Use Stokes’ theorem to solve the following integral (the curve is oriented counterclockwise when viewed from above.) $$\int_C(x+ 2y)dx+ (2z+ 2x)dy+ (z+y)dz$$ where $C$ is the intersection of the cylinder $x^2 + y^2 = 2y$ and the plane $y = z$
I don't fully understand how to use Stokes' theorem, can someone show me how to go about solving this example.
Stokes tells you that your integral is equal to $$ \iint_D \text{curl} F\cdot n\,dS, $$ where $D$ is any (good enough) surface that has $C$ as its boundary. We have $$ \text{curl}F=\begin{vmatrix} \mathrm i&\mathrm j&\mathrm k\\ \partial_x&\partial_y&\partial_z\\ x+2y&2z+2x&z+y \end{vmatrix} =(-1,0,0). $$ We can take $D$ to be the ellipse contained in the plane $y=z$ that has $C$ as a boundary. The cylinder is $x^2+(y-1)^2=1$, so the curve $C$ can be parametrized (as $z=y$) by $$ v(t)= ( \cos t, 1+\sin t, 1+\sin t),\ \ \ 0\leq t\leq2\pi. $$ The interior $D$ of the curve can be then parametrized by $$ u(r,t)= ( r\cos t, 1+r\sin t, 1+r\sin t),\ \ \ 0\leq r\leq1,\ \ 0\leq t\leq2\pi. $$ The normal vector is \begin{align} n=u_r\times u_t=\begin{vmatrix} \mathrm i&\mathrm j&\mathrm k\\ \cos t& \sin t&\sin t\\ -r\sin t&r\cos t&r\cos t \end{vmatrix}=(0,-r,r). \end{align} Then $\text{curl}F\cdot n=0$, and the integral is zero.