$\iiint _{V} \frac{dV}{r^2}= \iint_{S} \frac{\vec{r}\cdot \hat{n}}{r^2}\,dS$

Question

I have a question about a problem. It is triple integrals on surfaces. It is show $$\iiint _{V} \frac{dV}{r^2}= \iint_{S} \frac{\vec{r}\cdot \hat{n}}{r^2}\,dS$$ Some help?

Answer 1

As per Divergence theorem, $\, \displaystyle \iint_{S} \vec{F} \cdot \hat{n}\,d$S $\displaystyle = \iiint_V (div \,\vec{F}) \,dV$

We have, $\, \displaystyle \vec{F} = \frac{\vec{r}}{r^2}$

$\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}$

$r^2 = x^2 + y^2 + z^2$

$\displaystyle div \vec{F} = \nabla \cdot \vec{F} = \frac{\partial}{\partial x} (\frac{x}{r^2})+ \frac{\partial}{\partial y} (\frac{y}{r^2}) + \frac{\partial}{\partial z} (\frac{z}{r^2})$

$\displaystyle \frac{\partial}{\partial x} ( \frac{x}{r^2}) = \frac{1}{r^2} - \frac{2x^2}{r^4}$

Can you similarly find $F_y, F_z$ and add them? You will see $\displaystyle div\vec{F} = \frac{1}{r^2}$