MIT18_024S11_soln-pset10

MIT18_024S11_soln-pset10 - Solutions for PSet 10 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: ( t ) = ( x ( t ) , y ( t )). Then ( t ) = 1 thus n ( t ) = ( y ( t ) , x ( t )). We have l f n d s = ( P ( ( t )) , Q ( ( t ))) ( y ( t ) , x ( t )) d t = Q d x + P d y C C 2 Using Greens 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 Greens 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....

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