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Good night, i have a problem with this integral, please help me with the integration limits.

\begin{align} \int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{-\sqrt{1-x^{2}-y^{2}}}^{\sqrt{1-x^{2}-y^{2}}}\left(x^{2}+y^{2}+z^{2}\right)^{3/2}dzdydx &= \int_{0}^{2\pi}\int_{0}^{\pi}\int_{-\cos\alpha}^{\cos\alpha} (r^{5} sen\alpha) {\rm d}r {\rm d}\alpha {\rm d}\beta \\ &= 1/3\,\int_{0}^{2\,\pi }\!\int_{0}^{\pi }\! \left( \sin x \cos x \right) ^{6}\,{\rm d} x\,{\rm d}y \\ &= \frac{4\pi}{21} \\ \end{align}

But the answer is $\frac{2\pi}{3}$, please help me.

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  • $\begingroup$ The limit on the radial variable are from $0$ to $1$. What is $\alpha$? $\endgroup$ – Mark Viola May 19 '16 at 3:17
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Transforming to spherical coordinates, we find that

$$\begin{align} \int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{-\sqrt{1-x^{2}-y^{2}}}^{\sqrt{1-x^{2}-y^{2}}}\left(x^{2}+y^{2}+z^{2}\right)^{3/2}dzdydx &=\int_0^{2\pi}\int_0^\pi\int_0^1 r^5\,dr\sin(\theta)\,d\theta\,d\phi\\\\&=(2\pi)(2)\left(\frac16\right)\\\\ &=\frac{2\pi}{3}\end{align}$$

as expected!

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  • $\begingroup$ Thanks man but i have a question, how you can find the radial $0<r<1$? and $\alpha=\phi$ $\endgroup$ – Bvss12 May 19 '16 at 3:24
  • $\begingroup$ The integration region is a unit sphere $x^2+y^2+z^2=r^2=1$. $\endgroup$ – Mark Viola May 19 '16 at 3:29
  • $\begingroup$ And you're welcome. My pleasure. -Mark $\endgroup$ – Mark Viola May 19 '16 at 3:29

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