MIT18_024S11_soln-pset11

# MIT18_024S11_soln-pset11 - √ 2 r respectively...

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Solutions for PSet 11 1. (11.28:14) We substitute u = x y and v = x + y . the Jacobian of the transformation ( x, y ) ( u, v )is 1 1 1 1 ± This has determinant 2 and the vertices of the parallelogram S correspond to π<u<π,π<v< 3 π . Thus: ²² 3 π π y ) 2 sin 2 1 ( x ( x + y ) d x d y = ² ² u 2 sin 2 v d u d v S π π 2 This simpliﬁes to: 1 2 ³ ´ π v sin(2 v ) 3 π ³ u 3 ´ π 4 = 2 4 v = π 3 u = π 3 2. (11.28:16) (a) We can apply Fubini’s Theorem: r ²² r e ( x 2 + y 2 ) ² x 2 ² y 2 2 d x d y = e d x e d y =( I ( r )) R r r (b) C 1 R C 2 and the function e ( x 2 + y 2 ) > 0thus ²² ( x 2 + y 2 ) ²² ( x 2 + y 2 ) ²² ( x 2 + y 2 e d x d y e d x d y = e ) d x d y> 0 R C 1 R \ C 1 and ²² e ( x 2 + y 2 ) d x d y ²² e ( x 2 + y 2 ) d x d y = C 2 R ²² 2 e ( x + y 2 ) d x d y> 0 C 2 \ R Combining the two results: ²² ( x 2 + y 2 ) ²² ( x 2 + y 2 ) ²² ( x 2 2 e d x d y< e d x d y< e + y ) d x d y C 1 R C 2 thus proving the required statement. 1 ±

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(c) For a disc C of radius s 2 π s ( 2 + y 2 ) e x 2 d x d y = e ρ ρ d ρ d ϑ = C 0 0 s 2 1 2 2 π e u d u u = s u s 2 = π e = π (1 e ) 0 2 ± ² u =0 For the circles C 1 (inscribing square of side 2 r )and C 2 (circumscribing square of side 2 r ), the radii are r and
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Unformatted text preview: √ 2 r respectively. Substituting to (b) we get: π (1 − e − r 2 ) < I 2 ( r ) < π (1 − 2 e − 2 r ) as r → ∞ : π ≤ lim I 2 ( r ) ≤ π r →∞ Thus lim r I ( r ) exists and equals ³ ∞ − x 2 e d x = √ π . In other words, →∞ √ −∞ ³ ∞ − x 2 π e d x = . 2 2 MIT OpenCourseWare http://ocw.mit.edu 18.024 Multivariable Calculus with Theory Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms ....
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## This note was uploaded on 01/18/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024S11_soln-pset11 - √ 2 r respectively...

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