rev7 - Stewart Calculus ET 5e 0534393217;15. Multiple...

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± 0 1 ± y 1 x 3 +1 dxdy = ± 0 1 ± 0 x 2 x 3 +1 dydx = ± 0 1 x 3 +1 y y = x 2 y =0 dx = ± 0 1 x 2 x 3 +1 dx = 2 9 ( x 3 +1) 3/2 1 0 = 2 9 2 3/2 ² 1 ( ) ± 0 1 ± arcsin y ± /2 cos x 1+cos 2 x dxdy = ± 0 ± /2 ± 0 sin x cos x 1+cos 2 x dydx = ± 0 ± /2 cos x 1+cos 2 x y y =sin x y =0 dx = ± 0 ± /2 cos x 1+cos 2 x sin xdx Let u =cos x , du = ² sin xdx , dx = du /( ² sin x ) = ± 1 0 ² u 1+ u 2 du = ² 1 3 1+ u 2 ( ) 3/2 0 1 = 1 3 8 ² 1 ( ) = 1 3 2 2 ² 1 ( ) 44. 47. 1 Stewart Calculus ET 5e 0534393217;15. Multiple Integrals; 15.3 Double Integrals over General Regions
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