Question

Use Stokes' Theorem to evaluate |c F * dr where C is oriented counterclockwise as viewed from above.F(x, y, z) = xyi + 5zj + 7yk, where C is the curve of intersection of the plane x + z = 3 and the cylinder x2 + y2 = 144.

Answer 1

Compute the curl of $\vec F$:

$\vec F(x,y,z)=xy\,\vec\imath+5z\,\vec\jmath+7y\,\vec k\implies\nabla\times\vec F(x,y,z)=2\,\vec\imath-x\,\vec k$

By Stoke's theorem, the (line) integral of $\vec F$ along $C$ is equivalent to the (surface) integral (or flux) of $\nabla\times\vec F$ across $S$, the oriented surface with boundary $C$.

Take $S$ to be the part of the plane $x+z=3$ within the cylinder $x^2+y^2=144$. Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(3-u\cos v)\,\vec k$

with $0\le u\le12$ and $0\le v\le2\pi$. Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=u\,\vec\imath+u\,\vec k$

Then the integral is

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^{12}(2\,\vec\imath-u\cos v\,\vec k)\cdot(u\,\vec\imath+u\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^{12}(2u-u^2\cos v)\,\mathrm du\,\mathrm dv=\boxed{288\pi}$