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Find the volume between two intersecting cylinders x^2+y^2=r^2 and y^2+z^2=r^2 using polar coordinates and double integrals only.

Mathematics
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|dw:1352144686186:dw|
in the xy-plane|dw:1352144919394:dw|\[-\sqrt{r^2-y^2}\le x\le\sqrt{r^2-y^2}\]in the yz-plane|dw:1352145089843:dw|\[-\sqrt{r^2-y^2}\le z\le\sqrt{r^2-y^2}\]leaving\[-r\le y\le r\]hm...
in the xz-plane the intersection is square with sides=2r|dw:1352145471598:dw|

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Other answers:

I think the integral is\[\iint\limits_Dr\sin\theta dA=\int_0^{2\pi}\int_0^r r^2\sin\theta drd\theta\]can you check somewhow?
actually it would have to be\[\iint\limits_Dr\sin\theta dA=2\int_0^\pi\int_0^r r^2\sin\theta drd\theta\]to avoid getting zero
But the integrand should be r^2 * cos(theta) ?
since i am integrating z = sqrt(z^2-y^2)
so in polar i thought it was sqrt(r^2-y^2) = x = rcos(theta)
yeah I just came to that conclusion as well
I guess I was right the first time
but if I integrate with rcos(theta), I end up with zero in the end
I will get sin(2*pi) - sin(0) = 0.
Hm.. now I am not sure again. No way to check I presume...
@mahmit2012 double integral help

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