Chapter 1 Vector Analysis: Problem 1.54

Problem 1.54: Check the divergence theorem for the function
$$v=r^2\cos \theta \hat{r}+r^2\cos \varphi \hat{\theta }-r^2\cos \theta \sin \varphi \hat{\varphi }$$ ,
using as your volume one octant of the sphere of radius $R$ (Fig. 1.48). Make sure
you include the entire surface. 

 

 

Solution: 

Divergence Theorem States That for a verctor $\overrightarrow{v}$,

$${\int }_V(\overrightarrow{▽}\cdot \overrightarrow{v})d\tau ={\oint }_S\overrightarrow{v}\cdot d\overrightarrow{a}$$

Let's consider the octant where $r$ varies from $0$ to $R$, $\theta$ varies from $0$ to $\frac{\pi }{2}$ and $\varphi$ varies from $0$ to $\frac{\pi }{2}$ 

Given $$\overrightarrow{v}=r^{2}cos\theta \hat{r}+r^{2}cos\varphi \hat{\theta }-r^{2}cos\theta sin\varphi \hat{\varphi }$$ $$\overrightarrow{▽}\cdot \overrightarrow{v}=\frac{1}{r^{2}sin\theta }[\frac{\partial }{\partial r}(r^{2}sin\theta \cdot r^{2}cos\theta )+\frac{\partial }{\partial \theta }(rsin\theta \cdot r^{2}cos\varphi )+\frac{\partial }{\partial \varphi }(r\cdot (-r^{2}cos\theta sin\varphi ))]$$ $$\overrightarrow{▽}\cdot \overrightarrow{v}=\frac{1}{r^{2}sin\theta }[sin\theta cos\theta 4r^3+r^{3}cos\varphi cos\theta -r^{3}cos\theta cos\varphi ]$$ $$\overrightarrow{▽}\cdot \overrightarrow{v}=4rcos\theta$$ $${\int }_{vol.}(\overrightarrow{▽}\cdot \overrightarrow{v})d\tau ={\int }_{vol.}4rcos\theta \cdot r^{2}sin\theta drd\theta d\varphi$$ $${\int }_{vol.}(\overrightarrow{▽}\cdot \overrightarrow{v})d\tau =4\times {\int }_0^R{r}^{3}dr\times {\int }_0^{\frac{\pi }{2}}cos\theta sin\theta d\theta \times {\int }_0^{\frac{\pi }{2}}d\varphi$$ $${\int }_{vol.}(\overrightarrow{▽}\cdot \overrightarrow{v})d\tau =4\times \frac{R^4}{4}\times \frac{1}{2}\times \frac{\pi }{2}=\frac{1}{4}\pi R^4$$ Divide the surface into 4 parts where $S_1$ represents the part on the on the $xy$ plane. Hence $\theta =\pi /2$ $${\int }_{S_1}\overrightarrow{v}\cdot \overrightarrow{da}={\int }_{S_1}(r^{2}cos\varphi \hat{\theta })\cdot (rdrd\varphi \hat{\theta })$$ $${\int }_{S_1}\overrightarrow{v}\cdot \overrightarrow{da}={\int }_0^R{r}^{3}dr{\int }_0^{\pi /2}cos\varphi d\varphi$$ $${\int }_{S_1}\overrightarrow{v}\cdot \overrightarrow{da}=\frac{R^4}{4}\times 1=\frac{1}{4}{R}^4$$ $S_2$ represents the part on the on the $yz$ plane. Hence $\varphi =\pi /2$ $${\int }_{S_2}\overrightarrow{v}\cdot \overrightarrow{da}={\int }_{S_2}(r^{2}cos\theta \hat{r}-r^{2}cos\theta \hat{\varphi })\cdot (rdrd\theta \hat{\varphi })$$ $${\int }_{S_2}\overrightarrow{v}\cdot \overrightarrow{da}={\int }_{S_2}-r^{3}cos\theta drd\theta$$ $${\int }_{S_2}\overrightarrow{v}\cdot \overrightarrow{da}=-{\int }_0^R{r}^{3}dr{\int }_0^{\pi /2}cos\theta d\theta$$ $${\int }_{S_2}\overrightarrow{v}\cdot \overrightarrow{da}=-\frac{R^4}{4}\times 1=-\frac{1}{4}{R}^4$$ $S_3$ represents the part on the on the $xz$ plane. Hence $\varphi =0$ $${\int }_{S_3}\overrightarrow{v}\cdot \overrightarrow{da}={\int }_{S_3}(r^{2}cos\theta \hat{r}+r^{2}cos\varphi \hat{\theta })\cdot (-rdrd\theta \hat{\varphi })=0$$ $S_4$ represents the curved part of the sphere . Hence $r=R$ $${\int }_{S_4}\overrightarrow{v}\cdot \overrightarrow{da}={\int }_{S_4}(R^{2}cos\theta \hat{r}+R^{2}cos\varphi \hat{\theta }-R^{2}cos\theta sin\varphi \hat{\varphi })\cdot (R^{2}sin\theta d\theta d\varphi \hat{r})$$ $${\int }_{S_4}\overrightarrow{v}\cdot \overrightarrow{da}={\int }_{S_4}(R^{4}cos\theta sin\theta d\theta d\varphi )$$ $${\int }_{S_4}\overrightarrow{v}\cdot \overrightarrow{da}=R^4{\int }_0^{\pi /2}cos\theta sin\theta d\theta {\int }_0^{\pi /2}d\varphi$$ $${\int }_{S_4}\overrightarrow{v}\cdot \overrightarrow{da}=R^4\times \frac{1}{2}\times \frac{\pi }{2}=\frac{1}{4}\pi R^4$$ Hence,$${\oint }_S\overrightarrow{v}\cdot \overrightarrow{da}={\int }_{S_1}\overrightarrow{v}\cdot \overrightarrow{da}+{\int }_{S_2}\overrightarrow{v}\cdot \overrightarrow{da}+{\int }_{S_3}\overrightarrow{v}\cdot \overrightarrow{da}+{\int }_{S_4}\overrightarrow{v}\cdot \overrightarrow{da}$$ $${\oint }_S\overrightarrow{v}\cdot \overrightarrow{da}=\frac{1}{4}\pi R^4$$ Therefore, $${\int }_{vol.}(\overrightarrow{▽}\cdot \overrightarrow{v})d\tau ={\oint }_S\overrightarrow{v}\cdot \overrightarrow{da}$$ 

Hence divergence theorem is verified

 

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