Surface area of $x^2+z^2=a^2$ inside of $x^2+y^2 = 2ay$ and in first octant

Question

The questions is

What is the surface area of $x^2+z^2=a^2$ inside of $x^2+y^2 = 2ay$ and in first octant?

My attempt

The second equation can be rewritten as $x^2 + (y-a)^2=a^2$ to make it easier to work with. After this I tried parametrizing the first cylinder with $x= \operatorname{acos}\theta$, $z=\operatorname{asin}\theta$ and $y=y$.

This is where I get stuck. The area integral should be $A= \iint||T_{\theta} \times T_y||dS$ = $\iint a dyd\theta$. I'm not sure how to place the bounds on $\theta$ and $y$. My first guess was to let $0 \leq\theta \leq \pi/2$, but I'm still not sure what to do with $y$.

Any help would be greatly appreciated.

Answer 1

The bounds on your second variable can be taken from the equation for the second cylinder:

$$x^2+(y-a)^2=a^2\implies (y-a)^2=a^2-x^2\implies|y-a|=\sqrt{a^2-x^2}$$

The part of the cylinder $x^2+z^2=a^2$ that is bounded by the second cylinder then consists of those points $(x,y,z)$ where $-\sqrt{a^2-x^2}\le y-a\le\sqrt{a^2-x^2}$.

Then parameterize this part of the cylinder $x^2+z^2=a^2$ additionally bounded by the coordinate axes by

$$\sigma(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\langle a\cos v,u,a\sin v\rangle$$

with $0\le v\le\frac\pi2$ and $a-|a|\sin v\le u\le a+|a|\sin v$. Then the desired area is

$$\begin{align*} \int_{v=0}^{v=\pi/2}\int_{u=a-|a|\sin v}^{u=a+|a|\sin v}\|\sigma_u\times\sigma_v\|\,\mathrm du\,\mathrm dv&=|a|\int_{v=0}^{v=\pi/2}\int_{u=a-|a|\sin v}^{u=a+|a|\sin v}\mathrm du\,\mathrm dv\\[1ex] &=2a^2\int_{v=0}^{v=\pi/2}\sin v\,\mathrm dv\\[1ex] &=2a^2 \end{align*}$$