Let $\mathbf{F}(x,y,z)=x^{3}\mathbf{i}+y^{3}\mathbf{j}+z^{3}\mathbf{k}$. If $S$ is the surface of the spherical solid $E$ in the first octant enclosed by $z=\sqrt{ 4-x^{2}-y^{2}}$ and $z=0$, and $\mathbf{n}$ is the outer unit normal to $S$, use the Divergence Theorem to find $\iint\limits_S \mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS$.

Figure 69 $dV=\rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi$

Solution See Figure 69. Since ${\rm{div}}\mathbf{F} =3x^{2}+3y^{2}+3z^{2}=3(x^{2}+y^{2}+z^{2})$, we use the Divergence Theorem, obtaining $$\iint\limits_{S}\mathbf{F}\,{\bf\cdot}\, \mathbf{n}\,dS=3\iiint \limits_E(x^{2}+y^{2}+z^{2})\,dV$$

We use spherical coordinates to find the triple integral. Then $x^{2}+y^{2}+z^{2}=\rho ^{2}$ and $dV=\rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi$. The solid $E$ is given by $0\leq \rho \leq 2$, $0\leq \theta \leq \dfrac{\pi}{2}$, $0\leq \phi \leq \dfrac{\pi }{2}$. Then $$\begin{eqnarray*} \iint\limits_{S}\mathbf{F}\,{\cdot}\, \mathbf{n}\,dS &= &3\iiint\limits_E \rho ^{2}( \rho ^{2}\sin \phi \,d\rho \,d\theta \,d\phi ) = 3\int_{0}^{\pi /2}\int_{0}^{\pi /2 }\int_{0}^{2}\rho ^{4}\,\sin \phi \,d\rho \,d\theta \,d\phi \notag \\ &=&3\int_{0}^{\pi /2}\int_{0}^{\pi /2}\left[ \dfrac{\rho ^{5}}{5}\right] _{0}^{2}\sin \phi\, d\theta \,d\phi =3\int_{0}^{\pi /2}\int_{0}^{\pi /2 }\dfrac{ 32}{5}\sin \phi\, d\theta \,d\phi \\ &=&\dfrac{96}{5}\int_{0}^{\pi /2}\big[ \theta \big] _{0}^{\pi /2}\sin \phi \,d\phi \notag \\ &=&\dfrac{48\pi }{5}\int_{0}^{\pi /2}\sin \phi \,d\phi =\dfrac{48\pi }{5} \big[ -\cos \phi \big] _{0}^{\pi /2}=\dfrac{48\pi }{5} \end{eqnarray*}$$