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Unformatted text preview: 94 BRUCE K. DRIVER 8. Surfaces, Surface Integrals and Integration by Parts De f nition 8.1. A subset M R n is a n 1 dimensional C k Hypersurface if for all x M there exists > an open set D R n and a C kdi f eomorphism : D B ( x , ) such that ( D { x n = 0 } ) = B ( x , ) M. See Figure 16 below. Figure 16. An embedded submanifold of R 2 . Example 8.2. Suppose V R n 1 and g : V C k R . Then M := ( g ) = { ( y, g ( y )) : y V } is a C k hypersurface. To verify this assertion, given x = ( y , g ( y )) ( g ) de f ne ( y, z ) := ( y + y , g ( y + y ) z ) . Then : { V y ) R C k V R di f eomorphism (( V y ) { } ) = { ( y + y , g ( y + y )) : y V y } = ( g ) . Proposition 8.3 (Parametrized Surfaces) . Let k 1 , D R n 1 and C k ( D, R n ) satisfy (1) : D M := ( D ) is a homeomorphism and (2) ( y ) : R n 1 R n is injective for all y D. (We will call M a C k parametrized surface and : D M a parametrization of M. ) Then M is a C khypersurface in R n . Moreover if f C ( W R d , R n ) is a continuous function such that f ( W ) M, then f C k ( W, R n ) i f 1 f C k ( U, D ) . Proof. Let y D and x = ( y ) and n be a normal vector to M at x , i.e. n Ran ( ( y )) , and let ( t, y ) := ( y + y ) + t n for t R and y D y , see Figure 17 below. Since D y (0 , 0) = ( y ) and t (0 , 0) = n / Ran ( ( y )) , (0 , 0) is invertible. so by the inverse function theorem there exists a neighborhood V of (0 , 0) R n such that  V : V R n is a C k di f eomorphism. PDE LECTURE NOTES, MATH 237AB 95 Figure 17. Showing a parametrized surface is an embedded hypersurface. Choose an > such that B ( x , ) M ( V { t = 0 } ) and B ( x , ) ( V ) . Then set U := 1 ( B ( x , )) . One f nds  U : U B ( x , ) has the desired properties. Now suppose f C ( W R d , R n ) such that f ( W ) M, a W and x = f ( a ) M. By shrinking W if necessary we may assume f ( W ) B ( x , ) where B ( x , ) is the ball used previously. (This is where we used the continuity of f. ) Then 1 f = 1 f where is projection onto { t = 0 } . Form this identity it clearly follows 1 f is C k if f is C k . The converse is easier since if 1 f is C k then f = ( 1 f ) is C k as well. 8.1. Surface Integrals. De f nition 8.4. Suppose : D R n 1 M R n is a C 1 parameterized hypersurface of R n and f C c ( M, R ) . Then the surface integral of f over M, R M f d, is de f ned by Z M f d = Z D f ( y ) det[ ( y ) y 1  , . . . , ( y ) y n 1  n ( y )] dy = Z D f ( y )  det[ ( y ) e 1  . . .  ( y ) e n 1  n...
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This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.
 Three '10
 Smith
 Integrals, Integration By Parts

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