Question

g Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 7yi + xzj + (x + y)k, C is the curve of intersection of the plane z = y + 9 and the cylinder x2 + y2 = 1. Exp

Answer 1

By Stokes' theorem,

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

where $S$ is any oriented surface with boundary $C$. We have

$\vec F(x,y,z)=7y\,\vec\imath+xz\,\vec\jmath+(x+y)\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-7)\,\vec k$

Take $S$ to be the ellipse that lies in the plane $z=y+9$ with boundary on the cylinder $x^2+y^2=1$. Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+9)\,\vec k$

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

$\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k$

Then we have

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

$=\displaystyle\int_0^{2\pi}\int_0^1\big((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+2)\,\vec k\big)\cdot\big(-u\,\vec\jmath+u\,\vec k\big)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{2\pi}\int_0^1(3u+u^2\sin v)\,\mathrm du\,\mathrm dv=\boxed{3\pi}$