GATE ME 2020 Set 1 — Question 37

The given vector field is $\vec{f}(x, y, z) = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2 + y^2 + z^2)^{3/2}}$.
Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then $r = |\vec{r}| = \sqrt{x^2 + y^2 + z^2}$.
So, the vector field can be written as $\vec{f} = \frac{\vec{r}}{r^3}$. This is an inverse square law field.We need to evaluate the surface integral $\iint \vec{f} \cdot d\vec{S}$ over the inner and outer surfaces of a spherical shell with internal radius $R_1 = 1$ and external radius $R_2 = 2$, centered at the origin.
This is a closed surface, which is the boundary of the volume $V$ of the spherical shell ($1 \le r \le 2$).By Gauss's Divergence Theorem, for a vector field $\vec{F}$ and a volume $V$ bounded by a closed surface $S$:
$\iint_S \vec{F} \cdot d\vec{S} = \iiint_V (\nabla \cdot \vec{F}) \, dV$.Let's calculate the divergence of $\vec{f} = \frac{\vec{r}}{r^3}$:
$\nabla \cdot (\frac{\vec{r}}{r^3}) = \nabla \cdot (r^{-3} \vec{r})$
Using the identity $\nabla \cdot (\phi \vec{A}) = (\nabla \phi) \cdot \vec{A} + \phi (\nabla \cdot \vec{A})$ where $\phi = r^{-3}$ and $\vec{A} = \vec{r}$.
We know $\nabla r^n = n r^{n-2} \vec{r}$. So, $\nabla r^{-3} = -3 r^{-5} \vec{r}$.
Also, $\nabla \cdot \vec{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1+1+1 = 3$.Substituting these into the divergence formula:
$\nabla \cdot (r^{-3} \vec{r}) = (-3 r^{-5} \vec{r}) \cdot \vec{r} + r^{-3} (3)$
$= -3 r^{-5} (\vec{r} \cdot \vec{r}) + 3 r^{-3}$
$= -3 r^{-5} (r^2) + 3 r^{-3}$ (since $\vec{r} \cdot \vec{r} = r^2$)
$= -3 r^{-3} + 3 r^{-3} = 0$.Thus, $\nabla \cdot \vec{f} = 0$ for $r \neq 0$. The origin $(0,0,0)$ is the only point where the divergence is undefined. However, the volume of the spherical shell ($1 \le r \le 2$) does not include the origin. Therefore, the divergence is zero throughout the volume $V$ of the spherical shell.According to Gauss's Divergence Theorem, since $\nabla \cdot \vec{f} = 0$ in the volume $V$, the surface integral over the closed surface $S$ (which consists of the inner and outer spherical surfaces) is zero.Alternatively, we can calculate the flux through each surface: On this surface, $r = R_2 = 2$. So $\vec{f} = \frac{\vec{r}}{R_2^3}$.
The elemental surface vector is $d\vec{S}_2 = \hat{n} dS_2 = \frac{\vec{r}}{R_2} dS_2$.
Flux through $S_2 = \iint_{S_2} \vec{f} \cdot d\vec{S}_2 = \iint_{S_2} \left(\frac{\vec{r}}{R_2^3}\right) \cdot \left(\frac{\vec{r}}{R_2} dS_2\right) = \iint_{S_2} \frac{\vec{r} \cdot \vec{r}}{R_2^4} dS_2 = \iint_{S_2} \frac{r^2}{R_2^4} dS_2$
Since $r=R_2$ on $S_2$, this becomes $\iint_{S_2} \frac{R_2^2}{R_2^4} dS_2 = \iint_{S_2} \frac{1}{R_2^2} dS_2 = \frac{1}{R_2^2} (\text{Area of } S_2) = \frac{1}{R_2^2} (4\pi R_2^2) = 4\pi$. On this surface, $r = R_1 = 1$. So $\vec{f} = \frac{\vec{r}}{R_1^3}$.
Flux through $S_1 = \iint_{S_1} \vec{f} \cdot d\vec{S}_1 = \iint_{S_1} \left(\frac{\vec{r}}{R_1^3}\right) \cdot \left(-\frac{\vec{r}}{R_1} dS_1\right) = -\iint_{S_1} \frac{\vec{r} \cdot \vec{r}}{R_1^4} dS_1 = -\iint_{S_1} \frac{r^2}{R_1^4} dS_1$
Since $r=R_1$ on $S_1$, this becomes $- \iint_{S_1} \frac{R_1^2}{R_1^4} dS_1 = - \iint_{S_1} \frac{1}{R_1^2} dS_1 = -\frac{1}{R_1^2} (\text{Area of } S_1) = -\frac{1}{R_1^2} (4\pi R_1^2) = -4\pi$.The total surface integral is the sum of the fluxes through the outer and inner surfaces:
Total integral $= 4\pi + (-4\pi) = 0$.Both methods confirm that the surface integral is $0$.