Problem: Find the surface area of the part of the cylinder $x^2+z^2 = a^2$ that is inside the cylinder $x^2+y^2 = 2ay\;$ and also in the positive octant $( x \ge 0, y\ge 0, z\ge 0$). Assume $a > 0$.
I completed the square to turn the second equation into $x^2 + (y-a)^2 = a^2$. So I have two cylinders with radius a. I believe that I should use polar coordinates to change the variables, but I'm not sure how to do this since one equation is in terms of $x$ and $z$ and the other is in terms of $x$ and $y$.
Since both equations equal $a^2$, I set them equal to get $z = y-a$. I then tried using the parameterization $x = rcos\theta, \;y = rsin\theta, \;z = rsin\theta -a$, but I did not get the right answer. Could someone show me how to parameterize this correctly? Thank you!
(The answer is $2a^2$.)