Volume of the Paraboloid

Problem

Use integration to derive the volume of a paraboloid of radius $r$ and height $h$. Compare the volume of the paraboloid to the volume of the cylinder with equal base and height.

Figure 1. Paraboloid

Solution

Figure 2. Half a parabola

Consider half a parabola where the interval of $x$ is $[0,h]$:

$r(x)=a{\sqrt {x}}$

where $a$ is constant.

At $x=0$, we have $r(0)=a{\sqrt {0}}=0$.

At $x=h$, we have $r(h)=a{\sqrt {h}}=r$.

Solving for the constant yields $a={\frac {r}{\sqrt {h}}}$; thus,

$r(x)={\frac {r}{\sqrt {h}}}{\sqrt {x}}$.

Use the disk method to calculate the volume:

${\begin{aligned}V&=\pi \int _{a}^{b}[r(x)]^{2}dx\\&=\pi \int _{0}^{h}{\frac {r^{2}}{h}}xdx\\&={\frac {\pi r^{2}}{h}}\int _{0}^{h}xdx\\&={\frac {\pi r^{2}}{h}}\left[{\frac {x^{2}}{2}}{\Biggr |}_{0}^{h}\right]\\&={\frac {\pi r^{2}h}{2}}.\end{aligned}}$

When compared with the volume the cylinder ${V}_{cylinder}=\pi r^{2}h$, the volume of the paraboloid is half the volume of the cylinder.