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I need to find the volume between the cylinder $x^2+y^2=4$ and the cone $z^2=x^2+y^2$. From the graph I can see they intersect when $z=-2$ and $z=2$. In the xy plane there seems to be a circle of radius 2. And since there's a full rotation around the circle, I'd say $\theta = 2 \pi$. But I don't know how to compute $r$, since the radius varies with height.

I understand the radius goes from the cone to the cylinder, however I'm not sure how to write that and how to set up this triple integral bounds.

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You are correct: the projection of the solid in the $xy$ plane is a disc of radius $2$. Then, $z$ ranges from the negative part of the cone ($z=-r$) all the way up to its positive part ($z=r$). Therefore: $$ V=\int_{\theta=0}^{2\pi}\int_{r=0}^2\int_{z=-r}^r r\;dzdrd\theta = \frac{32\pi}{3} $$

Alternatively, a slice of the solid at height $z$ is an annulus with small radius $r=z$ and big radius $r=2$, so its area is $ A(z)=\pi(2^2-z^2) $, and so the volume is $$ V=\int_{-2}^2A(z)\;dz = \frac{32\pi}{3} $$

Don't forget: if you choose $r$ as constants, $z$ must depend on $r$, or vice versa. You mixed up both options.

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