It is double integral $(x + y) dS$ where $S$ is the part of the cylinder $y^2 + z^2 = 4$ . With $x$ being between $0$ and $5$
First question, if we want to get the integral of the surface of a cylinder, I don't quite understand why $(x + y)$ is being multipled by $dS$? What is the applications of doing something like that?
In terms of just the surface being described, is my parametrization correct?
$x = v$
$y = cos (u)$
$z = sin (u)$
$0 < u < 2\pi$
$0 < v < 5$
This problem is so hard for me visualize and wrap my head around, I think it would be easier if instead of " $y^2 + z^2 = 4$ " it was "$x^2 + y^2 = 4$". IS it correct to treat the variables $y$ and $z$ as I would treat them if it was $x$ and $y$, which is what I did? It seems logically sound, but I could be missing something maybe.
Assuming my stuff is correct, I take the cross product of the partial derivatives of this vector valued function and integrate across the region right? This is not taking into account the $(x + y)$ that's being multiplied by the surface, of course.
What is so hard to visualize? $f(x,y)=x+y$ is just a plane through the origin (i.e. $z-x-y=0)$ and the cylindar $y^{2}+z^{2}=4$ is just oriented along the $x$-axis instead of the $z$-axis as you claim is easier to visualize (it's not, if you think about it for a second).
The integral is then just a surface integral of the scalar field $x+y$ over the portion of the cylindar bounded between the planes $x=0$ and $x=5$, that is $$\int_{S}(x+y)\;dS,$$ $dS$ being the induced surface measure over the cylindar. It is taught in your class of course that $dS=|\phi_{u}\times\phi_{v}|\;du\;dv$ where $\phi(u,v)=(v,2\cos u,2\sin u)$ is the parametrization of the cylindar, once the integration is pulled back onto the parameter domain of $\phi$, which is as you indicated the rectangle $[0,5]\times[0,2\pi]$.
We have $\partial_{u}\phi(u,v)=(0,-2\cos u,2\cos u)$ and $\partial_{v}\phi(u,v)=(1,0,0)$, and so $|\phi_{u}\times\phi_{v}|=|(0,2\cos u, 2\sin u)|=2.$
The integral now becomes $$2\int_{0}^{5}\int_{0}^{2\pi}(v+2\cos u)\;du\;dv.$$
Now finish the job.