How do you 'mirror' a function (i.e. a cone's intersect) to appear on both sides of the x-axes?

Question

I stumbled across this when trying to prove with integration that the volume of every cone is $\frac{r^2\pi m}{3}$. Now everything is cleared up except I don't know how to formulate the function to be integrated.
How could I construct a function to describe a cone's intersect?

Answer 1

The integral can be written in several ways. Suppose that the cone has the vertex at $(0,0,0)$, and the axis along $x$.

Method 1: Use disks parallel to the $yz$ plane, at a distance $x$ from the origin, with thickness $dx$. Then the radius of such disk is $R(x)=\frac{r}{m}x$, and the volume is $\pi R^2(x)dx$. You can then write $$V=\int_0^m\pi\left(\frac{r}{m}x\right)^2dx$$

Method 2: Use infinitesimally small cubes - this yields a triple integral. $x$ goes from $0$ to $m$, $y$ goes from $-xr/m$ to $xr/m$ and $z$ goes between $\pm\sqrt{\frac{x^2r^2}{m^2}-y^2}$ $$V=\int_0^mdx\int_{-xr/m}^{xr/m}dy\int_{-\sqrt{\frac{x^2r^2}{m^2}-y^2}}^{+\sqrt{\frac{x^2r^2}{m^2}-y^2}}dz$$

Method 3: you can use rings. This yields a double integral. I let you write that out.