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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 1.49

As suggested by example 1.16 (page 49), there are two ways to solve this problem.

1. Short way: Using the Dirac Delta

Let's start by recalling

$$\begin{array}{}\text{(Eq. 1.99, page 48)}& \mathrm{\nabla }\cdot \left(\frac{\hat{\mathbf{r}}}{r^2}\right)=4\pi {\delta }^3(\mathbf{r})\end{array}$$

Let's use this formula on the integrand:

$$\begin{array}{rl}J& ={\int }_{\mathcal{V}}{e}^{-r}(\mathrm{\nabla }\cdot \frac{\hat{\mathbf{r}}}{r^2})d\tau \\ & ={\int }_{\mathcal{V}}{e}^{-r}(4\pi {\delta }^3(\mathbf{r}))d\tau \end{array}$$

To evaluate this, we recall:

$$\begin{array}{}\text{(Eq. 1.98, page 48)}& \int f(\mathbf{r}){\delta }^3(\mathbf{r}-\mathbf{a})d\tau =f(\mathbf{a})\end{array}$$

So

$$\begin{array}{rl}J& ={\int }_{\mathcal{V}}{e}^{-r}(4\pi {\delta }^3(\mathbf{r}))d\tau \\ & =4\pi {\int }_{\mathcal{V}}{e}^{-r}{\delta }^3(\mathbf{r})d\tau \\ & =4\pi e^{-r}{|}_{r=0}\\ & =4\pi \end{array}$$

2. Long way: Integration by parts

Let's start by recalling

$$\begin{array}{}\text{(Eq. 1.59, page 36)}& {\int }_{\mathcal{V}}f\cdot (\mathrm{\nabla }\cdot \mathbf{A})d\tau =-{\int }_{\mathcal{V}}\mathbf{A}\cdot \mathrm{\nabla }fd\tau +{\oint }_{\mathcal{S}}f\mathbf{A}\cdot d\mathbf{a}\end{array}$$

That is, we will transfer $\mathrm{\nabla }$ from the vector function $\hat{\mathbf{r}}/r^2$ to the scalar function $e^{-r}$. This will avoid the use of the Dirac Delta, but it picks up a new term (a surface integral)!

In our current problem:

$$\begin{array}{rl}f& =e^{-r}\\ \mathbf{A}& =\frac{\hat{\mathbf{r}}}{r^2}\end{array}$$

To compute the integral, let's remind ourselves the required integration elements:

$$\begin{array}{}\text{(As seen on the cover of your book)}& d\tau & =r^2\sin \left(\theta \right)drd\theta d\varphi \text{(At constant radius)}& d\mathbf{a}& =R^2\sin \left(\theta \right)d\theta d\varphi \hat{\mathbf{r}}\end{array}$$

Important remark: The vector area element won't always be that one. See page 39 for more information.

To calculate the volume integral we will need the gradient of $f=e^{-r}$. We recall that in spherical coordinates:

$$\begin{array}{}\text{(As seen on the cover of your book)}& \mathrm{\nabla }f=\frac{\partial f}{\partial r}\hat{\mathbf{r}}+\frac{1}{r}\frac{\partial f}{\partial \theta }\hat{\theta }+\frac{1}{r\sin \theta }\frac{\partial f}{\partial \varphi }\hat{\varphi }\end{array}$$

It is clear that the last 2 terms vanish, and we are left with

$$\mathrm{\nabla }f=-e^{-r}\hat{\mathbf{r}}$$

We are now ready to evaluate the integral $J$:

$$\begin{array}{rl}J& ={\int }_{\mathcal{V}}{e}^{-r}(\mathrm{\nabla }\cdot \frac{\hat{\mathbf{r}}}{r^2})d\tau \\ & =-{\int }_{\mathcal{V}}\left(\frac{\hat{\mathbf{r}}}{r^2}\right)\cdot (-e^{-r}\hat{\mathbf{r}})d\tau +{\oint }_{\mathcal{S}}{e}^{-r}\left(\frac{\hat{\mathbf{r}}}{r^2}\right)\cdot d\mathbf{a}\\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }{\int }_{r=0}^R\frac{e^{-r}}{\cancel{r^2}}(\cancel{r^2}\sin \left(\theta \right))drd\theta d\varphi +{\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\frac{e^{-R}\cancel{R^2}}{\cancel{R^2}}\sin \left(\theta \right)d\theta d\varphi \\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }{\int }_{r=0}^R{e}^{-r}\sin \left(\theta \right)drd\theta d\varphi +e^{-R}{\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\sin \left(\theta \right)d\theta d\varphi \\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }(1-e^{-R})\sin \left(\theta \right)d\theta d\varphi +e^{-R}{\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\sin \left(\theta \right)d\theta d\varphi \\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\sin \left(\theta \right)d\theta d\varphi -\cancel{e^{-R}{\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\sin \left(\theta \right)d\theta d\varphi }+\cancel{e^{-R}{\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\sin \left(\theta \right)d\theta d\varphi }\\ & ={\int }_{\varphi =0}^{2\pi }{\int }_{\theta =0}^{\pi }\sin \left(\theta \right)d\theta d\varphi \\ & ={\int }_{\varphi =0}^{2\pi }2d\varphi \\ & =4\pi \end{array}$$ We get (as expected) the same answer as with method 1. Still, method 2 took considerably more steps than the first.

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