Finding the Mass of a Solid in 3d

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Homework Statement


Find the mass of the solid bounded by the cylinder x^2+y^2=2x and the cone z^2=x^2+y^2 if the density is \delta = \sqrt{x^2+y^2}

[b2. The attempt at a solution[/b]
I had some trouble looking at how to set up the limits on this integral. What I came up with was:
2 \int_0^2 \int_0^{\sqrt{2x^2+2x}} \int_0^{\sqrt{x^2+y^2}} \sqrt{x^2+y^2} dzdydx
= 2 \int_0^2 x^2\sqrt{x^2+y^2} + ((\sqrt{x^2+y^2})^3)/3 dx
And this is just an ugly integral. I tried doing it in cylindrical coordinates but it wasn't working out terribly well that way either. Any hints? Thanks!
 
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Nevermind, I got an answer. Can anyone confirm 6\pi?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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