# 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:

$V = \int_a^b dV = \int_a^b \pi \big(f(x)\big)^2 dx.$

## 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 $\begin{aligned} V &= \int_{0}^{h} dV \\&= \int_{0}^{h} \pi r^2 dx \\ & = \left[ \pi r^2 x \right]^{h}_{0} \\ & = \pi r^2 h. \ _\square \end{aligned}$

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 $\Delta V = \pi y^2 = \pi \big(r^2 - x^2\big) \Delta x$ and the volume of the sphere is $\begin{aligned} V &= \int_{-r}^{r} dV \\&= \int_{-r}^{r} \pi \big(r^2-x^2\big) dx \\ & = \left[ \pi r^2 x - \frac{\pi x^3}{3} \right]^{r}_{-r}\\ &= \pi r^2 \big(r-(-r)\big) - \left( \frac{\pi r^3}{3} - \frac{\pi (-r)^3}{3} \right)\\ & = 2 \pi r^3 - \left(\frac{2 \pi r^3}{3} \right)\\ & = \frac{4\pi r^3}{3}. \ _\square \end{aligned}$

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 $\begin{aligned} V &= \int_{0}^{h} dV \\&= \int_{0}^{h} \pi \frac{r^2}{h^2 }x^2 dx \\ & = \left[ \pi \frac{r^2}{h^2} \cdot \frac{1}{3} x^3\right]^{h}_{0}\\ & = \frac{ \pi r^2 h}{3}. \ _\square \end{aligned}$

### 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:

$\begin{aligned} V_\text{cone} & = \frac{1}{3} V_\text{cylinder}\\\\ V_\text{sphere} &= \frac{2}{3} V_\text{cylinder}. \end{aligned}$

## 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

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

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

Let there be a region $R:\{(x,y) \ | \ x^{1/4}+y^{4} \leq 1\}$. What is the volume of the solid generated when $R$ is rotated around the line $x=0?$ Give your answer to 3 decimal places.

$$

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 higher value exercise on the third Calculus I test.

$y = \ln(x)$ from $x = 1$ to $x = e$ about the $x$-axis.

A bullet is formed by revolving the area bounded by the the curveIt is then shot straight into a very thick wall (i.e. it does not pierce through 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.