MIT18_024S11_soln-pset9

MIT18_024S11_soln-pset9 - Solutions for PSet 9 1. (11.9:8)...

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Unformatted text preview: Solutions for PSet 9 1. (11.9:8) Using Fubinis Theorem (we assumed that the double integral exists): e tx y t t e tx y d x d y = d x d y = 3 3 [0 ,t ] [1 ,t ] y 1 y 2 t y t t t e y tx e 1 y 3 y d y = d y = 2 x =0 ty 1 t 1 t 1 2 t 1 1 1 1 = e t 1 t 2 e y + + e t 3 ty t 2 y t t 3 t 3 =1 2. (11.15:2) 1 (1 + x ) sin y d x d y = S 1 1+ x (1 + x ) sin y d y d x = x 1 (1 + x )(1 cos(1 + x )) d x = ( x + 2) ( x + 1) sin( x + 1) cos( x + 1) 2 3 = + cos 1 + sin 1 cos 2 2 sin 2 2 3. (11.15:6) The volume can be computed as the double integral of the function 6 2 y f ( x, x y ) = over region S = { ( x, y ) | x 6 , y (6 : 3 x ) / 2 } 6 6 x x 2 y 2 y d y d x = 6 2 6 x d y S 3 d x = 3 6 6 x 6 6 6 x y 2 2 (6 x ) 2 (6 y x ) 3 3 d x = 3 y =0 d x = = 6 12 36 4. (11.15:13) The domain we integrate over is given as x 2 4 S = { 6 x 2 , y 2 x 4 } 1 Observe the points of...
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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-pset9 - Solutions for PSet 9 1. (11.9:8)...

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