Surface Area of Two Cylinders Calculus 3 Find the surface area of two cylinders $$y^2 + z^2 = 1$$ and $$x^2 + y^2 = 1$$
I have so far set the two equations to equal $$x= \pm z$$ and $$y= \sqrt{(1-z^2)}$$ I am a little confused on how to set up the integration problem. So far I have $$1/\sqrt{(1-z^2)}dy$$ from 0 to 1 and am not certain if that is the correct approach.
 A: The surface area is 
$$S=2\int\int_D \sqrt{1+f_x^2+f_y^2}dxdy$$ where $z=f=\sqrt{1-y^2}$ and so $f_y=\frac{-y}{\sqrt{1-y^2}}$ so
$$S=2\int\int_D \sqrt{1+\frac{y^2}{1-y^2}}dxdy=2\int_{-1}^1\int_{-\sqrt{1-y^2}}^\sqrt{1-y^2} \frac{1}{\sqrt{1-y^2}}dxdy
\\=2\int_{-1}^12dy=8 $$
A: Another way to find this is integrating on the lengths of the intersections of the surface of interest and the planes of the form $y=k$ for $-1\leq k \leq 1$. I haven't tried it in a while but here it goes.
The sections look like parenthesis joined by flat top and bottom. By parameterizing one of the cylinders with $\theta$; the length of the part that looks like parenthesis is $4\theta$ (here $\theta$ is in radians of course :)). The top and bottom segments add up to $4\cos(\theta)$.
Integrating from $\theta=0$ to $\theta=\frac{\pi}{2}$:
$$\int_0^\frac{\pi}{2} 8(\theta + \cos(\theta))d\theta=\pi^2+8$$
I am not 100% sure about this answer but it does compare nicely to the surface area of the unrestricted drum which is $6\pi$.
