# Volume of a frustum

SE users

I was given the problem:

Find the volume of a frustum of a right circular cone with height h, lower base radius R, and top radius r.

My working looks a bit too complicated and comes out with a really complicated solution. What method should I have used to avoid this? Just consider the difference between the volumes of two right cones, the first one having radius $R$ and height $\frac{Rh}{R-r}$, the second one having radius $r$ and height $\frac{rh}{R-r}$. Obviously the volume of the greates cone is just $\frac{R^3}{r^3}$ times the volume of the smallest, since all the dimensions are just multiplied by a factor $\frac{R}{r}$. This gives:

$$V = \left(1-\frac{r^3}{R^3}\right)\frac{\pi}{3} R^2 \frac{Rh}{R-r},$$ or, by writing $R^3-r^3$ as $(R-r)(R^2+Rr+r^2)$,

$$V = \frac{\pi\,h}{3}(R^2+Rr+r^2).$$

If you first simplify your expression for $x$, you get

$\displaystyle x=\frac{(R-r)(\frac{Rh}{R-r}-y)}{h}=R-\frac{R-r}{h}y,\;\;$ and then

$\displaystyle V=\pi\int_{0}^{h}(R^2-\frac{2R(R-r)}{h}y+\frac{(R-r)^2}{h^2}y^2) \;dy$

$\;\;\;\displaystyle=\pi\left[R^2h-\frac{R(R-r)}{h}\cdot h^{2}+\frac{(R-r)^2}{h^2}\cdot\frac{h^3}{3}\right]$

$\;\;\;\displaystyle=\pi\left[R^2h-R(R-r)h+\frac{1}{3}(R-r)^2h\right]$

$\;\;\;\displaystyle=\frac{\pi}{3}[3R^2h-3R(R-r)h+(R-r)^{2}h]$

$\;\;\;\displaystyle=\frac{\pi}{3}[R^2+Rr+r^2]h$.

During derivation look only at essentials making up the differential volume element.

The slope of slant generator

$$m= \frac{R-r}{h}\; ;$$

Equation of cone generator

$$x = m\, y ;$$

Volume of cone frustum

$$\int_{ y=r/m} ^{y=R/ m} \pi x^2 dy,$$

integrate and simplify.

Verify the result you got. V = $\dfrac{\pi\,h}{3}(R^2+Rr+r^2).$ For $r=0$ it should be the formula for a full cone , and for $r=R$ for a cylinder.