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{\bf Question}

Evaluate the triple integral

$$\int \!\!\! \int \!\!\! \int \sqrt{x^2+y^2} \,d(x,y,z)$$

where S is the solid sphere $x^2+y^2+z^2\leq 1$.



{\bf Answer}

Transform the spherical polar coordinates $(r,0,\phi)$ where
$x=r\sin \theta \cos \phi,\ y=r\sin \theta \sin \phi,\ z=r\cos
\theta$ and the sphere is described by the inequalities:
$\left.\begin{array}{ccccl} 0 & \leq & r & \leq & 1\\ 0 & \leq &
\theta & \leq & \pi\\ 0 & \leq & \phi & \leq & 2\pi \end{array}
\right\}$ $\sqrt{x^2+y^2} = \sqrt{r^2\sin^2 \theta(\cos^2
\theta+\sin^2 \phi)}=r\sin \theta$

and $d(x,y,z)=r^2\sin \theta \,dr\,d \theta\,d\phi$.

\begin {eqnarray*} I & = & \int_{\phi=0}^{\phi=2\pi} \! \int_{\theta=0}^{\theta=\pi}
\! \int_{r=0}^{r=1} (r\sin \theta)(r^2 \sin \theta) \,dr \,d\theta
\,d\phi\\ & {} & \rm{and\ since\ the\ limits\ are\ indenpendent\
of}\ r, \theta\ \rm{and}\ \phi: \\ & = & \int_0^{2\pi} \,d\phi \!
\int_0^1 r^3 dr \! \int_0^{\pi} \sin^2 \theta \,d\theta \\ & = &
[\phi]_0^{2\pi}\left[\frac{r^4}{4}\right]_0^1
\int_0^{\pi}\frac{1}{2}(1-cos 2\theta)\,d \theta\ \ [{\rm using}\
2\sin^2 \theta=1-\cos 2\theta]\\ & = &
\frac{\pi}{4}\left[\theta-\frac{1}{2}\sin 2\theta \right]_0^{\pi}
= \frac{\pi}{4}(\pi-0) =\frac{\pi^2}{4}.\end{eqnarray*}



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