Use Stoke’s theorem to evaluate $$ \int_C F dr = \iint \operatorname{curl} F dS = \iint (-Pg_x-Qg_y+R) dA $$
where $F =\langle 2z, 4x, 5y \rangle$ and $C$ is the intersection of $z = x + 4$ with the cylinder $x^2 + y^2 = 4$.
I calculated the $\operatorname{curl} F = \langle 5i+2j+4k\rangle$
$dS$: The projection of the surface onto a plane, in this case, because $z=g(x,y)$, it's onto the xy plane.
$$ z = g(x,y) = x+4 $$
$$g_x = 1$$
$$g_y = 0$$
$$ \iint (-1) dA $$
Switching to polar
$$ \iint (-1) rdrd \theta $$
where $$0≤\theta≤2\pi$$
$$0≤r≤2$$
I calculated the integral and got an answer of $-4\pi$. I haven't completed many Stokes' Theorem problems though so I'm not sure if my process, or answer, are correct.