# Problem with integration limits using spherical substitution

\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.

• The limit on the radial variable are from $0$ to $1$. What is $\alpha$? – Mark Viola May 19 '16 at 3:17

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