# Volumes of Solids of Revolution

> Spin a curve around an axis to make a solid, then find its volume by adding up thin circular slices with an integral.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/volumes-of-revolution/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The potter's wheel
A potter presses her thumb into a lump of spinning clay. As the wheel turns, the outline of her hand sweeps round and a smooth vase appears. Every vase made on a wheel is round when you cut it straight across.

That gives a neat way to find its volume. Cut the vase into very thin round slices, like a stack of coins. Each slice is a circle with a tiny thickness. Add them all up with an integral and you have the volume.

**In short:** A solid of revolution is made by spinning a curve around an axis, and its volume is found by adding up thin discs. Spun around the x-axis, $V = \pi\int_a^b y^2\,dx$, and around the y-axis, $V = \pi\int_c^d x^2\,dy$.

## Starting point: what you already know

- **Area of a circle:** $A = \pi r^2$. A thin coin of radius $r$ and thickness $h$ has volume $\pi r^2 h$.
- **Definite integrals as sums** (Mathematics Advanced): $\displaystyle\int_a^b f(x)\,dx$ adds up thin strips of height $f(x)$ and width $dx$.
- **Integrating powers:** $\displaystyle\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$.
- **Making $x$ the subject:** for example $y = x^2$ gives $x^2 = y$.

## Slicing into discs

Take the area under $y = f(x)$ from $x = a$ to $x = b$ and spin it once around the x-axis. Slice the solid at right angles to the axis.

- Each slice is a disc of radius $y$ and thickness $\delta x$.
- Its volume is about $\pi y^2\,\delta x$.
- Adding the slices and letting $\delta x \to 0$ turns the sum into an integral.

### Key idea: Volume formulas
About the x-axis: $\displaystyle V = \pi\int_a^b y^2\,dx$

About the y-axis: $\displaystyle V = \pi\int_c^d x^2\,dy$

*Diagram: The curve y equals root x from 0 to 4, its mirror image below the x-axis, and two elliptical slices showing the solid made by spinning the curve around the x-axis.*
*Spinning y = √x around the x-axis. Each slice is a disc whose radius is the height y of the curve.*

### Worked example: A bowl shape
Spin $y = \sqrt{x}$, from $x = 0$ to $x = 4$, around the x-axis.

$y^2 = x$, so
$$V = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi \text{ cubic units}$$

### Worked example: Proving the cone formula
A cone of radius $r$ and height $h$ is made by spinning the line $y = \dfrac{r}{h}x$ from $x = 0$ to $x = h$.

$$V = \pi\int_0^h \frac{r^2}{h^2}x^2\,dx = \pi\frac{r^2}{h^2}\left[\frac{x^3}{3}\right]_0^h = \frac{1}{3}\pi r^2 h$$

The formula you learned in Year 9 falls straight out.

### Calculator tip: Check a volume
If your calculator has an integral key, check $\pi\displaystyle\int_a^b y^2\,dx$ by typing $y^2$ inside the integral and multiplying by `π` outside. Use radian mode when $y$ involves trig functions.

## Spinning around the y-axis

Slices are now horizontal, with radius $x$ and thickness $\delta y$. Write $x^2$ in terms of $y$, and use $y$ limits.

### Worked example: A y-axis solid
Spin $y = x^2$, from $y = 0$ to $y = 4$, around the y-axis.

$x^2 = y$, so $V = \pi\displaystyle\int_0^4 y\,dy = \pi\left[\frac{y^2}{2}\right]_0^4 = 8\pi$.

### Watch out: Square y, not the integral
The formula squares $y$ before integrating: $\pi\displaystyle\int y^2\,dx$. It is not $\pi\left(\displaystyle\int y\,dx\right)^2$.

**Mind map**

```text
Volumes of revolution
  Idea
    Slice into discs
    Disc volume πr²δx
    Add with an integral
  x-axis
    V = π∫y² dx
    x limits
  y-axis
    V = π∫x² dy
    y limits
  Checks
    Cone: πr²h/3
    Sphere: 4πr³/3
  Hollow solids
    Outer minus inner
```

### Year 11: Proving the sphere formula
Spin the semicircle $y = \sqrt{r^2 - x^2}$, from $x = -r$ to $x = r$, around the x-axis.

$$V = \pi\int_{-r}^{r} (r^2 - x^2)\,dx = \pi\left[r^2x - \frac{x^3}{3}\right]_{-r}^{r} = \pi\left(\frac{2r^3}{3} + \frac{2r^3}{3}\right) = \frac{4}{3}\pi r^3$$

### Year 12: Hollow solids (washers)
The region between $y = x$ (top) and $y = x^2$ (bottom), from $x = 0$ to $x = 1$, is spun around the x-axis. Each slice is a washer: a big disc with a smaller hole.

$$V = \pi\int_0^1 \left(x^2 - x^4\right)dx = \pi\left(\frac{1}{3} - \frac{1}{5}\right) = \frac{2\pi}{15}$$

Square each curve first, then subtract. Using $(x - x^2)^2$ gives the wrong answer.

### Year 12: Using a trig identity
Spin one arch of $y = \sin x$, $0 \le x \le \pi$, around the x-axis:
$$V = \pi\int_0^{\pi} \sin^2 x\,dx = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}$$
using the result from Integrating with Trig Identities.

**Other standard integrals.** $y = \sec x$ on $\left[0, \frac{\pi}{4}\right]$ gives $\pi\displaystyle\int_0^{\pi/4}\sec^2 x\,dx = \pi$. $y = \dfrac{1}{\sqrt{1 + x^2}}$ on $[0, 1]$ gives $\pi\displaystyle\int_0^1 \frac{dx}{1 + x^2} = \pi\tan^{-1}1 = \frac{\pi^2}{4}$. $y = \frac{1}{x}$ on $[1, 2]$ gives $\pi\displaystyle\int_1^2 x^{-2}\,dx = \frac{\pi}{2}$.

**About the $y$-axis with logs.** For the region between $y = \ln x$, the $y$-axis, $y = 0$ and $y = 2$, write $x = e^y$, so $x^2 = e^{2y}$: $V = \pi\displaystyle\int_0^2 e^{2y}\,dy = \frac{\pi}{2}(e^4 - 1)$.

## Quick summary

- A solid of revolution is cut into thin discs of volume $\pi r^2\,\delta x$.
- About the x-axis: $V = \pi\displaystyle\int_a^b y^2\,dx$. About the y-axis: $V = \pi\displaystyle\int_c^d x^2\,dy$.
- Square first, then integrate. Use limits that match the variable.
- For a hollow solid, subtract the inner squared radius from the outer one.

## Quiz

Try these questions from the chapter quiz.

1. A thin disc has radius 3 and thickness 0.1. Its volume is:
   - A. $0.3\pi$
   - B. $0.9\pi$
   - C. $0.6\pi$
   - D. $9\pi$
   - Answer: B. $0.9\pi$. $\pi r^2 h = \pi \times 9 \times 0.1$.

2. The volume formula for spinning $y = f(x)$ around the x-axis is:
   - A. $\displaystyle\int_a^b \pi y\,dx$
   - B. $\displaystyle\pi\int_a^b y^2\,dx$
   - C. $\displaystyle 2\pi\int_a^b y\,dx$
   - D. $\displaystyle\pi\left(\int_a^b y\,dx\right)^2$
   - Answer: B. $\displaystyle\pi\int_a^b y^2\,dx$. Each slice is a disc of radius $y$.

3. Spin $y = x$ from $x = 0$ to $x = 3$ around the x-axis. The volume is:
   - A. $3\pi$
   - B. $9\pi$
   - C. $27\pi$
   - D. $\dfrac{9\pi}{2}$
   - Answer: B. $9\pi$. $\pi\displaystyle\int_0^3 x^2\,dx = \pi \cdot 9$.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/volumes-of-revolution/#quiz