Exercises: Divergence Theorem and Stokes Theorem

Transcription

1 Exercises: ivergence heorem and tokes heorem Problem 1. his exercise allows you to see the main idea behind the proof of the ivergence heorem. uppose that is a closed region in R 3 whose boundary surface can be divided into xy-monotone surfaces: 1 and 2, whose projections onto the xy-plane are the same region. (For example, the ball x 2 + y 2 + z 2 1 is such a region because we can divide its boundary into two xy-monotone surfaces (i 1 : x 2 +y 2 +z 2 1 with z 0, and (ii 2 : x 2 +y 2 +z 2 1 with z 0. Let f(x,y,z be a function that is continuous on. Orient by taking its outer side. Prove that f z dxdydz f dxdy. Proof: Let 1 be described by z φ 1 (x,y and 2 by z φ 2 (x,y, with (x,y. f dxdy f dxdy + f dxdy 1 2 (by the orientation of 1, 2 f(x,y,φ 1 (x,ydxdy f(x,y,φ 2 (x,ydxdy. (1 On the other hand: f z dxdydz ( φ1 (x,y φ 2 (x,y f z dz dxdy f(x,y,φ 1 (x,y f(x,y,φ 2 (x,ydxdy right hand side of (1. Problem 2. onsider the cylinder x 2 + y 2 1 and 0 z 1. Let be the boundary of the cylinder; see below. Use the ivergence heorem to calculate xydydz +y2 dxdz +zdxdy olution. Introduce f 1 xy,f 2 y 2, and f 3 z. Let be the cylinder. By ivergence heorem, 1

2 we have: xydydz +y 2 dxdz +zdxdy f 1 x + f 2 y + f 3 z dxdydz y +2y +1dxdydz 3y +1dxdydz. (2 learly, 1dxdydz π (volume of the cylinder. On the other hand, letting be the disc x 2 +y 2 1 in the xy-plane, we have: 1 ( ydxdydz ydxdy dz 0 ydxdy. o evaluate the above, we represent using polar coordinates: x rcosθ y rsinθ where r [0,1] and θ [0,2π]. Let R be the set of all such (r,θ. hen: ydxdy rsinθ rdrdθ R 1 ( 2π r 2 sinθdθ dr 0. herefore, we know that (2 equals π. 0 Problem 3. his exercise allows you to derive another popular form of the ivergence theorem. Let be a closed region in R 3 that is bounded by a surface, which is the union of a finite number of smooth surfaces 1,..., k. efine f(x,y,z to be a vector function that is continuous on each i (1 i k. For each point p (x,y,z, define n(x,y,z to be the unit vector of at p pointing towards the outside of. Prove: div f dxdydz f nda. 0 Proof. We can write n as [cosα,cosβ,cosγ], where α is the angle between n and i [1,0,0]; β is the angle between n and j [0,1,0]; γ is the angle between n and k [0,0,1]. 2

3 Hence: f nda (taking the outer side of (f 1 cosα+f 2 cosβ +f 3 cosγda f 1 dydz +f 2 dxdz +f 3 dxdy f 1 x + f 2 y + f 3 z dxdydz div f dxdydz. Problem 4. Let f [e x,e y,e z ]. Let be the boundary of the cube with x 1, y 1, and z 1. Let n be defined as in the previous problem. alculate f nda. olution. Let be the region enclosed by the cube. We have: f nda e x +e y +e z dxdydz. (3 Focusing on the e z term, we have: 1 ( 1 ( 1 e z dxdydz e z dz ( 1 (e 1/e dx 4(e 1/e. 1 dy 1 dy dx imilarly, ex dxdydz ey dxdydz 4(e 1/e. Hence, (3 equals 12(e 1/e. Problem 5. Let be the curve that is the intersection of x 2 +y 2 2z z 2 esignate the direction of as passing points (2,0,2, (0,2,2, and ( 2,0,2 in this sequence. Use the tokes theorem to calculate ydx xzdy +yz2 dz. olution. Let be the surface satisfying the following two equations simultaneously: x 2 +y 2 4 z 2. We can regard as the boundary of, when is oriented with its upper side taken. z y x 3

4 Introduce functions f 1 y,f 2 xz, and f 3 yz 2. hen, f 3 y f 2 z f 1 z f 3 x f 2 x f 1 y z 2 +x 4+x 0 By tokes heorem, we know: ydx xzdy +yz 2 dz z 1 3 (4+xdydz 3dxdy. (4 Note that is perpendicular to the yz-plane. Hence, (4 + xdydz 0. On the other hand, 3 dxdy is clearly 3 (4π 12π. herefore, (4 equals 12π. Problem 6. his exercise allows you to see an alternative form of the tokes theorem. Let be a piecewise surface and its boundary curve, both oriented in the way described in the tokes theorem (see lecture notes. Also, let f 1,f 2,f 3 be functions that have continuous partial derivatives on each smooth surface that constitutes. p s p efine f(x,y,z [f 1,f 2,f 3 ], and n(x,y,z be the unit normal vector of at point (x,y,z, emanating from the side of chosen. Fix any point p on. Given any point p on, denote by s the length of the curve from p to p, following the direction of. Let r(s [x(s,y(s,z(s] be a parametric form of. Prove: curlf nda f r (sds. Proof. By tokes theorem, we have curlf nda f 1 dx+f 2 dy +f 3 dz f dr f dr ds ds f r (sds. Problem 7. Let be the surface z x 2 with 0 x 2 and 0 y 1. Orient by taking its upper side. efine f [e y,e z,e x ]. alculate curlf nda, where n is as defined in the previous problem. alculate curlf nda. 4

5 z 1 y x olution. Let be the boundary curve of, directed as shown above. consists of 4 curves: 1, 2,..., 4. By tokes theorem, we have: curlf nda e y dx+e z dy +e x dz. (5 Next, we focus on each of the above terms separately. 4 e y dx e y dx i1 i e y dx+ e y dx 2 4 e 1 dx+e 0 dx 2 4 (note the directions of 2, 4 2e+2. 4 e z dy e z dy i1 i e z dy + e z dy 1 3 e 4 dy +e 0 dy 1 3 e x dz e e x dz i1 i e x dz + e x dz 2 4 e xdz 2 dx dx+ e xdz 4 dx dx 0 2 2x e x dx x e x dx 0. (6 5

6 herefore, (5 equals e 4 2e+1. Problem 8. Fix a vector function f(x,y,z [f 1,f 2,f 3 ]. Prove that if curlf 0, then the class of line integrals f 1dx+f 2 dy +f 3 dz is path independent. Proof. It suffices to prove that for any closed piecewise smooth curve, it holds that f 1dx+ f 2 dy+f 3 dz 0. Find an arbitrary piecewise smooth surface whose boundary curve is. Orient according to the direction of. By the tokes theorem, it holds that f 1 dx+f 2 dy +f 3 dz curlf nda 0. We thus complete the proof. 6