Use Stokes' theorem to calculate the curve integral $\int_{\gamma} x dx + (y + z) dy + (y + z) dz$ where $\gamma$ is the intersection between the cylinder $x^2 + y^2 = 4$ and the plane $x + y + z = 0$.
I didn't calculate the curl beforehand, if I had done it I wouldn't have wasted time. In any case, lets just assume I still needed to do the calculation to prove that the answer is 0, is this the right way to use Stoke's theorem. Feel free to correct my solution.
The intersection of the plane and the cylinder is given by where $z = -x-y$
Now I will change coordinates to: $${ \begin{cases}{x=r\cos(t)} \\ {y = r\sin(t)} \\ {z=-r\cos(t)-r\sin(t)} \end{cases} }$$ $$0\leq t \leq 2\pi, 0\leq r \leq 2$$
Then I will find the normal vector of the plane.
$$r(t,r) = <r\cos(t), r\sin(t), -r\cos(t) - r\sin(t)>$$ $$r_{r} = <\cos(t), \sin(t), -(\cos(t) + \sin(t))>$$ $$r_{t} = <-r\sin(t), r\cos(t), r\sin(t) - r\cos(t))>$$ $$\vec{n} = r_{t} \times r_{r} = (r,r,r)$$
The curl of F is:
$$\nabla \times \vec{F} = (0,0,0)$$
So I got to the conclusion:
$$\iint_{D} (0,0,0) \cdot (r,r,r) dt dr = \iint_{D} 0 dr dt = 0$$
So what I really want to know is when I changed from cartesian coordinates to polar coordinates, usually I have to multiply with some sort of value that scales the surface in this case $r$. Do I still need to do it here? Why or why not? Also assuming I'm right, is there an easier way to find the normal of the plane?
So, the way you did things was perhaps a little bit backwards as you seemed to notice, but it is definitely good that you realized you should have calculated the curl before proceeding to save time. Here's what you can notice about the integral you wrote down, $\int_D (\nabla \times F) \cdot (r, r, r) dr dt = \int_D (\nabla \times F) \cdot \left(\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}\right) \sqrt{3} r dr dt = \int_D (\nabla \times F) \cdot n dS$. Where $n$ is the unit normal to your surface (notice that it is just the normal to the plane) and $dS$ is the surface area element which has that extra factor of $\sqrt{3}$ because your surface is slanted relative to the disc you used to parametrize this portion of the surface. Essentially, the manner in which you did the computation "hid" the familiar form of the expression that I believe you were inquiring about, but you did everything correctly. Hope this helps.