F3k, namely, F F F (10.7.1) x y z

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1 10.7 Divergence theorem of Gauss riple integrals can be transformed into surface integrals over the boundary surface of a region in space and conversely. he transformation is done by the divergence theorem, which involves the divergence of a vector function F= F1i+ F2 j+ Fk, namely, F F F F = + + (10.7.1) x y z 1 2 div. heorem 1: Divergence theorem of Gauss Let be a closed bounded region in space whose boundary is a piecewise smooth orientable surface. Let F ( x, yz, ) be a vector function that is continuous and has continuous first partial derivatives in some domain containing. hen, ò divfdv = F nda. (10.7.2) In terms of components, where F F1i F2 j Fk and the outer unit normal vector of is n cos i cos j cos k, eqn (10.7.2) becomes, ò æ F1 F2 F ö + + dx dy dz = ( F1cos + F2cos + Fcos ) da ç è x y z ø ( F dy dz F dz dx F dx dy) = (10.7.2*) he proof follows after Example 1. Example 1: Evaluation of a surface integral by the divergence theorem Before proving the divergence theorem we consider a typical application. Evaluate, 2 2 ( ), I = x dydz+ x ydzdx+ x zdxdy where is the closed surface in Fig. 249 consisting of the cylinder x + y = a,(0 z b) and the circular disks z = 0 and z= b,( x + y a ). July 10, /5

2 2 2 olution: Here 2 F1 x, F2 x y, F x z. Hence, div F = x + x + x = 5x. he form of the surface suggests that we introduce polar coordinates r, defined by x rcos, y rsin (thus cylindrical coordinates r,, z). hen the volume element is dx dy dz rdr d dz, and we obtain, b 2 a ò ò ò ò I = 5 xdxdydz= (5r cos ) rdrd dz, z= 0 = 0 r= 0 b a b 2 a 5 4 = 5ò cos d dz 5 dz a b. z= 0ò = = 0 4 ò = z= Proof: We prove the divergence theorem, beginning with eqn (10.7.2*.1). his equation is true if and only if the integrals of each component on both sides are equal; i.e., 1 ò ç = ( cos 1 ), (10.7.) æ F ö ç dx dy dz F da çè x ø ò 2 ç ( cos 2 ), (10.7.4) æ F ö ç dx dy dz = F da ç è y ø ò ç = ( cos ). (10.7.5) æ F ö ç dx dy dz F da çè z ø We first prove eqn (10.7.5) for a special region that is bounded by a piecewise smooth orientable surface and has the property that any straight line parallel to any one of the coordinate axes and intersecting has at most one segment (or a single point) in common with. his implies that can be represented in the form, gxy (, ) z hxy (, ), (10.7.6) where ( x, y ) varies in the orthogonal projection of in the xy -plane. Clearly, z= g( x, y) represents the "bottom" 2 of (Fig. 250), whereas z= h( x, y) represents the "top" 1 of, and there may be a remaining vertical portion of. (he portion may degenerate into a curve, as for a sphere.) July 10, /5

3 o prove eqn (10.7.5), we use eqn (10.7.6). ince F is continuously differentiable in some domain containing, we have, ò æ ö é ù èç ê ú F h( x, y) F dx dy dz = dz dx dy z ø ò g( x, y) z ë û. (10.7.7) Integration of the inner integral gives triple integral in (10.7.7) equals, F[ x, y, h( x, y)] - F[ x, y, g( x, y)]. Hence the F[ x, y, h( x, y)] dx dy - F[ x, y, g( x, y)] dx dy. (10.7.8) But the same result is also obtained by evaluating the right side of eqn (10.7.5); that is (see also eqn (10.7.2*.2), F cos da = F dx dy, =+ F [ x, y, h( x, y)] dx dy - F [ x, y, g( x, y)] dx dy, where the first integral over gets a plus sign because cos > 0 on in Fig. 250 (as in eqn (10.6.5")), and the second integral gets a minus sign because cos < 0 on 2. his proves eqn (10.7.5). he relations (10.7.) and (10.7.4) now follow by merely relabeling the variables and using the fact that, by assumption, has representations similar to eqn (10.7.6), namely, g ( y, z) x h ( y, z) and gzx (, ) y hzx (, ). his proves the first equation in eqn (10.7.2*) for special regions. It implies eqn (10.7.2) because the left side of eqn (10.7.2*) is just the definition of the divergence, and the right sides of eqn (10.7.2) and of the first equation in eqn (10.7.2*) are equal, as was shown in the first line of eqn (10.6.4). Finally, equality of the right sides of eqn (10.7.2) and last line of eqn (10.7.2*), is seen from eqn (10.6.5). his establishes the divergence theorem for special regions. For any region that can be subdivided into finitely many special regions by means of auxiliary surfaces, the theorem follows by adding the result for each part separately; this procedure is analogous to that in the proof of Green's theorem in ec he surface integrals over the auxiliary surfaces cancel in pairs, and the sum of the remaining surface integrals is the surface integral over the whole boundary surface of ; the triple integrals over the parts of add up to the triple integral over. he divergence theorem is now proved for any bounded region that is of interest in practical problems. 1 July 10, /5

4 Example 2: Verification of the divergence theorem Evaluate ( 7x i- zk) nda over the sphere : x + y + z = 4, (a) by using eqn (10.7.2), and (b) directly. ( ) olution: (a) divf= div 7xi-zk = 6. Answer: (b) We can represent by eqn (10.5.) with (see eqn (10.6.*). Accordingly, 6 (4/) 2 64 =. : r= 2 cos v cosui+ 2 cosv sin uj+ 2sin vk. hen r =- 2cosvsin ui+ 2cosvcos uj, u r =-2sinvcosui- 2sinvsinuj+ 2cos vk, v u v a = 2 and we shall use nda = Ndu dv 2 2 4cos vcosu 4cos vsinu 4cosvsin v. N= r r = i+ j+ k Now on we have x = 2 cosv cos u, z= 2sin v, so that F= 7x i-z k on, becomes F( ) = 14 cos v cosui - 2sin vk, and 2 2 F( ) N = 56 cos v cos u-8cosv sin v. On we have to integrate over u from 0 to 2. his gives, 2 (56 cos v) -2 (8cosv sin v). 2 he integral of cos vsin v equals (sin v) /, and that of cos v equals sin v- (sin v) /. On we have - / 2 v / 2, so that by substituting these limits we get 64, as hoped for. he usefulness of the divergence theorem of Gauss is seen by comparing the effort needed for (a) and (b). Coordinate invariance of the divergence: he divergence eqn (10.7.1) is defined in terms of coordinates, but we can use the divergence theorem to show that divf has a meaning independent of coordinates. For this purpose we first note that triple integrals have properties quite similar to those of double integrals in ec In particular, the mean value theorem for triple integrals asserts that for any continuous function f ( xyz,, ) in a bounded and simply connected region there is a point Q:( x0, y0, z0) in such that, ò f ( xyzdv,, ) = f( x0, y0, z0) V ( ), (10.7.9) July 10, /5

5 where V ( ) is the volume of. In this formula we interchange the two sides, divide by V ( ), and set f = divf. hen by the divergence theorem we obtain for the divergence an integra l over the boundary surface ( ) of, 1 1 div F ( x0, y0, z0) = div dv da. V ( ) ò F = V ( ) F n ( ) ( ) e now choose a point P:( x, y, z ) in and let shrink down onto P so that the maximum W distance d ( ) of the poin ts of from P goes to zero. hen Q:( x0, y0, z 0) must approach P. Hence eqn ( ) becomes, 1 F P F n da, ( ) = div ( ) lim d( ) 0V ( ) ( ) which proves the following theorem. heorem 2: Invariance of the divergence he divergence of a vector f unction F with continuous first partial derivatives in a region is independent of the particular choice of Cartesian coordinates. For any point P in it is given by eqn ( ). Equation ( ) is sometimes used as a definition of the divergence. hen the representation eqn (10.7.1) in Cartesian coordinates can be derived from eqn ( ). July 10, /5