Question

Use Stokes' Theorem to evaluate S curl F · dS. F(x, y, z) = x2 sin(z)i + y2j + xyk, S is the part of the paraboloid z = 9 − x2 − y2 that lies above the xy-plane, oriented upward.

Answer 1

The vector field

$\vec F(x,y,z)=x^2\sin z\,\vec\imath+y^2\,\vec\jmath+xy\,\vec k$

has curl

$\nabla\times\vec F(x,y,z)=x\,\vec\imath+(x^2\cos z-y)\,\vec\jmath$

Parameterize $S$ by

$\vec s(u,v)=x(u,v)\,\vec\imath+y(u,v)\,\vec\jmath+z(u,v)\,\vec k$

where

$\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=(9-u^2)\end{cases}$

with $0\le u\le3$ and $0\le v\le2\pi$.

Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

Then by Stokes' theorem we have

$\displaystyle\int_{\partial S}\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^3(\nabla\times\vec F)(\vec s(u,v))\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^3u^3(2u\cos^3v\sin(u^2-9)+\cos^3v\sin v+2u\sin^3v+\cos v\sin^3v)\,\mathrm du\,\mathrm dv$

which has a value of 0, since each component integral is 0:

$\displaystyle\int_0^{2\pi}\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin v\cos^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\sin^3v\,\mathrm dv=0$

$\displaystyle\int_0^{2\pi}\cos v\sin^3v\,\mathrm dv=0$