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Unformatted text preview: 3. Let n denote the outward unit normal vector to the closed curve C parametrized by r ( t ) = ( x ( t ) ,y ( t )), a ≤ t ≤ b , such that the area inside C is on the left when one traverses C . Then n is given by n = ( y ( t ) , x ( t )) k ( y ( t ) , x ( t )) k . Therefore the flux of F = r across C is given by I C F · n ds = Z b a ( x i + y j ) · y i x j p ( y ) 2 + ( x ) 2 p ( x ) 2 + ( y ) 2 dt = Z b a x dy dt y dx dt dt = I C y dx + xdy = ZZ D ∂ ∂x x ∂ ∂y ( y ) dA (by Green’s theorem, where D is region enclosed by C ) = ZZ D 2 dA = 2(area inside C ) . 4. (a) Since ∂N ∂x = 2 x cos y = ∂M ∂y , F is conservative. To find a potential f for F , we solve for ∂f ∂x = 2 x sin y ∂f ∂y = x 2 cos y . Integrating the first equation wrt x , we obtain: f = Z 2 x sin y dx + g ( y ) = x 2 sin y + g ( y ) for some function g ( y ). Then from the second equation we have x 2 cos y = ∂f ∂y = x 2 cos y + g ( y ) , which implies g ( y ) = 0, or g ( y ) = C . By taking C = 0 we find a potential f for F as f = x 2 sin y . (b) Since ∇ × F = k 6 = , we conclude that F is not conservative....
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This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.
 Spring '11
 wang
 Math

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