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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 1.54

Let's start by recalling the divergence theorem (also known as The Fundamental Theorem for Divergences):

$${\int }_{\mathcal{V}}(\mathrm{\nabla }\cdot \mathbf{v})d\tau ={\oint }_{\mathcal{S}}\mathbf{v}\cdot d\mathbf{a}$$

In other words, if we want to integrate the divergence of a field over a volume, we could equally compute the (closed) surface integral of the field, where the closed surface is the surface that encloses the aforementioned volume.

We will integrate over an octant of a sphere of radius $R$. Feel free to move around the following 3D render of the octant to get familiar with its geometry.

To begin with the computations, let's take the divergence of $\mathbf{v}$:

$$\begin{array}{rl}\mathrm{\nabla }\cdot \mathbf{v}& =\frac{1}{r^2}\frac{\partial }{\partial r}(r^2{v}_r)+\frac{1}{r\sin \left(\theta \right)}\frac{\partial }{\partial \theta }(\sin \left(\theta \right)v_{\theta })+\frac{1}{r\sin \left(\theta \right)}\frac{\partial v_{\varphi }}{\partial \varphi }\\ & =\frac{1}{r^2}\frac{\partial }{\partial r}(r^2{r}^2\cos \left(\theta \right))+\frac{1}{r\sin \left(\theta \right)}\frac{\partial }{\partial \theta }(\sin \left(\theta \right)r^2\cos \left(\varphi \right))+\frac{1}{r\sin \left(\theta \right)}\frac{\partial }{\partial \varphi }(-r^2\cos \left(\theta \right)\sin \left(\varphi \right))\\ & =\frac{1}{r^2}(4r^3\cos \left(\theta \right))+\frac{1}{r\sin \left(\theta \right)}(r^2\cos \left(\varphi \right)\cos \left(\theta \right))+\frac{1}{r\sin \left(\theta \right)}(-r^2\cos \left(\theta \right)\cos \left(\varphi \right))\\ & =4r\cos \left(\theta \right)+r\cot (\theta )\cos \left(\varphi \right)-r\cot (\theta )\cos \left(\varphi \right)\\ & =4r\cos (\theta )\end{array}$$

The volume integral is then (recall that $d\tau =r^2\sin \left(\theta \right)drd\theta d\varphi$):

$$\begin{array}{rl}{\int }_{\mathcal{V}}(\mathrm{\nabla }\cdot \mathbf{v})d\tau & ={\int }_{\varphi =0}^{\pi /2}{\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^R(4r\cos \left(\theta \right))r^2\sin \left(\theta \right)drd\theta d\varphi \\ & =\frac{\pi R^4}{4}\end{array}$$

If you are not sure how to do the $\theta$-integral, consider using $u$-substitution with $u=\sin \left(\theta \right)$, which will turn your integral into $\int udu$.

Now, we have 4 surfaces to take care of to compute the surface integral.

Let's start by the surface at a constant radius (blue surface in the 3D render above). For it, we have that $d\mathbf{a}=r^2\sin \left(\theta \right)d\theta d\varphi \hat{\mathbf{r}}$. Thus this first surface integral has the value:

$$\begin{array}{rl}{S}_1& ={\int }_{\varphi =0}^{\pi /2}{\int }_{\theta =0}^{\pi /2}\mathbf{v}\cdot (R^2\sin \left(\theta \right)d\theta d\varphi \hat{\mathbf{r}})\\ & ={\int }_{\varphi =0}^{\pi /2}{\int }_{\theta =0}^{\pi /2}(R^2\cos \left(\theta \right))(R^2\sin \left(\theta \right))d\theta d\varphi \\ & ={\int }_{\varphi =0}^{\pi /2}{\int }_{\theta =0}^{\pi /2}{R}^4\cos \left(\theta \right)\sin \left(\theta \right)d\theta d\varphi \\ & =R^4{\int }_{\varphi =0}^{\pi /2}{\int }_{\theta =0}^{\pi /2}\cos \left(\theta \right)\sin \left(\theta \right)d\theta d\varphi \\ & =R^4\frac{1}{2}{\int }_{\varphi =0}^{\pi /2}d\varphi \\ & =\frac{R^4\pi }{4}\end{array}$$

For the surface on the $xy$-plane, we have $d\mathbf{a}=rdrd\varphi \hat{\theta }$, so:

$$\begin{array}{rl}{S}_2& ={\int }_{\varphi =0}^{\pi /2}{\int }_{r=0}^R\mathbf{v}\cdot (rdrd\varphi \hat{\theta })\\ & ={\int }_{\varphi =0}^{\pi /2}{\int }_{r=0}^R{r}^3\cos \left(\varphi \right)drd\varphi \\ & =\frac{R^4}{4}\end{array}$$

We are left with two surfaces at constant $\varphi$, one at $\varphi =0$ and the other at $\varphi =\pi /2$. In both cases, we have $d\mathbf{a}=rdrd\theta \hat{\varphi }$. Let's do the one at $\varphi =0$ first: $$\begin{array}{rl}{S}_3& ={\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^R\mathbf{v}\cdot (rdrd\theta \hat{\varphi })\\ & ={\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^R-r^3\cos \left(\theta \right)\sin \left(0\right)drd\theta \\ & ={\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^{R}0drd\theta \\ & =0\end{array}$$

And the other one:

$$\begin{array}{rl}{S}_4& ={\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^R\mathbf{v}\cdot (rdrd\theta \hat{\varphi })\\ & ={\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^R-r^3\cos \left(\theta \right)\sin \left(\pi /2\right)drd\theta \\ & ={\int }_{\theta =0}^{\pi /2}{\int }_{r=0}^R-r^3\cos \left(\theta \right)drd\theta \\ & =-\frac{R^4}{4}\end{array}$$

Now we can finally write:

$$\begin{array}{rl}{\oint }_{\mathcal{S}}\mathbf{v}\cdot d\mathbf{a}& =S_1+S_2+S_3+S_4\\ & =\frac{R^4\pi }{4}+\frac{R^4}{4}-\frac{R^4}{4}+0\\ & =\frac{R^4\pi }{4}\end{array}$$

Which agrees with the value obtained by integrating the divergence over the octant's volume.

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