Area and Volume Formulas — Question 9

Question 9 - Solution

We derive the formula using cross-sectional disks and integration.

Place the cone so that its tip is at the origin and its base lies in the plane $x=h$. The axis of the cone lies along the $x$-axis.

At a distance $x$ from the tip, where $0\le x\le h$, the radius of the cone is determined by similar triangles.

Since the radius grows linearly from $0$ to $R$ over the height $h$, the radius at position $x$ is $$r(x)=\frac{R}{h}x.$$

A cross-section perpendicular to the $x$-axis at position $x$ is a circle of radius $r(x)$. Its area is $$A(x)=\pi [r(x)]^2 =\pi\left(\frac{R}{h}x\right)^2 =\pi\frac{R^2}{h^2}x^2.$$

An infinitesimal slice of thickness $dx$ has volume $$dV=A(x)\,dx =\pi\frac{R^2}{h^2}x^2\,dx.$$

Integrate from $x=0$ to $x=h$ to obtain the total volume: $$V=\int_0^h \pi\frac{R^2}{h^2}x^2\,dx.$$

Factor out the constants: $$V=\pi\frac{R^2}{h^2}\int_0^h x^2\,dx.$$

Evaluate the integral: $$\int_0^h x^2\,dx=\left[\frac{x^3}{3}\right]_0^h=\frac{h^3}{3}.$$

Substitute back: $$V=\pi\frac{R^2}{h^2}\cdot\frac{h^3}{3} =\frac{1}{3}\pi R^2 h.$$

$$\boxed{V=\frac{1}{3}\pi R^2 h}$$