Final_98_sol

Final_98_sol - 1998 Final solutions of math 100 Solutions:...

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1998 Final solutions of math 100 Solutions: (1) (a) By Green’s Theorem, Z C ( e x sin y - 2 y ) dx + ( e x cos y - 2 x ) dy = Z Z R ∂x ( e x cos y - 2 x ) - ∂y ( e x sin y - 2 y ) dA = Z Z R ( e x cos y - 2) - ( e x cos y - 2) dA = 0 (b) We want the integral to be independent of path, so 2 ∂x ( e ( x n + y n ) y ) = ∂y ( e ( x n + y n ) x ) nx n - 1 e ( x n + y n ) y = ny n - 1 e ( x n + y n ) x n = 2 Now, for n = 2, φ ( x, y ) = Z ( e ( x 2 + y 2 ) y ) dy = 1 2 e ( x 2 + y 2 ) + k ( x ) φ x ( x, y ) = e ( x 2 + y 2 ) x + k 0 ( x ) k 0 ( x ) = 0 k ( x ) = c where c is a constant. The potential function is 1 2 e ( x 2 + y 2 ) + c . (2) (a) The surface is z = 4 - x 2 . Then, s ( ∂z ∂x ) 2 + ( ∂z ∂y ) 2 + 1 = s x 2 4 - x 2 + 1 = 2 4 - x 2 S = Z Z R s ( ∂z ∂x ) 2 + ( ∂z ∂y ) 2 + 1 dA = Z Z R 2 4 - x 2 dA = Z 2 - 2 Z 4 - x 2 - 4 - x 2 2 4 - x 2 dydx = Z 2 - 2 4 4 - x 2 4 - x 2 dx = 16 1
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2 (b) Z 3 - 3 Z 9 - y 2 - 9 - y 2 Z 9 - x
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This note was uploaded on 09/30/2010 for the course MATH 100 taught by Professor Qt during the Fall '09 term at HKUST.

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Final_98_sol - 1998 Final solutions of math 100 Solutions:...

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