Question

Compute the flux of curl(F) through the part of the paraboloid z = x 2 + y 2 that lies below the plane z = 4 with upward-pointing unit normal vector and F = h3z,5x,−2yi.

Answer 1

Parameterize this surface (call it S) by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+u^2\,\mathbf k$

with $0\le u\le2$ and $0\le v\le2\pi$.

The normal vector to S is

$\mathbf n=\dfrac{\partial\mathbf s}{\partial u}\times\dfrac{\partial\mathbf s}{\partial v}=-2u^2\cos v\,\mathbf i-2u^2\sin v\,\mathbf j+u\,\mathbf k$

Compute the curl of F :

$\nabla\times\mathbf F=-2\,\mathbf i+3\,\mathbf j+5\,\mathbf k$

So the flux of curl(F) is

$\displaystyle\iint_S(\nabla\times\mathbf F)\cdot\mathrm d\mathbf S=\int_0^{2\pi}\int_0^2(\nabla\times\mathbf F)\cdot\mathbf n\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^2(5u+4u^2\cos v-6u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{20\pi}$

Alternatively, you can apply Stokes' theorem, which reduces the surface integral of the curl of F to the line integral of F along the intersection of the paraboloid with the plane z = 4. Parameterize this curve (call it C) by

$\mathbf r(t)=2\cos t\,\mathbf i+2\sin t\,\mathbf j+3\,\mathbf k$

with $0\le t\le2\pi$. Then

$\displaystyle\iint_S(\nabla\times\mathbf F)\cdot\mathrm d\mathbf S=\int_0^{2\pi}\mathbf F\cdot\mathrm d\mathbf r$

$=\displaystyle\int_0^{2\pi}(20\cos^2t-24\sin t)\,\mathrm dt=\boxed{20\pi}$