Find the surface area of the solid obtained by intersecting the solid cylinder y$^2$+z$^2$ ≤ 4 with the solid cylinder x $^2$+z $^2$ ≤ 4.
So, I calculated it via symmetry: You can multiply the surface area of x$^2$ + z$^2$ = 4 by the area lying over disk y$^z$ + z$^2$ ≤ 4 by 8. How did you guys/would you guys do it?