Calc III - 5.4-6.2 - Solutions.pdf - Calc III Assignment 10 Solutions 5.4 2 Change the order of integration and evaluate Z 1Z 1 sin(x2 dx dy 0 y

Calc III - 5.4-6.2 - Solutions.pdf - Calc III Assignment 10...

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Calc III Assignment 10 Solutions 5.4 2. Change the order of integration and evaluate: Z 1 0 Z 1 y sin( x 2 ) dx dy. Solution. The region we are integrating over is the set of points ( x, y ) such that 0 y 1 and y x 1. We can instead write the region as 0 y x 1. Here we see that x ranges from 0 to 1 and that y ranges from 0 to x . Since this range for x does not depend at all on y , we can use these as our bounds when we change the order of integration. After changing the order of integration, we get Z 1 0 Z x 0 sin( x 2 ) dy dx. We evaluate the inner integral first. Since sin( x 2 ) is constant in y , we get Z 1 0 x sin( x 2 ) dx. Performing the substitution u = x 2 yields 1 2 Z 1 0 sin( u ) du = 1 2 ( - cos(1) + cos(0)) = 1 2 (1 - cos(1)) . 5.5 16. Evaluate the integral Z 1 0 Z x 0 Z y 0 ( y + xz ) dz dy dx Solution. We evaluate the integrals one at a time, inner to outer. Integrating with 1
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respect to z then y then x gives Z 1 0 Z x 0 Z y 0 ( y + xz ) dz dy dx = Z 1 0 Z x 0 yz + 1 2 xz 2 y z =0 dy dx = Z 1 0 Z x 0 y 2 + 1 2 xy 2 dy dx.
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