91Ó°ÊÓ

Write the given integral.

$J={∫}_{\mathrm{v}}{e}^{-\mathrm{r}}(∇·\frac{\hat{\mathrm{r}}}{{\mathrm{r}}^2})d\mathrm{τ}$,

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$\mathbf{∫}\left(∇·v\right)\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}{\mathbf{âˆ\circledR }}_{\mathbf{s}}\mathit{v}\mathbf{·}\mathit{d}\mathit{s}$. The product rule is given by $\mathit{f}\left(∇·A\right)\mathbf{=}\mathbf{∇}\mathbf{·}\mathit{f}\mathit{A}\mathbf{-}\mathit{A}\mathbf{·}\mathbf{∇}\mathit{f}$.

The second method is by using the Dirac delta function.

In the integral $J={∫}_v{e}^{-r}\left(∇·\frac{\hat{r}}{r^2}\right)d\mathrm{Ï„},e^{-r}\left(∇·\frac{\hat{r}}{r^2}\right)$is in the form of $f\left(∇·A\right)$. On applying the rule, ${∫}_{v}f\left(∇·A\right)d\mathrm{Ï„}=-{∫}_{v}A·\left(∇f\right)d\mathrm{Ï„}+{âˆ\circledR }_{s}fA·da.$.

Apply the product rulelocalid="1657518121843" $f\left(∇·A\right)=∇fA-A·∇f$ to ${∫}_{v}f\left(∇·A\right)d\mathrm{Ï„}=$$-{∫}_{v}A·\left(∇f\right)d\mathrm{Ï„}+{âˆ\circledR }_{s}fA·da$as follows:

$$\begin{gathered} {∫}_v\left(∇·fA-A·∇f\right)d\mathrm{Ï„}=-{∫}_{v}A·\left(∇f\right)d\mathrm{Ï„}+{âˆ\circledR }_{s}fA·da \\ {∫}_v∇·fAd\mathrm{Ï„}-{∫}_{v}A·∇fd\mathrm{Ï„}=-{∫}_{v}A·\left(∇f\right)d\mathrm{Ï„}+{âˆ\circledR }_{s}fA·da \\ {∫}_v∇·fAd\mathrm{Ï„}={âˆ\circledR }_{s}fA·da \end{gathered}$$

Apply the Gauss divergence theorem to ${∫}_v∇·fAd\mathrm{Ï„}={âˆ\circledR }_{s}fA·da$.

$$\begin{gathered} J={∫}_v\left\{\frac{\hat{r}}{r^2}\left(∇·e^{-r}\right)\right\}d\mathrm{Ï„}+âˆ\circledR \left(\frac{e^{-r}}{r^2}\hat{r}\right)·da \\ ={∫}_v\frac{1}{r^2}\hat{r}·\left(-e^{-r}\right)\hat{r}d\mathrm{Ï„}+âˆ\circledR \left(\frac{e^{-r}}{r^2}\hat{r}\right)·da \end{gathered}$$

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

$d\mathrm{τ}=r^2\sin \mathrm{θ}drd\mathrm{θ}d\mathrm{ϕ}$and $da=r^2\sin \mathrm{θ}drd\mathrm{θ}d\mathrm{ϕ}\hat{r}$.

The integral can now be calculated as follows:

$$\begin{gathered} J={∫}_v\left\{\frac{\hat{r}}{r^2}\left(∇·e^{-r}\right)\right\}{r}^2\sin \mathrm{θ}drd\mathrm{θ}d\mathrm{Ï•}+âˆ\circledR \left(\frac{e^{-r}}{r^2}\hat{r}\right)·r^2\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\hat{r} \\ ={∫}_v\left\{\frac{\hat{r}}{r^2}\left(∇·e^{-r}\right)\right\}{r}^2\sin \mathrm{θ}drd\mathrm{θ}d\mathrm{Ï•}+{âˆ\circledR }_s{e}^{-r}\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•} \\ ={∫}_0^R{e}^{-r}dr{∫}_0^{\mathrm{Ï€}}\sin \mathrm{θ}d\mathrm{θ}{∫}_0^{2\mathrm{Ï€}}d\mathrm{Ï•}+\left(\begin{array}{c}{e}^{-R}{\left|-\cos \mathrm{θ}\right|}_0^{\mathrm{Ï€}}\end{array}{\left|\mathrm{Ï•}\right|}_0^{2\mathrm{Ï€}}\right) \\ =4\mathrm{Ï€}\left(1-{\mathrm{e}}^{-\mathrm{R}}\right)+4\mathrm{Ï€}\left({\mathrm{e}}^{-\mathrm{R}}\right) \end{gathered}$$

Solving further

$J=4\mathrm{Ï€}$

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

role="math" localid="1657519533540" $$\begin{gathered} J={∫}_v{e}^{-r}(∇·\frac{\hat{r}}{r^2})d\mathrm{τ} \\ ={∫}_v{e}^{-r}4\mathrm{π}{\mathrm{δ}}^3\left(r\right)d\mathrm{τ} \\ =4\mathrm{π}{∫}_v{e}^{-r}{\mathrm{δ}}^3\left(r\right)d\mathrm{τ} \\ =4\mathrm{π} \end{gathered}$$

Thus, the value of integral $J={∫}_v{e}^{-r}(∇·\frac{\hat{r}}{r^2})d\mathrm{τ}$is $4\mathrm{π}$.