Question

Use the Divergence Theorem to calculate the flux of −→F (x, y, z) = hx 3 , y3 , −3z 2 i through the boundary of the solid T given by x 2 + y 2 ≤ 4, 0 ≤ z ≤ 2. Assuming −→F is the velocity field of a fluid, is the net flow into the solid T, or out of T?

Answer 1

I suppose

$\vec F(x,y,z)=\langle x^3,y^3,-3z^2\rangle$

whose divergence is

$\mathrm{div}\vec F=3x^2+3y^2-6z$

By the divergence theorem, the flux of $\vec F$ across the boundary of $T$ is equal to the integral of $\mathrm{div}\vec F$ over $T$:

$\displaystyle\iint_{\partial T}\vec F\cdot\mathrm d\vec S=\iiint_T\mathrm{div}\vec F\,\mathrm dV$

The volume integral is easily computed in cylindrical coordinates; take

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dV=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$

$\displaystyle\iiint_T\mathrm{div}\vec F\,\mathrm dV=\int_0^{2\pi}\int_0^2\int_0^2(3r^2-6\zeta)r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle6\pi\int_0^2\int_0^2(r^3-2r\zeta)\,\mathrm d\zeta\,\mathrm dr$

$=\displaystyle12\pi\int_0^2(r^3-2r)\,\mathrm dr=\boxed0$