Question

Let F=(2x,2y,2x+2z)F=(2x,2y,2x+2z). Use Stokes' theorem to evaluate the integral of FF around the curve consisting of the straight lines joining the points (1,0,1), (0,1,0) and (0,0,1). In particular, compute the unit normal vector and the curl of FF as well as the value of the integral

Answer 1

Stokes' theorem equates the line integral of $\vec F$ along the curve to the surface integral of the curl of $\vec F$ over any surface with the given curve as its boundary. The simplest such surface is the triangle with vertices (1,0,1), (0,1,0), and (0,0,1).

Parameterize this triangle (call it $T$) by

$\vec s(u,v)=(1-v)((1-u)(1,0,1)+u(0,1,0))+v(0,0,1)$

$\vec s(u,v)=((1-u)(1-v),u(1-v),1-u+uv)$

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

$\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}=(0,1-v,1-v)$

Divide this vector by its norm to get the unit normal vector. Note that this assumes a "positive" orientation, so that the boundary of $T$ is traversed in the counterclockwise direction when viewed from above.

Compute the curl of $\vec F$:

$\vec F=(2x,2y,2x+2z)\implies\mathrm{curl}\vec F=(0,-2,0)$

Then by Stokes' theorem,

$\displaystyle\int_{\partial T}\vec F\cdot\mathrm d\vec r=\iint_T\mathrm{curl}\vec F\cdot\mathrm d\vec S$

where

$\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\,\mathrm dS$

$\mathrm d\vec S=\dfrac{\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}}{\left\|\frac{\partial\vec s}{\partial u}\times\frac{\partial\vec s}{\partial v}\right\|}\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv$

$\mathrm d\vec S=\left(\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right)\,\mathrm du\,\mathrm dv$

The integral thus reduces to

$\displaystyle\int_0^1\int_0^1(0,-2,0)\cdot(0,1-v,1-v)\,\mathrm du\,\mathrm dv=\int_0^12(v-1)\,\mathrm dv=\boxed{-1}$