Math 6A Practice Problems III

Transcription

1 Math 6A Practice Problems III Written by Victoria Kala H 63u Office Hours: R 1:3 1:3pm Last updated 6//16 Answers π. 5. a) 8π b) 8π π 3 1

2 etailed olutions 1. Use the transformation x u + v, y u + v to evaluate the integral x 3y)dA, where R is the R triangular region with vertices, ),, 1), 1, ). olution. The equations of the lines edges) of the triangle are y x, y 1 x, y 3 x. Plug the given transformations into these lines. When y x: u + v u + v) u + v u + v u. u + v 1 u + v) u + v u + 1 v v. When y 3 x: u + v 3 u + v) u + v 3 u v 3u + 3v 3 u + v 1. The region that defines u, v can be described as {u, v) : u 1, v 1 u}, or {u, v) : v 1, u 1 v}. We now calculate the Jacobian: Therefore x 3y)dA R 3 3 J 1 1 u x u y u x v y v [u + v 3u + v)] 3 dvdu 3 uv 5 ) 1 u v du 3 v 3 u + u 5 ) du u u 5v)dvdu u1 u) 5 1 u) ) du 1 u3 + u 5 u ) 1. valuate the surface integral x yzd where is the part of the plane z 1 + x + 3y that lies above the rectangle [, 3] [, ]. olution. x yzd x yz n da. We need to find a parametrization r to find the normal n and the region. tart off with r x, y, z. We need to narrow this down to two variables. Our surface is the plane z 1 + x + 3y, plug this for z in our parametrization: rx, y) x, y, 1 + x + 3y. We are given that we are above the rectangle [, 3] [, ], which is x 3, y. This is our region. Now we calculate the normal vector n r x r y and take the magnitude: r x 1,, r y, 1, 3 i j k n r x r y 1, 3, n ) + 3)

3 Therefore x yzd x y1 + x + 3y) 1dydx 1 x y ) + x 3 y + x 3 y 3 ) x3 + x y 3 dx 1 3 x y + x 3 y + 3x y )dydx 1x + x 3 )dx 3. valuate F d where Fx, y, z) yi+xj+zk and is the boundary of the solid region enclosed by the paraboloid z 1 x y and the plane z. olution. You can also evaluate this surface integral using ivergence Theorem, but we will instead calculate the surface integral directly. We have two surfaces: the paraboloid, call it 1, and the plane z, call it. We need to calculate the surface integral for each surface then add them together at the end: F d F d + 1 F d. We will assume that has positive outward) orientation. This means that 1 has upward orientation and has downward orientation. Recall that F d F nda where n is the normal of our parametrization vector r. 1 : tart off with the parametrization r x, y, z We need to narrow this down to two variables. Our surface is the paraboloid z 1 x y, plug this in for z in our parametrization: rx, y) x, y, 1 x y. The region is the range of x, y, which we can find when the paraboloid intersects with the plane z. When z, we have the circle x + y 1, which can be described as {r, θ) : r 1, θ π}. Now we calculate the normal vector n r x r y and take the magnitude: r x 1,, x r y, 1, y i j k n r x r y 1 x x, y, 1 1 y Therefore F d 1 y, x, z x, y, 1 da 1 r r cos θ sin θ + 1 r )rdrdθ ) r sin θ + r 1 dθ xy + z)da 1 xy + 1 x y )da r 3 sin θ + r r 3 )drdθ sin θ + 1 ) cos θ dθ + θ π π. 3

4 : We can use the parametrization r x, y, to find the normal vector, but we can also do so by observation. ince has downward orientation, the normal vector to z is n k,, 1. Then F d y, x, z,, 1 da da. Thus F d F d + F d π 1 + π.. Use tokes Theorem to evaluate curl F d where F x z i + y z j + xyzk and is the part of the paraboloid z x + y that lies inside the cylinder x + y. olution. We need to find a parametrization rt) to evaluate F dr. The closed curve of intersection of the paraboloid and cylinder is the circle x + y at z, which can be parametrized by Then we have F dr Frt)) r t)dt rt) cos t, sin t, ), t π. 18 sin t cos t + 18 sin t cos t)dt 18 6 cos t, 6 sin t, 16 cos t sin t) sin t, cos t, )dt 1 3 cos3 t + 1 ) π 3 sin3 t. 5. onsider the vector field Fx, y, z) yzi + xzj + e xy k, where is circle x + y 16, z 5 oriented counterclockwise when viewed from above. a) alculate F dr by finding an appropriate parametrization vector rt). olution. The parametrization for the circle x + y 16, z 5 is given by Then F dr Frt)) r t)dt rt) cos t, sin t, 5), t π. 8 sin t + 16 cos t)dt sin t, cos t, e 16 cos t sin t ) sin t, cos t, )dt + 1 cos t)dt t + 6 sin t) 1 cos t) cos t))dt π 8π. b) alculate F dr using tokes Theorem, and verify it is equal to your solution in part a). olution. We need to evaluate curl F d. The normal to the surface is the normal to z 5, which is n k. The curl of F is given by i j k curl F x y z yz xz e xy i y z xz e xy j x z yz e xy + k x y yz xz ixe xy x) jye xy y) + kz z).

5 Then curl F n z, but on our surface z 5, hence curl F d zda 5dA 5 π 8π. 6. Verify that the ivergence Theorem is true for the vector field Fx, y, z) 3xi + xyj + xzk where is the cube bounded by the planes x, x 1, y, y 1, z, z 1. Note: to verify the theorem is true you need to show that F d div FdV ; that is, you need to calculate both integrals and show they are equal. olution. We will evaluate F d on each of the different faces of the cube then take their sum. On x, n i, {y, z) : y 1, z 1} and F d 3xdA da. On x 1, n i, {y, z) : y 1, z 1} and F d 3xdA On y, n j, {x, z) : x 1, z 1} and F d xyda On y, n j, {x, z) : x 1, z 1} and F d xyda xda On z, n k, {x, y) : x 1, z 1} and F d xzda On z 1, n k, {x, y) : x 1, z 1} and Therefore F d xzda xda 3dA 3. da. 1 1 da. 1 1 F d xdxdz 1. xdxdy 1. Now we will evaluate div FdV where {x, y, z) : x 1, y 1, z 1}. The divergence of F is given by Therefore div FdV div F x 3x) + y xy) + xz) 3 + x + x 3x + 3. z 3x + 3)dV x + 3)dxdydz ) 3 1 x + 3x

6 7. Use the ivergence Theorem to calculate the surface integral F d; that is, calculate the flux of F across where Fx, y, z) cos z + xy )i + xe z j + sin y + x z)k, and is the surface of the solid bounded by the paraboloid z x + y and the plane z. olution. We will evaluate div FdV where {x, y, z) : x + y z } {r, θ, z) : r, θ π, r z }. The divergence of F is given by Then div FdV div F x cos z + xy ) + y xe z ) + z sin y + x z) y + x. π 3π 3. x + y )dv r 3 r )dr π r r rdzdrdθ π r 3 r 5 )dr π r 1 6 r6 ) r 3 dzdr π π r 3 z ) z dr zr 6