Looking for where I went wrong: Finding the volume of a solid that lies within both a cylinder and sphere

“Isn’t $r^2$ always 1?”

Yes, on the surface of the cylinder.

But you aren’t integrating on the surface of the cylinder. You’re integrating over a three-dimensional region, namely the one bounded by both the cylinder and the sphere. This region is described in cylindrical coordinates as the set with

$$r\in[0,1]$$ $$\theta\in[0,2\pi)$$ $$z\in\left[-\sqrt{4-r^2}, \sqrt{4-r^2}\right]$$


No, $r$ is not always $1$. We're in the integral; $r$ is a variable we're going to integrate over later, and it doesn't have any fixed value.

The cylinder equation $r=1$ represents part of the boundary. It's certainly not true for every point inside the solid.


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

Tags:

Volume

Integration

Calculus

Multivariable Calculus

Multiple Integral

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