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$$ \\ \int_0^2\int_0^y\int_0^{\sqrt{4-y^2}}2xdxdzdy\\ \int_0^2\int_0^y4-y^2dzdy\\ \int_0^2(4-y^2)ydy\\ -\frac{1}{2}\left( \left.4y-\frac{1}{3}y^3\right|_0^2\right )\\ =-\frac{8}{3} $$

I have worked over this problem several times and I cannot find the step where I went wrong.

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1 Answer 1

up vote 3 down vote accepted

The step where you made mistake is in evaluating the integral $\displaystyle \int_0^2 (4-y^2)ydy$ $$\int_0^2 (4-y^2)ydy = \int_0^2 \left( 4y - y^3\right)dy = \left(2y^2 - y^4/4 \right)_{y=0}^{y=2} = 8 - 2^4/4 = 8 - 4 =4$$

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Thank you very much. It's always the stupid mistakes that get me. This is probably a case of too little sleep and too much studying. Thank you again! –  user1405177 Oct 28 '12 at 4:18

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