# Disc Method

#### Contents

## Summary

Given a region under a curve \(y=f(x)\) for the interval \(a \leq x \leq b\), revolving this region around the \(x\)-axis gives a three-dimensional figure called a **solid of revolution**.

To find the volume of this solid of revolution, consider a thin vertical strip with thickness \(\Delta x\) and height \(y=f(x)\) and consider revolving this thin strip around the \(x\)-axis. This strip generates a thin circular disk with radius \(y=f(x)\) and thickness \( \Delta x\), which has volume

\[ \Delta V = \pi y^2 \Delta x = \pi (f(x))^2 \Delta x.\]

The **disk method** calculates the volume of the full solid of revolution by summing the volumes of these thin circular disks from the left endpoint \(a\) to the right endpoint \(b\) as the thickness \( \Delta x \) goes to \(0\) in the limit. This gives the volume of the solid of revolution:

\[ V = \int_a^b dV = \int_a^b \pi (f(x))^2 dx.\]

## Basic Examples

Volume of a cylinder: A cylinder with height \(h\) and base radius \(r\) can be thought of as the solid of revolution obtained by revolving the line \(y=r\) around the \(x\)-axis. Using the disk method, find the volume of the cylinder of height \(h\) and base radius \(r\).

Using the disk method, we revolve the line \(y=r\) around the \(x\)-axis from \(x=0\) to \(x=h\). Then \[ \Delta V = \pi y^2 = \pi r^2 \Delta x\] and the volume of the right circular cylinder is \[ \begin{align} V &= \int_{0}^{h} dV = \int_{0}^{h} \pi r^2 dx \\ & = \left[ \pi r^2 x \right]^{h}_{0} \\ & = \pi r^2 h. \ _\square \end{align} \]

Volume of a sphere: A sphere can be thought of as the solid of revolution obtained by revolving a semicircle around the \(x\)-axis. Using the disk method, find the volume of the sphere of radius \(r\).

We can consider the semicircle to be centered at the origin with radius \(r\), which has equation \(y = \sqrt{r^2 - x^2}\). Then \[ \Delta V = \pi y^2 = \pi (r^2 - x^2) \Delta x\] and the volume of the sphere is \[ \begin{align} V &= \int_{-r}^{r} dV = \int_{-r}^{r} \pi (r^2-x^2) dx \\ & = \left[ \pi r^2 x - \frac{\pi x^3}{3} \right]^{r}_{-r}\\ &= \pi r^2 (r-(-r)) - \left( \frac{\pi r^3}{3} - \frac{\pi (-r)^3}{3} \right)\\ & = 2 \pi r^3 - \left(\frac{2 \pi r^3}{3} \right)\\ & = \frac{4\pi r^3}{3}. \ _\square \end{align}\]

Volume of a right circular cone: A right circular cone can be thought of as the solid of revolution obtained by revolving a right triangle around the \(x\)-axis. Using the disk method, find the volume of the right circular cone of height \(h\) and base radius \(r\).

The hypotenuse of the right triangle we revolve around the \(x\) axis is given by \(y = \frac{r}{h}x \). Then \[ \Delta V = \pi y^2 = \pi \left( \frac{r}{h}x \right)^2 \Delta x\] and the volume of the right circular cone is \[ \begin{align} V &= \int_{0}^{h} dV = \int_{0}^{h} \pi \frac{r^2}{h^2 }x^2 dx \\ & = \left[ \pi \frac{r^2}{h^2} \cdot \frac{1}{3} x^3\right]^{h}_{0}\\ & = \frac{ \pi r^2 h}{3}. \ _\square \end{align}\]

### As shown in the examples above, we have the following relationships between the volumes of the cone and sphere as fractional parts of the volumes of their respective circumscribed cylinders:

\[ \begin{align} V_\text{cone} & = \frac{1}{3} V_\text{cylinder}\\ V_\text{sphere} &= \frac{2}{3} V_\text{cylinder}. \end{align}\]

## Intermediate Examples

## When the curve \(\displaystyle{y=\frac{1}{x}}\ \ (1\le x \le \infty)\) is revolved about the \(x\)-axis, a funnel-shaped surface is formed. What is the volume of that revolution?

Using the disk method gives

\[ \Delta V = \pi y^2 = \pi\cdot \frac{1}{x^2} \Delta x,\]

and the volume of the revolution is

\[ \begin{align} V &= \int_{1}^{\infty} dV = \int_{1}^{\infty} \pi\cdot\frac{1}{x^2}dx \\ & =\left.-\pi\cdot\frac{1}{x}\right|_{1}^{\infty}\\ &= -\pi(0-1)\\ & = \pi. \end{align}\]

It is interesting to note that the volume is finite. \(_\square\)

Note: You may use a calculator for the final step of your calculation.

###### This question was posed by my calculus teacher as the last and of higher value exercise on its third Calculus I test.

It is then shot straight into a very thick wall (i.e. it does not pierce the other side at all) making a closed cylindrical hole until it stops moving. Then the bullet is carefully extracted without affecting the hole at all, leaving an empty hole with a pointy end where the bullet once was.

The length of the entire hole is \(e+1\). If the volume of the hole can be expressed as \[\pi i e\] where \(i\) is a constant, find the value of \(i\).

The above figure consists of a rectangle with a semicircle cut out of one end and added to the other end, where \(L\) is the width of the rectangle, and the curved length of a semicircle is \( \pi r \).

To calculate the area of the shaded figure, Svatejas applies the disc method as follows:

Consider the axis of integration to be the semicircular arc, which has length \( \pi r \). For each horizontal strip, we have an area element (technically length element) of \(L \). Hence, the area is

\[ \int_{R} L \, dR = \pi r \times L \]

**What is the area of the shaded figure**?

Inspiration, see solution comments.