Question

Calculate the circulation, F · dr, C in two ways, directly and using Stokes' Theorem. F = y i + z j + xk and C is the boundary of S, the paraboloid z = 4 − (x2 + y2), z ≥ 0 oriented upward. (Hint: Use polar coordinates.) F · dr C =

Answer 1

$C$, the boundary of $S$, is a circle in the $x,y$ plane centered at the origin and with radius 2, hence we can parameterize it by

$\vec r(t)=2\cos t\,\vec\imath+2\sin t\,\vec\jmath$

with $0\le t\le2\pi$. Then the line integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}(2\sin t\,\vec\imath+2\cos t\,\vec k)\cdot(-2\sin t\,\vec\imath+2\cos t\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^{2\pi}-4\sin^2t\,\mathrm dt$

$=\displaystyle-2\int_0^{2\pi}(1-\cos2t)\,\mathrm dt=\boxed{-4\pi}$

By Stokes' theorem, the line integral of $\vec F$ along $C$ is equal to the surface integral of the curl of $\vec F$ across $S$:

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(4-u^2)\,\vec k$

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

$\vec s_u\times\vec s_v=2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k$

The curl is

$\nabla\times\vec F=-\vec\imath-\vec\jmath-\vec k$

Then the surface integral is

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^2(-\vec\imath-\vec\jmath-\vec k)\cdot(2u^2\cos v\,\vec\imath+2u^2\sin v\,\vec\jmath+u\,\vec k)\,\mathrm du\,\mathrm dv$

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