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

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

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3 years ago