69 π 64. o our final answer is the sum of the two flux integrals: F d )64π.