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I have to evaluate this integral:

$$ \int_{0}^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}}\int_{0}^{\sqrt{4-x^2-y^2}} \sqrt{x^2+y^2+z^2}dzdxdy $$

in spherical coordinates. I see that the region in the xy plane is a circular sector bound by $y=x$ and $y=\sqrt2$ with a radius of 2, I have found that the region in three dimensions becomes complicated to evaluate because of the plane that cuts the spherical sector at y=sqrt(2). I am having trouble finding an expression for $\rho$ or r that describes both the spherical part and the planar part, as well as an $\phi$ that works as well, is see that $ \frac{\pi }{4}\leq \theta \leq \pi $.

Thanks

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    I found a solution: I take the integral of the spherical sector, then I subtract out the area cut off by the plane to get the answer.2012-04-06
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    Perhaps I'm miscalculating but: In the $x$-$y$ plane you have $0\le y\le\sqrt2$ and $y\le x\le\sqrt{4-y^2}$. This region is a quarter-circle in the first quadrant of the $x$-$y$ plane of radius 2 with one side on the positive $x$-axis. No?2012-04-06
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    @David: I think you mean an eighth circle?2012-04-06
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    @joriki Yes, indeed; thanks.2012-04-06

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