Math 293 prelim1solution

Math 293 prelim1solution - Math 293 1 In the integral...

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Math 293 Solutions to Prelim #1 Spring 2000 y = x x 1 (1,1) y 3 1 . In the integral R 1 0 R 1 y 1 / 3 1 x 4 +1 dx dy the limits of integration say that x goes from x = 0 to x = y 1 3 , and then y goes from 0 to 1. Solving x = y 1 3 for y gives y = x 3 . So the region of integration is as shown in the figure. Reversing the order of integration, we get R 1 0 R x 3 0 1 x 4 +1 dy dx . This integral is easier to evaluate than the first one. We get R 1 0 R x 3 0 1 x 4 +1 dy dx = R 1 0 ( y x 4 +1 fl fl x 3 0 ) dx = R 1 0 x 3 x 4 +1 dx = 1 4 ln( x 4 + 1) fl fl 1 0 = 1 4 ln 2. x z y 1 2 . It would be better not to integrate dz first since this would require two separate integrals, one for the volume below the flat top and the other for the volume below the sloping piece. So let’s try the order dx dy dz . (Or dy dx dz would also work.) We get R 1 0 R 2 - z 0 R 2 - y - z 0 dx dy dz = R 1 0 R 2 - z 0 (2 - y - z ) dy dz = R 1 0 ( 2 y - y 2 2 - yz fl fl 2 - z 0 ) = R 1 0 (2 - 2 z + z 2 2 ) dz = 2 z - z 2 + z 3 6 fl fl 1 0 = 7 6 . x z y 3 . (a) We can compute the volume for the upper half of the solid and then double this.
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