VC5.49

$\iint_S(x^3dy dz+y^3 dz dx+z^3 dx dy)\,$

= $\iiint_V[\frac{\partial}{\partial x}(x^3)+\frac{\partial}{\partial y}(y^3)+\frac{\partial}{\partial z}(z^3)]dx dy dz\,$ by divergence theorm.

= $3\iiint_V(x^2+y^2+z^2)dx dy dz\,$ where V is the volume bounded by the given sphere.

We use the spherical polar coordinats $(r,\theta,\phi)\,$ so that

$x=r\sin\theta\cos\phi,y=r\sin\theta\sin\phi,z=r\cos\theta\,$

and $dx dy dz=dV=(dr)(r d\theta)(r\sin\theta d\theta)=r^2\sin\theta dr d\theta d\phi\,$

Also, $x^2+y^2+z^2=r^2\,$

The limits are r=0 to r=1, $\theta=0\,$ to $\pi\,$ and those of $\phi\,$ are from 0 to $2\pi\,$

Therefore, $\iiint_V(x^2+y^2+z^2)dx dy dz\,$

= $\int_0^1\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi} r^2(r^2\sin\theta)dr d\theta d\phi\,$

= $\int_0^1\int_{\theta=0}^{\pi}r^4\sin\theta[\phi]_{0}^{2\pi}dr d\theta\,$

= $2\pi\int_0^1\int_{0}^{\pi}r^4\sin\theta dr d\theta\,$

= $2\pi\int_0^1 r^4[-\cos\theta]_{0}^{\pi}dr=4\pi\int_0^1 r^4 dr=4\pi[\frac{r^5}{5}]_0^1=\frac{4\pi}{5}\,$

From (1), $\iint_S(x^3dy dz+y^3 dz dx+z^3 dx dy)=3[\frac{4\pi}{5}]=\frac{12\pi}{5}\,$