Answer to Question #167657 in Electricity and Magnetism for Taruk

Question

Using Gauss’ theorem, calculate the flux of the vector field A= x^3 i + x^2z j + yz k

through the surface of a cube of side 2 units

Answer 1

Flux is defined in vector notation as the integral of the field intensity vector over the needed area.

$\qquad\qquad \begin{aligned} \small \text{Flux}&=\small \int_s \bold{F}d\bold{S} \end{aligned}$

Gauss's theorem says that the surface integral can also be written as a volume integral of the divergence of that particular vector field.

In notation,

$\qquad\qquad \begin{aligned} \small \int_s\bold{F}\bold{dS}&= \small \iiint_v div \bold{F}.dv \end{aligned}$

Therefore, replacing F with the given A, flux can be calculated

$\qquad\qquad \begin{aligned} \small div\bold{A}&= \small \nabla\bold{A}\\ &=\bigg(i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}\bigg).(x^3i+x^2j+yzk)\\ &=\small \frac{\partial(x^3)}{\partial x}+\frac{\partial(x^2z)}{\partial y}+\frac{\partial(yz)}{\partial z}\\ &= \small 3x^2+0+y\\ &=\small 3x^2+y \end{aligned}$

And $dv$ is the infinitesimal volume which can be written as $dv=dxdydz$

Then applying all these,

$\qquad\qquad \begin{aligned} \small \text{Flux}&= \small \iiint(3x^2+y)dxdydz \end{aligned}$

To proceed further, limits of xyz variables are needed. To define limits, define a coordinate system from the bottom vertex of the box such as it is bounded by x=0, x=2, y=0, y=2, z=0 & z=2.

Then having those limits input, that can be integrated with respect to one variable at a time.

$\qquad\qquad \begin{aligned} &=\small \iint dxdy \int_0^2 (3x^2+y) dz\\ &= \small \iint dxdy (3x^2z+yz\big|_0^2\\ &=\small \int dx \int_0^2 (6x^2+2y)dy\\ &=\small \int dx(6x^2y+y^2\big|_0^2\\ &=\small \int_0^2 (12x^2+4)dx\\ &=\small (4x^3+4x\Big|_0^2\\ \small \text{Flux}&=\small \bold{40} \end{aligned}$