Solids of Revolution — Disk Method
Spin a curve around the $x$-axis and a solid appears — the kind of shape a lathe produces. The volume of that solid is found by stacking thin circular disks along the axis. Each disk has radius $f(x)$ and thickness $dx$, giving volume $\pi[f(x)]^2\,dx$. Sum infinitely many and you have the disk method formula.
Rotate $y = \sqrt{x}$ about the $x$-axis from $x = 0$ to $x = 4$. Without computing — sketch the resulting solid and estimate whether its volume is closer to $8\pi$, $16\pi$, or $25\pi$ cubic units.
When we rotate a curve $y = f(x) \geq 0$ about the $x$-axis, we generate a solid of revolution. To find its volume, slice the solid perpendicular to the $x$-axis. Each slice is a disk (a thin cylinder) with:
- Radius $r = f(x)$ — the height of the curve at position $x$
- Thickness $dx$ — infinitely thin
- Volume $= \pi r^2 \cdot dx = \pi[f(x)]^2\,dx$
Integrating these infinitesimally thin disks from $x = a$ to $x = b$ gives the total volume:
$V = \pi\displaystyle\int_a^b [f(x)]^2\,dx$
This is the disk method. Note: $[f(x)]^2$ is the square of the radius — not just $f(x)$.
Key facts
- Disk method about $x$-axis: $V = \pi\displaystyle\int_a^b [f(x)]^2\,dx$
- Each disk has volume $\pi[f(x)]^2\,dx$
- Limits $a$ and $b$ are $x$-values (or $y$-values for rotation about the $y$-axis)
Concepts
- Why slicing perpendicular to the axis of rotation produces circular disks
- How the formula $V = \pi\int [f(x)]^2\,dx$ arises from the Riemann sum of disk volumes
- The connection between disk area ($\pi r^2$) and the integrand $\pi[f(x)]^2$
Skills
- Set up and evaluate $V = \pi\displaystyle\int_a^b [f(x)]^2\,dx$ for standard curves
- Identify and correctly square the radius function
- Express the final volume exactly in terms of $\pi$
To find the volume of a solid formed by rotating $y = f(x)$ about the $x$-axis from $x = a$ to $x = b$:
- Sketch the curve and the solid. Draw a representative disk at position $x$, labelling its radius $r = f(x)$ and thickness $dx$.
- Write the disk volume: $dV = \pi[f(x)]^2\,dx$.
- Integrate: $V = \pi\displaystyle\int_a^b [f(x)]^2\,dx$.
- Evaluate the integral and state the answer in terms of $\pi$ cubic units.
To find the volume of a solid formed by rotating $y = f(x)$ about the $x$-axis from $x = a$ to $x = b$:
Pause — copy the disk method formula $V=\pi\int_a^b[f(x)]^2\,dx$ and the four-step setup (identify $f$, square, set limits, integrate) into your book.
Quick check: Which integral gives the volume when $y = x^2$ is rotated about the $x$-axis from $x = 0$ to $x = 2$?
We just saw the disk method formula $V=\pi\int_a^b[f(x)]^2\,dx$ and the four-step setup: identify $f(x)$, write $[f(x)]^2$, state limits, integrate. That raises a question: applying this to $y=\sqrt{x}$ on $[0,4]$, the integrand becomes $[\sqrt{x}]^2=x$ — how does the squaring simplify the computation, and what is the exact volume? This card answers it → $V=\pi\int_0^4 x\,dx=\pi[x^2/2]_0^4=8\pi$.
Example: Find the volume of the solid formed when $y = \sqrt{x}$ is rotated about the $x$-axis from $x = 0$ to $x = 4$.
Step 1: $f(x) = \sqrt{x}$, so $[f(x)]^2 = x$.
Step 2: $V = \pi\displaystyle\int_0^4 x\,dx$.
The volume is $\mathbf{8\pi} \approx 25.1$ cubic units. The solid is a shape called a paraboloid — check your estimate!
Notice: When $f(x) = \sqrt{x}$, squaring gives $[f(x)]^2 = x$ — a huge simplification. Always square the radius function before integrating.
Example: Find the volume of the solid formed when $y = \sqrt{x}$ is rotated about the $x$-axis from $x = 0$ to $x = 4$.
Pause — copy the worked example $y=\sqrt{x}$ on $[0,4]$: integrand $=x$, $V=\pi\int_0^4 x\,dx=8\pi$ into your book.
Did you get this? True or false: when $y = x$ is rotated about the $x$-axis from $x = 0$ to $x = 3$, the integrand in the disk formula is $x^2$ (not $x$).
Worked examples · 3 in a row, reveal as you go
Find the volume of the solid formed when $y = x$ is rotated about the $x$-axis from $x = 0$ to $x = 3$.
Find the volume of the solid formed when $y = x^2$ is rotated about the $x$-axis from $x = 0$ to $x = 2$.
Find the volume of the solid formed when $y = \sin x$ is rotated about the $x$-axis from $x = 0$ to $x = \pi$.
Fill the gap: When $y = x$ is rotated about the $x$-axis from $x = 0$ to $x = 3$, the integrand in the disk formula is $[f(x)]^2 = $ and the volume equals $\pi$ cubic units.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: to evaluate $\pi\displaystyle\int_0^{\pi}\sin^2 x\,dx$, you should first replace $\sin^2 x$ with $\dfrac{1 - \cos 2x}{2}$.
Activities · practice with the ideas
Find the volume of the solid formed when $y = 2x$ is rotated about the $x$-axis from $x = 0$ to $x = 1$.
Find the volume of the solid formed when $y = \sqrt{x}$ is rotated about the $x$-axis from $x = 1$ to $x = 9$.
Find the volume of the solid formed when $y = x^3$ is rotated about the $x$-axis from $x = 0$ to $x = 2$.
Find the volume of the solid formed when $y = \cos x$ is rotated about the $x$-axis from $x = 0$ to $x = \dfrac{\pi}{2}$. (Use the double-angle identity.)
The curve $y = \sqrt{r^2 - x^2}$ (upper semicircle of radius $r$) is rotated about the $x$-axis from $x = -r$ to $x = r$. Show that the volume is $\dfrac{4}{3}\pi r^3$.
Odd one out: Three of these volume integrals are set up correctly. Which one is NOT?
Earlier you estimated the volume when $y = \sqrt{x}$ is rotated about the $x$-axis from $x = 0$ to $x = 4$.
The exact answer is $V = 8\pi \approx 25.1$ cubic units. The key insight: squaring $\sqrt{x}$ gives $x$, turning a potentially messy integral into $\pi\displaystyle\int_0^4 x\,dx = \pi\cdot 8 = 8\pi$. Recognising when squaring simplifies the function is an important exam skill.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the volume of the solid formed when $y = 3x$ is rotated about the $x$-axis from $x = 0$ to $x = 2$. (2 marks)
Q2. Find the volume of the solid formed when $y = x^2$ is rotated about the $x$-axis from $x = 1$ to $x = 3$. (3 marks)
Q3. The upper semicircle $y = \sqrt{9 - x^2}$ is rotated about the $x$-axis. By applying the disk method, find the volume of the resulting sphere and verify it equals $\dfrac{4}{3}\pi r^3$ with $r = 3$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $y = 2x$: $[f]^2 = 4x^2$; $V = \pi\displaystyle\int_0^1 4x^2\,dx = \pi\left[\dfrac{4x^3}{3}\right]_0^1 = \dfrac{4\pi}{3}$ cu units.
2. $y = \sqrt{x}$, $[1,9]$: $V = \pi\displaystyle\int_1^9 x\,dx = \pi\left[\dfrac{x^2}{2}\right]_1^9 = \pi\cdot\dfrac{81-1}{2} = 40\pi$ cu units.
3. $y = x^3$, $[0,2]$: $[f]^2 = x^6$; $V = \pi\left[\dfrac{x^7}{7}\right]_0^2 = \dfrac{128\pi}{7}$ cu units.
4. $y = \cos x$: $\cos^2 x = \tfrac{1+\cos 2x}{2}$; $V = \tfrac{\pi}{2}\left[x + \tfrac{\sin 2x}{2}\right]_0^{\pi/2} = \tfrac{\pi}{2}\cdot\dfrac{\pi}{2} = \dfrac{\pi^2}{4}$ cu units.
5. $y = \sqrt{r^2 - x^2}$: $[f]^2 = r^2 - x^2$; $V = \pi\left[r^2 x - \dfrac{x^3}{3}\right]_{-r}^{r} = \pi\left[\left(r^3 - \dfrac{r^3}{3}\right) - \left(-r^3 + \dfrac{r^3}{3}\right)\right] = \pi\cdot\dfrac{4r^3}{3} = \dfrac{4}{3}\pi r^3$ ✓.
Q1 (2 marks): $[f(x)]^2 = 9x^2$ [1]. $V = \pi\displaystyle\int_0^2 9x^2\,dx = \pi\left[3x^3\right]_0^2 = 24\pi$ cu units [1].
Q2 (3 marks): $[f(x)]^2 = x^4$ [1]. $V = \pi\displaystyle\int_1^3 x^4\,dx = \pi\left[\dfrac{x^5}{5}\right]_1^3 = \pi\cdot\dfrac{243-1}{5} = \dfrac{242\pi}{5}$ cu units [2].
Q3 (3 marks): $[f(x)]^2 = 9 - x^2$ [1]. $V = \pi\left[9x - \dfrac{x^3}{3}\right]_{-3}^{3} = \pi\left[(27-9) - (-27+9)\right] = \pi(18 + 18) = 36\pi$ cu units [1]. Check: $\dfrac{4}{3}\pi(3)^3 = \dfrac{4}{3}\pi\cdot 27 = 36\pi$ ✓ [1].
Five timed disk-method questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering disk-method questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.