Lecture Notes: Divergence Theorem and Stokes Theorem

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1 Lecture Notes: ivergence heorem and tokes heorem Yufei ao epartment of omputer cience and Engineering hinese University of Hong Kong In this lecture, we will discuss two useful theorems on surface integrals. 1 ivergence heorem heorem 1 (ivergence heorem). 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, 2,.., k. Let f 1,f 2, and f 3 be functions of x,y,z that have continuous partial derivatives on each i (1 i k). If we orient by taking its outer side, then it holds that x + f 2 + f 3 dxdydz = f 1 dydz +f 2 dxdz +f 3 dxdy. We omit a proof for the theorem (which follows the same idea as our proof of the Green s theorem, and is a good exercise for you). he theorem is also called Gauss heorem. Example 1. alculate the volume of the ball x 2 +y 2 +z 2 1. olution. Let be the target ball, and be its boundary x 2 +y 2 +z 2 = 1, oriented by having its outer side taken. Introduce f 1 = x, f 2 = y, and f 3 = z. hen, by heorem 1, we know that 1dxdydz = zdxdy. (1) enote by 1 the upper half of satisfying z 0, and 2 the lower half of satisfying z 0. We thus have: zdxdy = zdxdy + zdxdy. (2) 1 2 Let us first calculate 1 zdxdy. Note that 1 is oriented with its upper side taken. Hence: 1 zdxdy = zdxdy. (3) Let us represent 1 in a parametric form: r(u,v) = [x(u,v),y(u,v),z(u,v)] where x(u,v) = cosu sinv y(u,v) = sinu sinv z(u,v) = cosv 1

2 where u [0,2π] and v [0,π/2]. Let R be the collection of all such (u,v). Now we change the integral variables in (3) to u and v by the Jacobian rule: zdxdy = cosv sinvcosvdudv = 2π/3. R imilarly, we also have 2 zdxdy = 2π/3. hus, we know from (1) and (2) that 1dxdydz = 4π/3. Example 2. Let be the cube as shown below, and be its boundary surface with its outer side taken. z y (1,1,1) x alculate y(x z)dydz +x 2 dzdx+(y 2 +xz)dxdy. olution. By heorem 1, we have: y(x z)dydz +x 2 dzdx+(y 2 +xz)dxdy = We know xdxdydz = 1 0 ( 1 ( 1 imilarly, ydxdydz = 1/2. Hence, (4) equals ) ) xdx dy dz = 1/2. x+ydxdydz. (4) Remark 1. You may be wondering why heorem 1 is called the ivergence heorem. In fact, if we define a vector function f(x,y,z) = [f 1,f 2,f 3 ], then divf = x + f 2 + f 3. Hence, the left hand side of the equation in heorem 1 can also be written as divf dxdydz. Remark 2. Recall that surface integrals are inherently 2d. Hence, the divergence theorem essentially reveals a relationship between a 2d integral and a 3d integral. Also recall that, in contrast, the Green s theorem reveals a relationship between a 1d integral and a 2d integral. 2 tokes heorem onsider to be a piecewise-smooth surface with a closed boundary curve. Note that, not all surfaces have a boundary curve, e.g., a sphere x 2 +y 2 +z 2 = 1 does not have a boundary curve. Intuitively, has a boundary curve if it is not closed, namely, it does not separate R 3 into an interior part and an exterior part. he following is an example of such an : 2

3 Fix a direction of. Let us then orient by choosing a side of it as shown in the above figure. Formally, imagine that a point moves around along the direction we have decided. Now, watch the point s movement from the side of we have chosen (i.e., allowing normal vectors of that side to shoot into our eyes). We should see that the point is moving in the counterclockwise direction. heorem 2. Let and be as described earlier (with the direction of fixed, and oriented). uppose that is the union of a finite number of smooth surfaces 1, 2,.., k. Let f 1,f 2, and f 3 be functions of x,y,z that have continuous partial derivatives on each i (1 i k). hen: f 1 dx+f 2 dy +f 3 dz = f 3 f 2 dydz + f 3 x dzdx+ f 2 x dxdy. Proof. We first prove the theorem for the special case where is all monotone, namely, it is simultaneously xy-monotone, xz-monotone, and yz-monotone. In particular, we will show that f 1 dx = dzdx dxdy (5) f 2 f 2 dx = x dxdy f 2 dydz f 3 f 3 dx = dydz f 3 x dzdx. hen, the theorem will follow from summing up both sides of the above three equations. ue to symmetry, it suffices to prove (5). For this purpose, suppose that is described by z = g(x,y), and that we oriented by taking its upper side. Introduce h(x,y,z) = z g(x,y). Hence, is also described by h(x,y,z) = 0. Let xy be the projection of onto the xy-plane, and be the region in the xy-plane enclosed by xy. f 1 dx = f 1 (x,y,z)dx xy (by Green s heorem) = (x,y,z) dxdy = + dxdy. (6) Note that h = [ h x, h, h ] is a normal vector of. Let γ be the angle between the directions of h and k = [0,0,1], and β be the angle between the directions of h and j = [0,1,0]. We have: cosβ = cosγ = h g g = g = g h g = g = 1 g. 3

4 herefore: We thus have: (6) = cosβ cosγ =. dxdy + = dxdy + cosβda = dxdy + dzdx. = right hand side of (5). cosβ cosγ dxdy. If is not all monotone, we can always cut into a set of disjoint surfaces each of which is all monotone. hen, we can obtain heorem 2 by applying what we have proved to each of those surfaces. he details are omitted and serve as a good exercise for you. Example 3. Use tokes heorem to calculate ydx+zdy +xdz where is the sequence of line segments: ABA with points A = (1,0,0), B = (0,1,0), and = (0,0,1). z = (0,0,1) O y B = (0,1,0) A = (1,0,0) x olution. Let be the triangle AB, oriented with its lower side taken. Introduce f 1 (x,y,z) = y, f 2 (x,y,z) = z, and f 3 (x,y,z) = x. We have: f 3 f 2 f 3 x f 2 x = 0 1 = 1 = 0 1 = 1 = 0 1 = 1. (7) 4

5 herefore, tokes heorem gives: ydx+zdy +xdz = Letting be the projection of onto the xy-plane, we know: 1dxdy = 1dxdy = 1/2. ( 1)dydz +( 1)dzdx+( 1)dxdy. (8) imilarly, we know 1dydz = 1dzdx = 1/2. herefore, (8) equals 3/2. he tokes heorem has another useful form. Let n = [cosα,cosβ,cosγ] be a unit normal vector of, coming out from the side of we have chosen. pecifically: α is the angle between the directions of n and i = [1,0,0]; β is the angle between the directions of n and j = [0,1,0]; γ is the angle between the directions of n and k = [0,0,1]. From our earlier discussion on surface integral by area, we know: ( f3 f ) 2 f 3 cosαda = f 2 dydz ( f1 f ) 3 cosβda = x f 3 x dzdx ( f2 x f ) 1 f 2 cosγda = x dxdy. herefore, from heorem 2, we get: f 1 dx+f 2 dy +f 3 dz ( f3 = f 2 ) cosα+ ( f1 f ) ( 3 f2 cosβ + x x f ) 1 cosγda. Let us introduce a vector function f = [f 1,f 2,f 3 ]. In an earlier lecture, we have defined: [ f3 curlf = f 2, f 3 x, f 2 x f ] 1. We thus obtain the following concise form of heorem 2: f 1 dx+f 2 dy +f 3 dz = curlf nda. (9) Example 4. Let be the hemisphere x 2 +y 2 +z 2 = 1 with z 0. Orient by taking its upper side. Let f = [ y,x,z 2 ]. alculate curlf nda. olution. We could solve this question by using the methods we learned in the lecture surface integral by area. However, due to the special nature of the integrand function, we can apply tokes heorem to convert the surface integral into a line integral, which simplifies calculation 5

6 significantly. he boundary curve of is the circle x 2 + y 2 = 1 in the xy-plane, directed counterclockwise. Hence, by (9), we have: curlf nda = ydx+xdy +z 2 dz. (as is in the xy-plane) = ydx+xdy (by Green s heorem) = 2 dxdy = 2π where is the disc x 2 +y 2 1 in the xy-plane. Remark 3. Note that the tokes theorem reveals a relationship between a 1d integral and a 2d integral. Namely, it has the same nature as the Green s theorem (which in fact is a special case of the tokes theorem). 6

MATH 52 FINAL EXAM SOLUTIONS

MATH 52 FINAL EXAM SOLUTIONS MAH 5 FINAL EXAM OLUION. (a) ketch the region R of integration in the following double integral. x xe y5 dy dx R = {(x, y) x, x y }. (b) Express the region R as an x-simple region. R = {(x, y) y, x y }