Question

Use the Divergence Theorem to calculate the surface integral S F · dS; that is, calculate the flux of F across S. F(x, y, z) = (x3 + y3)i + (y3 + z3)j + (z3 + x3)k, S is the sphere with center the origin and radius 3.

Answer 1

By the divergence theorem, the flux of $\vec F$ across S is equal to the volume integral of $\mathrm{div}(\vec F)$ over the interior of S.

We have

$\vec F(x,y,z) = (x^3+y^3)\,\vec\imath + (y^3+z^3)\,\vec\jmath + (z^3+x^3)\,\vec k \\\\ \implies \mathrm{div}(\vec F) = \dfrac{\partial(x^3+y^3)}{\partial x} + \dfrac{\partial(y^3+z^3)}{\partial y} + \dfrac{\partial(z^3+x^3)}{\partial z} = 3(x^2+y^2+z^2)$

so that

$\displaystyle \iint_S \vec F(x,y,z)\cdot\mathrm d\vec s = \iiint_T \mathrm{div}(\vec F)\,\mathrm dV = 3 \iiint\limits_{x^2+y^2+z^2\le3} (x^2+y^2+z^2)\,\mathrm dx\,\mathrm dy\,\mathrm dz$

To compute the volume integral, convert to spherical coordinates with

x = ρ cos(θ) sin(ϕ)

y = ρ sin(θ) sin(ϕ)

z = ρ cos(ϕ)

so that

ρ ² = x ² + y ² + z ²

dx dy dz = ρ ² sin(ϕ) dρ dϕ dθ

The region T is the interior of the sphere S, given by the set

$T = \left\{(\rho,\theta,\phi) \mid 0\le\rho\le3 \text{ and } 0\le\phi\le\pi \text{ and }0\le \theta\le2\pi\right\}$

So we have

$\displaystyle 3 \int_0^{2\pi} \int_0^\pi \int_0^3 \rho^4 \sin(\phi) \,\mathrm d\rho \,\mathrm d\phi \,\mathrm d\theta \\\\ = 6\pi \left(\int_0^\pi \sin(\phi)\,\mathrm d\phi\right) \left(\int_0^3 \rho^4 \,\mathrm d\rho\right) = \boxed{\frac{2916\pi}5}$