Further Maths/Lessons/Volumes of Revolution/Volumes of Revolution: x- and y-axis
Core Pure · Volumes of Revolution · Year 1

Volumes of Revolution: x- and y-axis

Spin a curve full circle about an axis and the region beneath it sweeps out a solid — sliced into infinitely many thin discs, its volume becomes a single integral.

13 min read 3 worked examples

The disc method

When the region under a curve $y=f(x)$, between $x=a$ and $x=b$, is rotated $360°$ about the $x$-axis, it sweeps out a solid of revolution. Slicing this solid into thin vertical discs of thickness $\delta x$ and radius $y$ gives each disc volume $\pi y^2\,\delta x$ (a cylinder's volume, $\pi r^2h$). Summing every disc and letting $\delta x\to0$ turns the sum into an integral.

Theorem 4.1 — volume of revolution about the x-axis

$$V = \pi\int_a^b y^2\,\mathrm dx$$

Diagram showing a curve above the x-axis with a thin vertical strip of width delta x and height y, and the solid of revolution formed by rotating the region about the x-axis, sliced into thin discs of radius y
Fig. 1 — rotating a thin strip of height y and width δx about the x-axis sweeps out a disc of radius y and thickness δx, volume πy²δx; summing and integrating gives the total volume.

Rotating about the $y$-axis instead works identically, with the roles of $x$ and $y$ swapped: thin horizontal strips of width $\delta y$ and length $x$ sweep out discs of radius $x$.

Theorem 4.2 — volume of revolution about the y-axis

$$V = \pi\int_c^d x^2\,\mathrm dy$$

Example 1 · Rotating about the x-axis
The region bounded by $y=\sqrt{x}$, the $x$-axis, and the line $x=4$ is rotated $360°$ about the $x$-axis. Find the volume of the solid formed.

$y^2=x$, so:

$$V = \pi\int_0^4 x\,\mathrm dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi(8) = 8\pi$$
Example 2 · Rotating about the y-axis
The region bounded by $y=x^2$, the $y$-axis, and the line $y=4$ (for $x\geq0$) is rotated $360°$ about the $y$-axis. Find the volume of the solid formed.

Since $y=x^2$, we have $x^2=y$ directly, so:

$$V = \pi\int_0^4 y\,\mathrm dy = \pi\left[\frac{y^2}{2}\right]_0^4 = \pi(8) = 8\pi$$
Example 3 · A more involved curve
The region bounded by $y=3-x^2$ and the $x$-axis is rotated $360°$ about the $x$-axis. Find the exact volume of the solid formed.

The curve meets the axis where $3-x^2=0$, i.e. $x=\pm\sqrt3$. Then $y^2=(3-x^2)^2=9-6x^2+x^4$.

$$V = \pi\int_{-\sqrt3}^{\sqrt3}(9-6x^2+x^4)\,\mathrm dx = \pi\left[9x-2x^3+\frac{x^5}{5}\right]_{-\sqrt3}^{\sqrt3}$$

The antiderivative is an odd function of $x$, so the definite integral is twice its value at $x=\sqrt3$:

$$9\sqrt3-2(3\sqrt3)+\frac{9\sqrt3}{5} = 9\sqrt3-6\sqrt3+1.8\sqrt3 = 4.8\sqrt3$$

So $V = \pi\times2\times4.8\sqrt3 = 9.6\sqrt3\,\pi = \dfrac{48\sqrt3\,\pi}{5}$.

Common mistake

The formula uses $y^2$, not $y$ — always square the function (or the given $y^2$ expression) before integrating, and don't forget the factor of $\pi$ out the front, which is easy to drop when the algebra inside the integral gets busy.

Exam-style questions

Question 1 5 marks

The region bounded by $y=e^x$, the $x$-axis, the $y$-axis and the line $x=\ln2$ is rotated $360°$ about the $x$-axis. Find the exact volume of the solid formed.

$V = \pi\displaystyle\int_0^{\ln2} e^{2x}\,dx = \pi\left[\dfrac{e^{2x}}{2}\right]_0^{\ln2}$ (M1, A1)

$= \dfrac{\pi}{2}\left(e^{2\ln2}-1\right) = \dfrac{\pi}{2}(4-1) = \dfrac{3\pi}{2}$ (M1, A1, A1)

Question 2 4 marks

The region bounded by the curve $y^2=4x$, the $x$-axis, and the line $x=1$ is rotated $360°$ about the $x$-axis. Find the volume of the solid formed.

$V = \pi\displaystyle\int_0^1 4x\,dx = \pi\left[2x^2\right]_0^1 = 2\pi$ (M1, A1, M1, A1)

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