Find flux of the vector field $\vec A$ through the surface of a sphere with radius R and center on the origin

1.2k Views Asked by At

The vector field is given by the spherical coordinates $\vec A = kr^3\,\vec {e}_r$, where $k$ is a constant and $r=\sqrt {x^2+y^2+z^2}$.

I thought about using the Gauss-Ostrogradski theorem, where the flow is given by: $$ \iiint_V \operatorname{div}{\!\vec A} \; dV $$

But i got lost in the math and couldn't find a proper answer. Do you guys have any tips or ideas on how to solve this problem?

A little insight on my math: I got confused on how to calculate the divergent of $\vec A$ because of the base vector $\vec {e}_r$, since I am not used to it. But, my idea was to use the following intervals on the triple integral: ($\sqrt {R^2-x^2-y^2}$,$-\sqrt {R^2-x^2-y^2}$) and integrate in relation to dz;($\sqrt {R^2-x^2}$,$-\sqrt {R^2-x^2}$) and integrate in relation to dy; ($R$,$-R$) and integrate in relation to dx. I have no idea if it's correct or not. If anyone can please help or solve this for me, it would be really helpfull.

1

There are 1 best solutions below

0
On

It is easier if you do the whole thing without resorting to cartesian coordinates. Let $\phi$ denote the flux. Then \begin{align} \phi&=\oint_{S}\vec E\cdot \vec{dS}\\ &=\oint_{S}kr^3\vec{e_r}\cdot dS\vec{e_r}\\ &=kr^3\oint_{S}dS\\ &=kr^3\times 4\pi r^2\\ &=4\pi k r^5 \end{align} `