Spin a curve around an axis and it traces out a three-dimensional solid — a vase, a rocket nose cone, or a wine glass. By slicing that solid into infinitely thin disks and adding up their volumes, we can find the exact volume of almost any rotational shape. This is the power of volumes of solids of revolution.
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Imagine spinning the line $y = x$ around the $x$-axis from $x = 0$ to $x = 3$. The shape you create is a cone. If you sliced this cone into many thin circular disks perpendicular to the $x$-axis, how might you use the radius of each disk to approximate the total volume? What would happen to your approximation as the disks become thinner?
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Wrong: Integration and differentiation produce identical results.
Right: Integration is the reverse of differentiation, but includes an arbitrary constant (+C) for indefinite integrals.
📚 Core Content
When a curve $y = f(x)$ is rotated $360^\circ$ about the $x$-axis, it sweeps out a solid. If we slice this solid perpendicular to the $x$-axis, each slice is approximately a circular disk with:
Summing these disks from $x = a$ to $x = b$ gives the total volume:
$$V = \pi \int_a^b \bigl(f(x)\bigr)^2 \, dx$$
Similarly, rotating $x = g(y)$ about the $y$-axis from $y = c$ to $y = d$ gives:
$$V = \pi \int_c^d \bigl(g(y)\bigr)^2 \, dy$$
If the region being rotated is bounded by two curves, the cross-section is a washer — a disk with a hole in the middle. The volume of each washer is the volume of the outer disk minus the volume of the inner disk:
$$V = \pi \int_a^b \Bigl(\bigl(R(x)\bigr)^2 - \bigl(r(x)\bigr)^2\Bigr) \, dx$$
where $R(x)$ is the outer radius (distance from axis to outer curve) and $r(x)$ is the inner radius (distance from axis to inner curve).
Sometimes a curve is given as $y = f(x)$ but you need to rotate around the $y$-axis. In that case, rearrange to express $x$ in terms of $y$ before integrating.
🧮 Worked Examples
🧪 Activities
1 Find the volume when $y = x$ is rotated about the $x$-axis from $x = 0$ to $x = 3$.
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2 Find the volume when $y = x^2$ is rotated about the $x$-axis from $x = 0$ to $x = 2$.
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3 The region bounded by $y = \sqrt{x}$, the $y$-axis, and $y = 2$ is rotated about the $y$-axis. Find the volume.
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4 Find the volume when the region between $y = x$ and $y = x^2$ is rotated about the $x$-axis.
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1 Explain why the disk method formula contains $\pi r^2$ and not just $r^2$.
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2 A student is finding the volume when $y = x^2$ from $x = 0$ to $x = 3$ is rotated about the $y$-axis. They integrate $\pi \int_0^3 (x^2)^2 \, dx$ with respect to $x$. What is wrong with their setup?
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Earlier you thought about slicing a cone into thin disks.
Each disk has radius $y = x$ and thickness $dx$, so its volume is $\pi x^2 \, dx$. Integrating from $0$ to $3$ gives $\pi \int_0^3 x^2 \, dx = 9\pi$. Remarkably, this matches the geometric formula for a cone: $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3)^2 (3) = 9\pi$. The disk method is simply calculus verifying geometry — but it works for shapes far too complex for any simple geometric formula.
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5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Find the volume of the solid formed when each curve is rotated about the given axis. Leave answers in terms of $\pi$. 4 MARKS
(a) $y = 2x$ about the $x$-axis from $x = 0$ to $x = 1$.
(b) $y = x^2$ about the $y$-axis from $y = 0$ to $y = 4$.
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7. Find the volume of the solid formed when the region bounded by $y = x + 1$, $y = 0$, $x = 0$, and $x = 2$ is rotated about the $x$-axis. 3 MARKS
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8. The region enclosed by $y = x^2$ and $y = 2x$ is rotated about the $x$-axis. 4 MARKS
(a) Sketch the region and the solid formed.
(b) Find the volume of the solid using the washer method. Show all working.
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1. $V = \pi \int_0^3 x^2 \, dx = \pi \bigl[\frac{x^3}{3}\bigr]_0^3 = 9\pi$
2. $V = \pi \int_0^2 x^4 \, dx = \pi \bigl[\frac{x^5}{5}\bigr]_0^2 = \frac{32\pi}{5}$
3. $x = y^2$, so $V = \pi \int_0^2 (y^2)^2 \, dy = \pi \int_0^2 y^4 \, dy = \pi \bigl[\frac{y^5}{5}\bigr]_0^2 = \frac{32\pi}{5}$
4. Washer method: $V = \pi \int_0^1 (x^2 - x^4) \, dx = \frac{2\pi}{15}$
1. Each slice is a circular disk. The area of a circle is $\pi r^2$, so the volume of each thin disk is $\pi r^2 \cdot \text{thickness}$. Without $\pi$, we would just be summing $r^2$ values, not actual volumes.
2. Rotation about the $y$-axis requires integration with respect to $y$, not $x$. We need $x = \sqrt{y}$ and limits from $y = 0$ to $y = 9$: $V = \pi \int_0^9 y \, dy$.
1. B — $V = \\pi int_0^2 x^2 , dx = \\pi \bigl[\frac{x^3}{3}\bigr]_0^2 = \frac{8pi}{3}$.
2. B — Rotation about the $y$-axis requires integrating with respect to $y$, expressing $x$ as a function of $y$.
3. B — Each disk has volume $\\pi r^2 , dx = \\pi y^2 , dx$, so $V = \\pi int_a^b y^2 , dx$.
4. B — Washer method: $V = \\pi int_0^1 (x^2 - x^4) , dx = \\pi(\frac{1}{3} - \frac{1}{5}) = \frac{2pi}{15}$.
5. B — Volume of washer = $\\pi R^2 - \\pi r^2 = \\pi(R^2 - r^2)$.
6. B — $V = \\pi int_0^4 (\\sqrt{x})^2 , dx = \\pi int_0^4 x , dx = \\pi \bigl[\frac{x^2}{2}\bigr]_0^4 = 8pi$.
7. B — $x = \\sqrt{y}$, so $V = \\pi int_0^4 y , dy = \\pi \bigl[\frac{y^2}{2}\bigr]_0^4 = 8pi$.
Q6 (4 marks): (a) $V = \pi \int_0^1 4x^2 \, dx = \pi \bigl[\frac{4x^3}{3}\bigr]_0^1 = \frac{4\pi}{3}$ [2]. (b) $x = \sqrt{y}$, so $V = \pi \int_0^4 y \, dy = \pi \bigl[\frac{y^2}{2}\bigr]_0^4 = 8\pi$ [2].
Q7 (3 marks): $V = \pi \int_0^2 (x + 1)^2 \, dx = \pi \int_0^2 (x^2 + 2x + 1) \, dx = \pi \bigl[\frac{x^3}{3} + x^2 + x\bigr]_0^2 = \frac{26\pi}{3}$ [3].
Q8 (4 marks): (a) Sketch: parabola and line intersecting at $(0,0)$ and $(2,4)$, rotated about $x$-axis [1]. (b) Washer method: $V = \pi \int_0^2 \bigl[(2x)^2 - (x^2)^2\bigr] \, dx = \pi \int_0^2 (4x^2 - x^4) \, dx = \pi \bigl[\frac{4x^3}{3} - \frac{x^5}{5}\bigr]_0^2 = \frac{64\pi}{15}$ [3].
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