Question

Find the flux of the vector field v(x, y, z) = 7xy2i + 4x2yj + z3k out of the unit sphere.

Answer 1

Denote the unit sphere by $\mathcal S$. If the sphere is closed, then the flux of $\mathbf v(x,y,z)$ across $\mathcal S$ is given by the divergence theorem to be

$\displaystyle\iint_{\mathcal S}\mathbf v\cdot\mathrm d\mathbf S=\iiint_{\mathcal B}\nabla\cdot\mathbf v\,\mathrm dV$

where $\mathcal B$ denotes the space with boundary $\mathcal S$. We have

$\nabla\cdot\mathbf v=\dfrac{\partial(7xy^2)}{\partial x}+\dfrac{\partial(4x^2y)}{\partial y}+\dfrac{\partial(z^3)}{\partial z}=7y^2+4x^2+3z^2$

So the flux across $\mathcal S$ is equivalent to

$\displaystyle\iiint_{\mathcal B}(7y^2+4x^2+3z^2)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

We convert to spherical coordinates to evaluate the integral:

$x=\rho\cos\theta\sin\varphi$
$y=\rho\sin\theta\sin\varphi$
$z=\rho\cos\varphi$
$\implies 4x^2+7y^2+3z^2=3\rho^2+x^2+4y^2$
$=3\rho^2+\rho^2\cos^2\theta\sin^2\varphi+4\rho^2\sin^2\theta\sin^2\varphi$

$=\rho^2(3+\sin^2\varphi(1+3\sin^2\theta))$

$\displaystyle\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^2(3+\sin^2\varphi(1+3\sin^2\theta))\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

$=\dfrac{56\pi}{15}$