I want to use Stokes' Theorem to evaluate the line integral $F\cdot dr$
$F = (-y^2, x, z^2)$ and $C$ is the curve of the intersection of the plane $y+z=2$ and the cylinder $x^2+y^2=1$. $C$ should be oriented anticlockwise when viewed from above.
I am completely lost as to how to solve this. Im not even sure how to solve the line integral. Any help is appreciated.
Many thanks!
If you want to use Stokes, $$ \int_C F\cdot dr=\int\!\!\!\int_S\text{curl} F\cdot dS, $$ where $S$ is a surface that has $C$ as boundary.
The advantage here is that $$ \text{curl}\,F=\begin{vmatrix}i&j&k\\ D_x& D_y& D_z\\-y^2&x&z^2\end{vmatrix} =(0,0,1+2y) $$ is simple.
If we take $S$ as the disc in the plane $y+z=2$ and bounded by the cylinder, we can parametrize it by $$ u(r,t)=(r\cos t, r\sin t, 2-r\sin t),\ \ 0\leq r\leq1,\ 0\leq t\leq 2\pi. $$ Then normal vector is $$ u_r\times u_t=\begin{vmatrix}i&j&k\\ \cos t& \sin t& -\sin t\\ -r \sin t& r\cos t&-r\cos t\end{vmatrix} =(0,r,r). $$ Then $$ \int_C F\cdot dr=\int\!\!\!\int_S\text{curl} F\cdot dS =\int_0^1\int_0^{2\pi}(0,0,1+2r\sin t)\cdot(0,r,r)\,dt\,dr\\ =\int_0^1\int_0^{2\pi}(r+2r^2\sin t)\,dt\,dr =\int_0^12\pi r\,dr=\pi. $$
In this case, it is also easy to solve the line integral directly: using the parametrization $c(t)=(\cos t, \sin t,2-\sin t)$, $0\leq t\leq 2\pi$, we get $$ \int_C F\cdot dr=\int_0^{2\pi}(-\sin^2t,\cos t,(2-\sin t)^2)\cdot(-\sin t,\cos t,-\cos t)\,dt=\int_0^{2\pi}(-\sin^3t+\cos^2 t-(2-\sin t)^2\cos t )\,dt=\pi. $$