Edit Calculate the surface integral to find the area of a cylinder $z^2=4-y^2$ between $x=y$ and $x=2$. In the first octant.
I don't know how to parametrize $x$ $y$ and $z$. And I don't understand what is the shape of the intersection between that cylinder and the planes
The parameterization that I thought is $x=x$, $y=2 \cos \theta$, $z=2 \sin \theta$ with $2≤x≤y$ and as we are in the first octant, $0≤ \theta ≤ \dfrac {\pi}{2}$ but as I said I'm not sure if this parameterization is correct so I would greatly appreciate the help and how to solve the rest of the exercise.
The equation for the cylinder is $z^2 +y^2=4$. Parametrize in cylindrical coordinates $x=x$, $y=r\cos\theta$ and $z=r\sin\theta$, with the limits $0\le r\le2$ and $0\le\theta\le 2\pi$.
Then, integrate $2-x=2-r\cos\theta$ over the circular region $r\le 2$ to obtain the volume,
$$\int_0^{2\pi}\int_0^2 ( 2- r\cos\theta) rdr d\theta=8\pi$$
Edit:
The surface area of the cylinder between the two planes can be integrated around the circumference of the circle, with the surface element given by $(2-2\cos\theta)2d\theta$, i.e.
$$\int_0^{2\pi} ( 2- 2\cos\theta) 2 d\theta=8\pi$$