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**Unformatted text preview: **Z C ··· = ZZ D 1 dA. Checking the region, using y = x 2 in the equation x = y 2 gives x = x 4 , so x 4-x = x ( x-1)( x 2 + x + 1) = 0 , with solutions x = 0 , 1 . Solving x = y 2 for y = √ x gives the iterated integral Z 1 Z √ x x 2 1 dydx = Z 1 ‡ [ y ] √ x x 2 · dx = Z 1 ( √ x-x 2 ) dx = h 2 3 x 3 2-1 3 x 3 i 1 = 2 3-1 3 = 1 3 . Week 13 Homework: 16.5 Curl and Divergence Problem 16.5.5: Find Div( ± F ) and Curl( ± F ) when ± F = < e x sin y, e x cos y, z > . Solution: For ± F = < P, Q, R >, Div( ± F ) = P x + Q y + R z , so Div( ± F ) = e x sin y-e x sin y + 1 = 1 . Using the cross product formula we have Curl( ± F ) = < R y-Q z , R x-P z , Q x-P y > = < , , > . Notice that if ± F = < f x , f y , f z > is the...

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- Fall '06
- YUKICH
- Integrals, Vector Calculus, Line integral, Stokes' theorem, 2 Week, 3 Week