Question

Let C be the unit circle in the xy-plane, oriented counterclockwise as seen from above. The divergence of the vector field F~ = (z, x, y) is zero, and as a result the flux through every surface with boundary C should be the same. Confirm that this is the case with the upper half of the unit sphere, the lower half of the unit sphere, and the unit disk in the xy-plane

Answer 1

Upper half of the unit sphere (call it $S_1$): parameterize by

$\vec s(u,v)=(\cos u\sin v,\sin u\sin v,\cos v)$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$. Take the normal vector to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=(\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v)$

Then the flux of $\vec F$ over this surface is

$\displaystyle\iint_{S_1}\vec F\cdot\mathrm d\vec S=\int_0^{\pi/2}\int_0^{2\pi}(\cos v,\cos u\sin v,\sin u\sin v)\cdot(\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}\cos u\sin^2v\cos v+\cos u\sin u\sin^3v+\sin u\cos v\sin^2v=\boxed{0}$

Lower half of the sphere (call it $S_2$): all the details remain the same as above, but with $\frac\pi2\le v\le\pi$. The flux is again $\boxed{0}$.

Unit disk (call it $D$): parameterize the disk by

$\vec s(u,v)=(u\cos v,u\sin v,0)$

with $0\le u\le1$ and $0\le v\le2\pi$. Take the normal vector to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,0,u)$

Then the flux across $D$ is

$\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^1(0,u\cos v,u\sin v)\cdot(0,0,u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1u^2\sin v\,\mathrm du\,\mathrm dv=\boxed{0}$