Volume of a cone given constraints on $z$

Question

This question is pretty straightforward, however, it's been a couple years since I've had to solve an integral like this!

Compute the volume bounded by $x^2 + y^2 \le z^2$ such that $c_1 \le z \le c_2$ where $c_1,c_2 \in \Bbb{R}_+$

I believe the first equation is a cone so I think the solution is something like $$\int_0^1\int_{0}^{2\pi}\int_{c_1}^{c_2}dzrd\theta dr$$ however, I'm not confident in the bounds I set.

Answer 1

No; that's not the volume of that cone; instead, it's the volume of the cylinder$$\left\{(x,y,z)\in\Bbb R^3\,\middle|\,x^2+y^2\leqslant1\wedge z\in[c_1,c_2]\right\}.$$Note that\begin{align}x^2+y^2\leqslant z^2&\iff r^2\leqslant z^2\\&\iff r\leqslant z,\end{align}since $z,r\geqslant0$. So, it should be$$\int_0^{2\pi}\int_{c_1}^{c_2}\int_0^zr\,\mathrm dr\,\mathrm dz\,\mathrm d\theta$$

Answer 2

You don't need the spherical coordinates in the particular case.

The volume is given by $$ V=\int_{c_1}^{c_2}\left(\iint_{x^2+y^2\le z^2} 1 dxdy\right)dz =\int_{c_1}^{c_2}\pi z^2dz=\frac{\pi}{3}z^3\big|_{c_1}^{c_2} $$