Answer to Question #192657 in Electricity and Magnetism for Abhishek Mehta

Question

Using guass theorem calculate the flux of the vector field A =x³i+x²zj+yzk through the square of a cube of a side 2 unit's

Answer 1

$\text{div} \vec{A}=3x^2\vec{i}+y\vec{k},$

$\Phi=\int _0^2dx\int_0^2dy\int _0^2(3x^2+y)dz=2\int_0^2dx\int_0^2(3x^2+y)dy=2\int_0^2(6x^2+2)dx=2(2x^3+2x)|_0^2=40~Wb.$