Evaluate the integral

$\mathit{J}\mathbf{=}{\mathbf{\int }}_{\mathbf{v}}{\mathit{e}}^{\mathbf{-}\mathbf{r}}\mathbf{(}\mathbf{\nabla }\mathbf{·}\frac{\hat{\mathbf{r}}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{)}\mathit{d}\mathit{\tau }$,

where V is a sphere of radius R centered at origin by two different methods as in Ex. 1.16..

Write the given integral.

$J={\int }_{\mathrm{v}}{e}^{-\mathrm{r}}(\nabla ·\frac{\hat{\mathrm{r}}}{{\mathrm{r}}^2})d\tau$,

Here, v is a sphere of radius Rcentred at the origin.

According to the Gauss divergence theorem, the integral of the derivative of a function $\mathit{f}\left(x,y,z\right)$over an open surface area is equal to the volume integral of the function$\mathbf{\int }\left(\nabla ·v\right)\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}{\mathbf{\oint }}_{\mathbf{s}}\mathit{v}\mathbf{·}\mathit{d}\mathit{s}$. The product rule is given by $\mathit{f}\left(\nabla ·A\right)\mathbf{=}\mathbf{\nabla }\mathbf{·}\mathit{f}\mathit{A}\mathbf{-}\mathit{A}\mathbf{·}\mathbf{\nabla }\mathit{f}$.

The second method is by using the Dirac delta function.

In the integral $J={\int }_v{e}^{-r}\left(\nabla ·\frac{\hat{r}}{r^2}\right)d\tau ,e^{-r}\left(\nabla ·\frac{\hat{r}}{r^2}\right)$is in the form of $f\left(\nabla ·A\right)$. On applying the rule, ${\int }_{v}f\left(\nabla ·A\right)d\tau =-{\int }_{v}A·\left(\nabla f\right)d\tau +{\oint }_{s}fA·da.$.

Apply the product rulelocalid="1657518121843" $f\left(\nabla ·A\right)=\nabla fA-A·\nabla f$ to ${\int }_{v}f\left(\nabla ·A\right)d\tau =$$-{\int }_{v}A·\left(\nabla f\right)d\tau +{\oint }_{s}fA·da$as follows:

$$\begin{gathered} {\int }_v\left(\nabla ·fA-A·\nabla f\right)d\tau =-{\int }_{v}A·\left(\nabla f\right)d\tau +{\oint }_{s}fA·da \\ {\int }_v\nabla ·fAd\tau -{\int }_{v}A·\nabla fd\tau =-{\int }_{v}A·\left(\nabla f\right)d\tau +{\oint }_{s}fA·da \\ {\int }_v\nabla ·fAd\tau ={\oint }_{s}fA·da \end{gathered}$$

Apply the Gauss divergence theorem to ${\int }_v\nabla ·fAd\tau ={\oint }_{s}fA·da$.

$$\begin{gathered} J={\int }_v\left\{\frac{\hat{r}}{r^2}\left(\nabla ·e^{-r}\right)\right\}d\tau +\oint \left(\frac{e^{-r}}{r^2}\hat{r}\right)·da \\ ={\int }_v\frac{1}{r^2}\hat{r}·\left(-e^{-r}\right)\hat{r}d\tau +\oint \left(\frac{e^{-r}}{r^2}\hat{r}\right)·da \end{gathered}$$

The differential volume and area for a sphere of radius R is

$d\tau =r^2\sin \theta drd\theta d\varphi$and $da=r^2\sin \theta drd\theta d\varphi \hat{r}$.

The integral can now be calculated as follows:

$$\begin{gathered} J={\int }_v\left\{\frac{\hat{r}}{r^2}\left(\nabla ·e^{-r}\right)\right\}{r}^2\sin \theta drd\theta d\varphi +\oint \left(\frac{e^{-r}}{r^2}\hat{r}\right)·r^2\sin \theta d\theta d\varphi \hat{r} \\ ={\int }_v\left\{\frac{\hat{r}}{r^2}\left(\nabla ·e^{-r}\right)\right\}{r}^2\sin \theta drd\theta d\varphi +{\oint }_s{e}^{-r}\sin \theta d\theta d\varphi \\ ={\int }_0^R{e}^{-r}dr{\int }_0^{\mathrm{\pi }}\sin \theta d\theta {\int }_0^{2\mathrm{\pi }}d\varphi +\left(\begin{array}{c}{e}^{-R}{\left|-\cos \theta \right|}_0^{\mathrm{\pi }}\end{array}{\left|\varphi \right|}_0^{2\mathrm{\pi }}\right) \\ =4\mathrm{\pi }\left(1-{\mathrm{e}}^{-\mathrm{R}}\right)+4\mathrm{\pi }\left({\mathrm{e}}^{-\mathrm{R}}\right) \end{gathered}$$

Solving further

$J=4\mathrm{\pi }$

The value of the integral can be computed using the result $\nabla ·\frac{\hat{r}}{r^2}={\delta }^3\left(r\right)$.

role="math" localid="1657519533540" $$\begin{gathered} J={\int }_v{e}^{-r}(\nabla ·\frac{\hat{r}}{r^2})d\tau \\ ={\int }_v{e}^{-r}4\pi {\delta }^3\left(r\right)d\tau \\ =4\pi {\int }_v{e}^{-r}{\delta }^3\left(r\right)d\tau \\ =4\pi \end{gathered}$$

Thus, the value of integral $J={\int }_v{e}^{-r}(\nabla ·\frac{\hat{r}}{r^2})d\tau$is $4\pi$.