TITLE: Looking for where I went wrong: Finding the volume of a solid that lies within both a cylinder and sphere
QUESTION [12 upvotes]: I'm currently working on this question:
Find the volume of the solid that lies within both the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$.
I decided to use polar coordinates so that the cylinder equation becomes $r^2=1$ and the sphere becomes $r^2+z^2=4$.
Solving for $z$, I get the inequality $-\sqrt{4-r^2}\leq z\leq \sqrt{4-r^2}$. Since I know what $r^2$ is, I plug that in to get the inequality where $z$ is between $-\sqrt{3}$ and $\sqrt{3}$. Combining that to make a triple integral, I get:
$$\int_0^{2\pi}\int_0^1\int_{-\sqrt{3}}^\sqrt{3}r\,\mathrm{d}z\,\mathrm{d}r\,\mathrm{d}\theta$$
However, Slader has a different answer where they didn't plug in $\sqrt{3}$ into the bounds. Why does plugging in the value for $r^2$ make the calculation wrong? Isn't $r^2$ always $1$?

REPLY [4 votes]: A more interesting and beautiful answer is that the answer is the same as the difference of two spheres. One is the sphere of diameter 4, described in the problem, and the other is the sphere with the same diameter as the height of the 'napkin ring' [1] which is the volume left after removing the cylinder intersection you want to find. Basic Pythagoras and volume of sphere equation gets you the answer - no need for calculus. 
I've repeated this answer a few times to similar questions before but it's simplicity is ignored as OP always thinks they MUST do it the complicated way because the question is being asked in a calculus course.
[1] https://en.wikipedia.org/wiki/Napkin_ring_problem