Question

Use stokes' theorem to compute z c ~f d~s where ~f(x; y; z) = y2~i + z2~j + x~k and the curve c is the triangle in 3-space with vertices at (1; 0; 0), (0; 1; 0), and (0; 0; 1). hints: (a) for the surface s needed in stokes' theorem, use the part of the plane x + y + z = 1 that lies in the rst octant. (b) that plane can be written parametrically as (x; y) = hx; y; 1 ???? x ???? yi. (c) make sure you have the upward pointing normal!

Answer 1

By Stoke's theorem, the line integral of $\mathbf f(x,y,z)=y^2\,\mathbf i+z^2\,\mathbf j+z\,\mathbf k$ along $C$ (presumably the *boundary* of the triangle, and not the triangle itself)

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r$

is given by the surface integral of $\nabla\times\mathbf f(x,y,z)$ along the surface with boundary $C$,

$\displaystyle\iint_S(\nabla\times\mathbf f)\cdot\mathrm d\mathbf S$

First, compute the curl of $\mathbf f$:

$\nabla\times\mathbf f=-2z\,\mathbf i-\,\mathbf j-2y\,\mathbf k$

Not sure what kind of parameterization you're given for $S$, but you can use

$\mathbf s(u,v)=(1-u)\,\mathbf k+u(1-v\,\mathbf i+v\,\mathbf j)=(u-uv)\,\mathbf i+uv\,\mathbf j+(1-u)\,\mathbf k$

where $(u,v)\in[0,1]\times[0,1]$. Then

$\mathbf s_u\times\mathbf s_v=u(\mathbf i+\mathbf j+\mathbf k)$

So the surface integral is equivalent to

$\displaystyle\iint_S(\nabla\times\mathbf f)\cdot\mathrm d\mathbf S=\int_{u=0}^{u=1}\int_{v=0}^{v=1}(-2z(u,v)\,\mathbf i-\,\mathbf j-2y(u,v)\,\mathbf k)\cdot u(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dv\,\mathrm du$
$\displaystyle=\int_{u=0}^{u=1}\int_{v=0}^{v=1}(-2(1-u)\,\mathbf i-\,\mathbf j-2uv\,\mathbf k)\cdot u(\mathbf i+\mathbf j+\mathbf k)\,\mathrm dv\,\mathrm du$
$\displaystyle=\int_{u=0}^{u=1}\int_{v=0}^{v=1}(2u^2(1-v)-3u)\,\mathrm dv\,\mathrm du$
$=-\dfrac76$