Finding volume of cone using triple integral

Question

I can't understand where I am wrong right here. If I have a cone $z=\sqrt{x^2+y^2}$ located between $z=0$ and $z=1$, its volume is NOT the following integral:

$$V=\int_{-1}^{1}dx\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}dy\int_{0}^{1}dz$$

Below you can see the graph of that cone

So I don't really know where is the mistake in this integral. I would appreciate your help.

Answer 1

The limits for the integral of the $dy$ part is not correct, as the $1$ in $1 - x^2$ should be $z^2$ instead. As such, the integral for the volume using Cartesian co-ordinates would then be

$$V=\int_{-1}^{1}dx\int_{-\sqrt{z^2 - x^2}}^{\sqrt{z^2 - x^2}}dy\int_{0}^{1}dz \tag{1}\label{eq1A}$$

This is because the boundary condition is

$$\begin{equation}\begin{aligned} z & = \sqrt{x^2 + y^2} \\ z^2 & = x^2 + y^2 \\ y^2 & = z^2 - x^2 \\ y & = \pm\sqrt{z^2 - x^2} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Note I would write the integral in the order required to do evaluations from innermost to outermost (i.e., $y$ and $x$ for each circular disk and then $z$ for the sums of volumes of the disks), so it would then instead be

$$V=\int_{0}^{1}\int_{-1}^{1}\int_{-\sqrt{z^2 - x^2}}^{\sqrt{z^2 - x^2}}dydxdz \tag{3}\label{eq3A}$$

Answer 2

The integral you wrote, with the inner integral corrected, equals, with polar (cylindrical in 3D) coordinates:

$$\int_0^{2\pi}\int_0^1\int_r^1 r\, dz\,dr\,d\theta=2\pi\int_0^1r(1-r)\,dr=2\pi\left(\frac12-\frac13\right)=\frac\pi3$$

This is after projecting on the $\;z=0\;$ plane.