Question

Use the disk method to verify that the volume of a right circular cone is \(\frac{1}{3} \pi r^{2} h,\) where \(r\) is the radius of the base and \(h\) is the height.

Short answer

The disk method confirms the formula for the volume of a right circular cone, which is \(\frac{1}{3} \pi r^{2} h\), derived by using integral calculus.

Step-by-step solution

Set up the disk method

Consider a cone as a collection of small disks stacked upon each other. The radius of each disk can be expressed as a function \(y(x) = r\frac{x}{h}\), where \(x\) is the distance from the vertex of the cone, \(r\) is the radius of the base, and \(h\) is the height of the cone. The small change in \(x\) is presented as \(dx\). The small volume \(dV\) of the tiny disk is therefore \(dV = \pi [y(x)]^2 dx = \pi [r\frac{x}{h}]^2 dx\).

Integrate to find the volume

Next, the integral of this expression from \(x = 0\) to \(x = h\) is taken to sum up all the small volumes \(dV\). This gives us the total volume \(V\) of the cone: \(V = \int_0^h \pi [r\frac{x}{h}]^2 dx\).

Evaluate the Integral

Evaluate the above integral. Upon doing so, we get: \(V = \pi \int_0^h (\frac{r^2x^2}{h^2}) dx\). Which simplifies to \(V = \frac{\pi r^2}{h^2} \int_0^h x^2 dx\). Evaluating the integral on the rhs, we get \(V = \frac{\pi r^2}{h^2} [\frac{x^3}{3}]_0^h = \frac{\pi r^2}{h^2}[\frac{h^3}{3}-0] = \frac{1}{3} \pi r^{2} h\). This matches with the formula for the volume of a cone.