Answers and olutions to ection 1.7 Homework Problems 1 19 (odd). F. Ellermeyer April 2, 24 1. The hemisphere and the paraboloid both have the same boundary curve, the circle x 2 y 2 4. Therefore, by tokes Theorem, both of the surface integrals are equal to the line integral of F around this common boundary curve.. The given vector field is Fx,y,z x 2 e yz i y 2 e xz j z 2 e xy k. The given surface is the hemisphere : x 2 y 2 z 2 4 z (oriented upward). This oriented surface has positively oriented boundary curve : rt 2costi 2sintj xti ytj t 2. Note that r t 2sinti 2costj yti xtj and note that Frt r t x 2 e yz i y 2 e xz j z 2 e xy k yi xj x 2 ye yz xy 2 e xz According to tokes Theorem, curlf d F dr 2 Frt r tdt 2 x 2 ye yz xy 2 e xz dt 2 8cos 2 tsint 8costsin 2 t dt. 5. The given vector field is Fx,y,z xyzi xyj x 2 yzk. The given surface,, is the top and the four sides of the cube with vertices at 1,1,1 oriented outward. This oriented surface has positively oriented boundary curve,, consisting of the line segments 1
According to tokes Theorem, 1 : rt i tj k, 1 t 1 2 : rt ti j k, 1 t 1 : rt i tj k, 1 t 1 4 : rt ti j k, 1 t 1 curlf d F dr F dr 1 F dr 2 F dr F dr 4 1 1 xy dt 1 xyz dt 1 1 1 xy dt 1 xyz dt 1. 7. The given vector field is Fx,y,z x y 2 i y z 2 j z x 2 k. Note that curlf x y z x y 2 y z 2 z x 2 2zi 2xj 2yk. The given curve is the triangle with vertices at 1,,,,1,, and,,1. To find a vector that is orthogonal to this triangle, we let a 1,1,, b 1,,1 and compute the cross product a b 1 1 1 1 i j k. A unit normal vector that agrees with the counterclockwise orientation of the triangle (as viewed from above) is thus n i j k. tokes Theorem says that the line integral of the tangential component of F around the triangle should be equal to the surface integral of the normal component of the curl of F over the surface,, which is the triangle. Observe that 2
curlf n 2zi 2xj 2yk i j k Thus 2 x y z. F dr curlf d 2 x y zd. However, note that the triangle lies in the plane x y z 1. We therefore have F dr 2 2 2 1. 1d area of triangle 1 2 2 2 9. The given vector field is Fx,y,x 2zi 4xj 5yk. Note that curlf x y z 2z 4x 5y 5i 2j 4k. The curve of intersection of the plane z x 4 and the cylinder x 2 y 2 4 is an ellipse that bounds the surface z gx,y x 4 inside the cylinder x 2 y 2 4. By tokes Theorem,
F dr curlf d 1 curlf n d 5 g x 2 g y 4 5 2 4 da 1 da da 1 area of circle x 2 y 2 4 4. 11. For Fx,y,z x 2 zi xy 2 j z 2 k, we have curlf x y z x 2 z xy 2 z 2 x2 j y2 k. The curve of intersection of the plane z 1 x y and the cylinder x 2 y 2 9 is an ellipse that bounds the part of the surface z gx,y 1 x y that lies inside the cylinder. By tokes Theorem F dr curlf d curlf n d x 2 g x x 2 y 2 da 2 r 2 rdrd y 2 g y 81 2. 1. For the vector field Fx,y,z y 2 i xj z 2 k, we have da 4
curlf x y z y 2 x z 2 1 2yk. If is the part of the paraboloid z x 2 y 2 that lies below the plane z 1 and we assume that is oriented upward, then curlf d curlf n d g x 1 2y da g y 2 1 1 2rsinrdrd 1 2y1. The positively oriented boundary curve of is the circle rt costi sintj k xti ytj ztk Thus r t yti xtj. We thus have F dr Frtr tdt We have shown that 2 y 2 i xj z 2 k yti xtjdt 2 y x 2 dt 2 cos 2 t sin t dt. F dr curlf d. 17. We want to calculate the work done by the force field Fx,y,z x x z 2 i y y x 2 j z z y 2 k on a particle that moves around the part of the edge of the hemisphere x 2 y 2 z 2 4 that lies in the first octant (in a counterclockwise direction as viewed from above). The work done is da 5
F dr where r is the path taken along the hemisphere. Note that we are not told the exact path taken, but we may assume that r lies in the xz plane that that r/2 lies in the yz plane. In particular, we may assume (from what is given) that r has the form rt xti ytj ztk where x 2 y 2 z 2 4 for all t,/2 dx for all t,/2 dt dy for all t,/2 dt y x 2. We will use tokes Theorem to evaluate this integral using the part of the disk of radius 2 centered at the origin that is determined by the curve r. The unit normal vector we use is thus n 1 2 xi 1 2 yj 1 2 zk. Also, note that curlf Thus, the work done is x y z x x z 2 y y x 2 z z y 2 2yi 2zj 2xk. 6
F dr curlf d curlf n d xy yz xz d xy yz xz r r da /2 /2 4cossinsin 2 4sindd /2 /2 4cossinsin4sindd /2 /2 4coscossin4sindd 16 16 16 16. 19. uppose that is the sphere x a 2 y b 2 z c 2 R 2. (This sphere has radius R and is centered at the point a,b,c.) According to tokes Theorem, we can take the equator of the sphere to be the curve 1 oriented counterclockwise when viewed from above and take 1 to be the upper hemisphere and conclude that 1 curlf d 1 F dr. Taking 2 to be the lower hemisphere, we also obtain by tokes Theorem that However, recall that We thus obtain 2 curlf d 1 F dr. 1 F dr 1 F dr. 7
curlf d curlf d curlf d 1 2 F dr 1 F dr 1 F dr 1 F dr 1. 8