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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 Fubini’s 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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MIT18_024S11_soln-pset9 - Solutions for PSet 9 1(11.9:8...

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