Check the divergence theorem for the function
v=r^2\sin \theta \hat{r}+4r^2\cos \theta \hat{\pmb{\theta} } +r^2\tan \theta \hat{\pmb{\phi} } ,
using the volume of the “ice-cream cone” shown in Fig. 1.52 (the top surface is spherical, with radius R and centered at the origin). [Answer: (\pi R^4/12)\left(2\pi +3\sqrt{3} \right) ]
∇\cdot v=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2r^2\sin \theta \right) +\frac{1}{r\sin \theta }\frac{\partial}{\partial \theta }\left(\sin \theta 4r^2\cos \theta \right)+\frac{1}{r\sin \theta }\frac{\partial}{\partial \phi }\left(r^2\tan \theta \right) .
=\frac{1}{r^2}4r^3\sin \theta +\frac{1}{r\sin \theta } 4r^2\left(\cos ^2\theta -\sin ^2\theta \right)=\frac{4r}{\sin \theta } \left(\sin ^2\theta +\cos ^2\theta -\sin ^2\theta \right)
=4r\frac{\cos ^2\theta }{\sin \theta } .
\int{}\left(∇\cdot v\right)d\tau =\int{}\left(4r\frac{\cos ^2\theta }{\sin \theta } \right)\left(r^2\sin \theta drd\theta d\phi \right)=\int\limits_{0}^{R}{}4r^3dr\int\limits_{0}^{\pi /6}{}\cos ^2\theta d\theta \int\limits_{0}^{2\pi }{d\phi }=\left(R^4\right)(2\pi )\left[\frac{\theta }{2}+\frac{\sin 2\theta }{4} \right] \mid ^{\pi /6}_{0}=2\pi R^4\left(\frac{\pi }{12}+\frac{\sin 60°}{4} \right)=\frac{\pi R^4}{6}\left(\pi +3\frac{\sqrt{3} }{2} \right) =\frac{\pi R^4}{12} \left(2\pi +3\sqrt{3} \right) .
Surface coinsists of two parts:
(1) The ice cream: r = R;\phi :0\rightarrow 2\pi ;\theta :0\rightarrow \pi /6;da = R^2\sin \theta d\theta d\phi \hat{r} ;v·da =\left(R^2\sin \theta \right)\left(R^2\sin \theta d\theta d\phi \right)=R^4\sin ^2\theta d\theta d\phi .
\int{} v·da = R^4\int\limits_{0}^{\pi /6}{}\sin ^2\theta d\theta \int\limits_{0}^{2\pi }{}d\phi = \left(R^4\right) \left(2\pi \right)\left[\frac{1}{2}\theta -\frac{1}{4} \sin 2\theta \right]^{\pi /6}_{0}=2\pi R^4 \left(\frac{\pi }{12}-\frac{1}{4}\sin 60° \right)=\frac{\pi R^4}{6}\left(\pi -3\frac{\sqrt{3} }{2} \right)(2) The cone: \theta =\frac{\pi }{6};\phi :0\rightarrow 2\pi ;r:0\rightarrow R;da=r\sin \theta d\phi dr\hat{\pmb{\theta} }=\frac{\sqrt{3} }{2}rdrd\phi \hat{\pmb{\theta} } ; v · da =\sqrt{3} r^3drd\phi
\int{} v · da =\sqrt{3}\int\limits_{0}^{R}{}r^3 dr \int\limits_{0}^{2\pi }{}d\phi =\sqrt{3}\cdot \frac{R^4}{4}\cdot 2\pi =\frac{\sqrt{3} }{2} \pi R^4 .
Therefore \int{}v · da =\frac{\pi R^4}{2} \left(\frac{\pi }{3}-\frac{\sqrt{3} }{2}+\sqrt{3} \right)=\frac{\pi R^4}{12} \left(2\pi +3\sqrt{3} \right) .