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Find the volume of a solid bounded by: $z=0$, $x^2+2y^2=2$, and $x+y+2z=2$
I got this triple integral: $$\int_{-1}^1\int_{-\sqrt{2-2y^2}}^\sqrt{2-2y^2}\int_0^{1-x/2-y/2}dzdxdy$$
I think it's wrong because I keep getting a negative value. I'd appreciate any help, thanks!

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Your integral is correct but there may be some computational mistakes (me too in the following). The integral reduces to a double integral $$\int_{-1}^1\int_{-\sqrt{2-2y^2}}^\sqrt{2-2y^2}(1-x/2-y/2)dxdy=\int\int_{D}(1-x/2-y/2)dxdy,$$ where $D: x^2+2y^2\leq 2$. To evaluate this integral we can make use of change of variables: first let $x=\sqrt{2}u,\, y=v$, so the Jacobian $J=\sqrt{2}$. Then $\int\int_{D}(1-x/2-y/2)dxdy=\displaystyle\int\int_{u^2+v^2\leq 1}\sqrt{2}\big(1-\sqrt{2}u/2-v/2\big)dudv$. Now we use the polar coordinates $u=r\cos\theta$, $v=r\sin\theta$, where $0\leq r\leq 1$ and $0\leq\theta\leq 2\pi$. Thus we have $\displaystyle\int\int_{u^2+v^2\leq 1}\sqrt{2}\big(1-\sqrt{2}u/2-v/2\big)dudv=\displaystyle\int_0^{2\pi}\int_0^1\sqrt{2}\big(1-\sqrt{2}r\cos\theta/2-r\sin \theta/2\big)rdrd\theta=\sqrt{2}\pi.$

I dont claim this is the shortest solution.

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  • $\begingroup$ Thanks a lot! I kept trying to solve this directly without the change of variables or use of polar coordinates. The above method is definitely better. $\endgroup$ – kunov Nov 10 '13 at 16:47

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