Question

The volume integral \(\displaystyle \int_V e^{-\left(\frac{r}{R}\right)^2} \vec{\nabla} \cdot \left(\frac{\hat{r}}{r^2}\right) d^3r\), where \(V\) is the volume of a sphere of radius \(R\) centered at the origin, is equal to:
 

• A. \(4\pi\)
• B. \(0\)
• C. \(\dfrac{4}{3}\pi R^3\)
• D. \(1\)

Hint:

In spherical symmetry problems, divergence of \(\frac{\hat{r}}{r^2}\) gives \(4\pi\delta(\vec{r})\), simplifying integrals involving point sources.

Answer 1

The Correct Option is A

Step 1: Evaluate divergence.
\[ \vec{\nabla} \cdot \left( \frac{\hat{r}}{r^2} \right) = 4\pi \delta^3(\vec{r}) \] where \(\delta^3(\vec{r})\) is the Dirac delta function in three dimensions.

Step 2: Substitute into integral.
\[ \int_V e^{-(r/R)^2} \, 4\pi \delta^3(\vec{r}) \, d^3r = 4\pi e^{-(0/R)^2} = 4\pi \]

Step 3: Conclusion.
Thus, the value of the volume integral is \(4\pi.\)