Question

Using gauss theorem calculate the flux of the vector field A = x³i^{^} + x²zj^{^} + yzk^{^} through the surface of a cube of side 2 units

xUsing gauss theorem calculate the flux of the vector field A =

Answer 1

$$div{\vec{A}}=3x^2+y\\\Phi=\int_0^2\int_0^2\int_0^2(3x^2+y)dxdydz \\ div{\vec{A}}=3x^2+y\\\Phi=2^2\int_0^2(3x^2)dx+2^2\int_0^2(y)dy\\\Phi=2^2(2^3)+2^2(0.5\cdot2^2)\\\Phi=40$$