21-356 Lecture 38

we have div f x y z 0 yz x 0

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Unformatted text preview: of the previous corollary (why?). We have div f (x, y, z ) = # # # (0) + (yz ) + (x) = 0 + 1z + 0, #x #y #z and so by the divergence theorem A A 2 f · $ dH = z dxdydz. "U U Using cylindrical coordinates x = r cos !, y = r sin !, z = z , we get r2 cos2 ! + r2 sin2 ! < z 2 , that is, r2 < z 2 , r2 cos2 ! + r2 sin2 ! + z 2 < 2r sin !, that is, r2 + z 2 < 2r sin !, and z > 0, hence, r2 < z 2 < 2r sin ! ' r2 , which implies that r2 < 2r sin ! ' r2 , or equivalently, r < sin !. In turn, sin ! should be positive, and so ! ! (0, " ). F G 2 W := (r, !, z ) ! R3 : 0 < ! < "...
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This document was uploaded on 03/31/2014.

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