MIT18_024S11_soln-pset10

MIT18_024S11_soln-pset10 - Solutions for PSet 10 1(E7:1,2 1...

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Unformatted text preview: Solutions for PSet 10 1. (E7:1,2) 1 Let l be the arclength of C , and parameterize C by its arclength: α ( t ) = ( x ( t ) , y ( t )). Then α ( t ) = 1 thus n ( t ) = ( y ( t ) , − x ( t )). We have l f · n d s = ( P ( α ( t )) , Q ( α ( t ))) · ( y ( t ) , − x ( t )) d t = − Q d x + P d y C C 2 Using Green’s Theorem for the function g = ( − Q, P ) ∂P ∂Q + d x d y = − Q d x + P d y R ∂x ∂y C But part(1) above gives us the value of the RHS. Combining, we have ∂P ∂Q + d x d y = f · n d s ∂x ∂y R C 2. (11.22:2) Q can be written as the sum of two functions Q = Q 1 + Q 2 , where 1 Q 1 = − x 2 ye − y 2 and Q 2 = . Thus, the integral to evaluate is x 2 + y 2 P d x + Q 1 d y + Q 2 d y. C Let R 1 be the square {| x | ≤ a, | y | ≤ a } and C its boundary. First note that ∂P ∂Q 1 = − 2 yxe − y 2 = ∂y ∂x Then, by Green’s theorem: ∂P ∂Q 1 P d x + Q 1 d y = − d x d y = 0 . C R 1 ∂y ∂x To compute the remaining part C Q 2 d y , first observe that the integral will certainly be zero along y = ± a as there one has dy ≡ 0. So, we can compute a a dt dt...
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MIT18_024S11_soln-pset10 - Solutions for PSet 10 1(E7:1,2 1...

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