How to show that the volume of a cone with an arbitrary base is $(1/3)Bh$?

Question

I know how to show, by means of integration, that the volume of a circular based cone is $(1/3)Bh$, where $B$ is the area of the base. I also know how to show that the volume of a squared base pyramid is given by the same formula. However, my Geometry book states (whithout proof) that that formula can also be used to find the volume of a cone with an arbitrary-shaped base. How can we prove that? Thanks.

Answer 1

If we make a slice through the cone at level $z$ from the top, we will get a similar figure, scaled by factor of $z/h$. Its area then will be $B\cdot z^2/h^2$. If the slice has a thickness of $dz$, its elementary volume will be $$dV = B\cdot z^2/h^2\cdot dz$$ Integrate it for $z$ from $0$ to $h$ and you wil get the volume:

$$V = \frac{B}{h^2}\int_{0}^{h}{z^2 dz} = \frac{B}{h^2}\frac{h^3}{3}=\frac{Bh}{3}$$

Answer 2

Place the vertex of the origin at height $0$.

The area of the base is $B$, that is when the height is $h$, the area is $B$.

The area of the cross section at height $r$ would be $\left( \frac{r}{h}\right)^2B$

Hence the volume would be $$\int_0^h \left( \frac{r}h \right)^2B\, dr=\frac13hB$$

Answer 3

It's even more general. The only requirement is that both Base dimensions contract equally and linearly with height. Then you can have an arbitrarily shaped base area which contracts proportional to the square of height. Integrating will then always give the factor $1/3$.

Answer 4

Place the cone "on its side", so that the vertex is at the origin and we have a line (the cone's lateral side) from the origin to the point $\;(h,r)\;$ , with $\;h=\;$ the cone's height and $\;r=\;$ its circular base's radius.

Define the function $\;f(x)=\frac rhx\;$ , so that our cone is obtained by revolving this line about the $\;x\,-$ axis, and thus its volume is

$$V=\pi\int_0^hf(x)^2dx=\pi\int_0^h\frac{r^2}{h^2}x^2\,dx=\left.\pi\frac{r^2}{3h^2}x^3\right|_0^h=\frac{\pi r^2h}3=\frac13Bh$$

since $\;B=\pi r^2\;$ ...

Answer 5

For a generic shape, assuming that the width $d(x)$ is linear with the height

$$d(x)=d_0-\frac{d_0}{h}x$$

we have then the area is quadratic

$$B(x)=kd^2(x)=kd_0^2-2\frac{kd_0^2}{h}x+\frac{kd_0^2}{h^2}x^2=B_0-2\frac{B_0}{h}x+\frac{B_0}{h^2}x^2$$

then

$$V=\int_0^h (B_0-2\frac{B_0}{h}x+\frac{B_0}{h^2}x^2) dx = \left[B_0x-\frac{B_0}{h}x^2+\frac{B_0}{3h^2}x^3\right]_0^h= B_0h-B_0h+\frac{B_0}{3}h=\frac13B_0h$$

Answer 6

Actually you just need Cavalieri's principle. It ensures that a cone and a pyramid with the same height and the same base area have the same volume, so it is enough to prove that the volume of a pyramid with height $h$ and base given by a square with side length $l$ is $\frac{1}{3}hl^2$. We may also assume that the projection of the apex on the base lies at the center of the base. And we may consider that a cube with side length $l$ can be dissected into six pyramids, having the center of the cube as common apex and their bases at the faces of the cube. This proves that if the height is $\frac{l}{2}$ and the base area is $l^2$ the volume is $\frac{l^3}{6}$. The general claim follows by applying a suitable dilation.