Core Pure · Volumes of Revolution · Year 1

Volumes of Revolution Around the x-Axis

Spin the region under a curve a full turn about the x-axis and it sweeps out a solid — stack infinitely many thin discs and integration tells you exactly how much space it fills.

19 min read 3 worked examples

The disc method

Theorem 14.1 — volume of revolution about the x-axis

When the region bounded by $y=f(x)$, the $x$-axis, and the lines $x=a$, $x=b$ is rotated $360°$ about the $x$-axis, the resulting solid has volume $$V=\pi\int_a^b y^2\,dx$$

Fig. 1 — rotating the region under y = f(x) about the x-axis sweeps out a solid built from thin circular discs of radius y and thickness dx. The highlighted strip and disc show one such slice as it sweeps from $a$ to $b$.

ProofDeriving the disc-method formula

Slice the solid into thin discs perpendicular to the $x$-axis, each of thickness $\delta x$.

Standard strategy for volumes of revolution: approximate with simple shapes, then take a limit.

Each disc is (approximately) a cylinder of radius $y=f(x)$ and thickness $\delta x$, with volume $\delta V\approx\pi y^2\,\delta x$.

Volume of a cylinder $=\pi r^2h$, with $r=y$ and $h=\delta x$.

Summing all discs from $x=a$ to $x=b$: $V\approx\displaystyle\sum\pi y^2\,\delta x$.

Total volume is the sum of all thin disc volumes.

Taking the limit as $\delta x\to0$ turns the sum into an integral: $V=\pi\displaystyle\int_a^by^2\,dx$.

Definition of the definite integral as a limit of a Riemann sum.

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Example 1 · Volume under a simple curve

Find the volume generated when the region bounded by $y=\sqrt{x}$, the $x$-axis, and the lines $x=0,x=4$ is rotated $360°$ about the $x$-axis.

$V=\pi\displaystyle\int_0^4(\sqrt x)^2\,dx=\pi\int_0^4x\,dx=\pi\left[\dfrac{x^2}{2}\right]_0^4=\pi\left(\dfrac{16}{2}-0\right)=8\pi$.

Example 2 · Volume under a quadratic curve

The region bounded by $y=x^2+1$, the $x$-axis, and the lines $x=1,x=3$ is rotated $360°$ about the $x$-axis. Find the volume generated.

$V=\pi\displaystyle\int_1^3(x^2+1)^2\,dx=\pi\int_1^3(x^4+2x^2+1)\,dx=\pi\left[\dfrac{x^5}{5}+\dfrac{2x^3}{3}+x\right]_1^3$.

At $x=3$: $\dfrac{243}{5}+\dfrac{54}{3}+3=\dfrac{243}{5}+18+3=\dfrac{243}{5}+21=\dfrac{243+105}{5}=\dfrac{348}{5}$.

At $x=1$: $\dfrac15+\dfrac23+1=\dfrac{3}{15}+\dfrac{10}{15}+\dfrac{15}{15}=\dfrac{28}{15}$.

$V=\pi\left(\dfrac{348}{5}-\dfrac{28}{15}\right)=\pi\left(\dfrac{1044}{15}-\dfrac{28}{15}\right)=\dfrac{1016\pi}{15}$.

Example 3 · A curve requiring the roots as limits

Find the volume generated when the region bounded by $y=4-x^2$ and the $x$-axis (for $y\geq0$) is rotated $360°$ about the $x$-axis.

Curve meets the $x$-axis where $4-x^2=0 \Rightarrow x=\pm2$, so the limits are $x=-2$ to $x=2$.

$V=\pi\displaystyle\int_{-2}^{2}(4-x^2)^2\,dx=\pi\int_{-2}^2(16-8x^2+x^4)\,dx$.

By symmetry (even integrand), $=2\pi\displaystyle\int_0^2(16-8x^2+x^4)\,dx=2\pi\left[16x-\dfrac{8x^3}{3}+\dfrac{x^5}{5}\right]_0^2$.

$=2\pi\left(32-\dfrac{64}{3}+\dfrac{32}{5}\right)=2\pi\left(\dfrac{480}{15}-\dfrac{320}{15}+\dfrac{96}{15}\right)=2\pi\cdot\dfrac{256}{15}=\dfrac{512\pi}{15}$.

Common mistake

Students frequently forget to square $y$ before integrating, or expand $(f(x))^2$ incorrectly. Always write out $y^2$ fully expanded as a polynomial (or simplified trig/exponential expression) before attempting to integrate term by term.

Sanity check — testing the formula on a known shape

Before trusting a volume calculation, it's worth checking $V=\pi\int_a^by^2\,dx$ against a shape whose volume you already know. Rotating the constant line $y=r$ between $x=0$ and $x=h$ about the $x$-axis should sweep out a cylinder of radius $r$ and height $h$: $$V=\pi\int_0^hr^2\,dx=\pi r^2\big[x\big]_0^h=\pi r^2h$$ which matches the standard cylinder formula exactly. Running this check on a simple case is a good habit whenever a formula is new to you.

Exam-style questions

Question 1 5 marks

The region bounded by $y=e^x$, the $x$-axis, and the lines $x=0,x=\ln2$ is rotated $360°$ about the $x$-axis. Find the exact volume generated, in terms of $\pi$.

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

$=\pi\left(\dfrac{e^{2\ln2}}{2}-\dfrac12\right)=\pi\left(\dfrac{4}{2}-\dfrac12\right)=\pi\left(2-\dfrac12\right)=\dfrac{3\pi}{2}$ (M1, A1, A1)

Question 2 7 marks

The region bounded by $y=3x-x^2$ and the $x$-axis is rotated $360°$ about the $x$-axis. Find the exact volume generated.

Limits: $3x-x^2=0 \Rightarrow x(3-x)=0 \Rightarrow x=0,3$ (M1, A1)

$V=\pi\displaystyle\int_0^3(3x-x^2)^2\,dx=\pi\int_0^3(9x^2-6x^3+x^4)\,dx$ (M1, A1)

$=\pi\left[3x^3-\dfrac{3x^4}{2}+\dfrac{x^5}{5}\right]_0^3=\pi\left(81-\dfrac{243}{2}+\dfrac{243}{5}\right)$ (M1, A1)

$=\pi\left(\dfrac{810}{10}-\dfrac{1215}{10}+\dfrac{486}{10}\right)=\pi\cdot\dfrac{81}{10}=\dfrac{81\pi}{10}$ (A1)

Question 3 5 marks

The region bounded by $y=\dfrac{1}{x}$, the $x$-axis, and the lines $x=1,x=3$ is rotated $360°$ about the $x$-axis. Find the exact volume generated.

$V=\pi\displaystyle\int_1^3\dfrac{1}{x^2}\,dx=\pi\left[-\dfrac1x\right]_1^3$ (M1, A1)

$=\pi\left(-\dfrac13-(-1)\right)=\pi\left(\dfrac23\right)=\dfrac{2\pi}{3}$ (M1, A1, A1)

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Linear Transformations of Roots

Volumes of Revolution Around the y-Axis