Question

Evaluate the line integral ∫ F.dr by evaluating the surface integral in​ Stokes' Theorem with an appropriate choice of S. Assume that C has a counterclockwise orientation when viewed from above and will spin clockwise when viewed from below.
F= (6y,-z,x)
C is circle x^2+y^2=20 in the plane z=0

Answer 1

Stokes' theorem allows for any choice of $S$ with $C$ as its boundary. The simplest region would be the disk $x^2+y^2\le20$ in the plane $z=0$, which can be parameterized using polar coordinates by

$\vec s(u,v)=(u\cos v,u\sin v,0)$

with $0\le u\le\sqrt{20}$ and $0\le v\le2\pi$. Take the normal vector to $S$ to be

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,0,u)$

Then by Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

where $\nabla\times\vec F$ denotes the curl of $\vec F$. The curl is

$\nabla\times\vec F=(1,-1,-6)$

so the line integral has a value of

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt{20}}(1,-1,-6)\cdot(0,0,u)\,\mathrm du\,\mathrm dv=-12\pi\int_0^{\sqrt{20}}u\,\mathrm du=\boxed{-120\pi}$