Volumes of revolutionEdexcel A-Level Further Maths: Revision notes

Rotate the region under $y=\mathrm{f}(x)$ about the $x$-axis. A thin slice of width $\delta x$ becomes a disc of radius $y$ and volume $\approx\pi y^2\delta x$. Adding the slices and letting $\delta x\to0$: $V=\pi\int_a^by^2\,\mathrm{d}x\quad\text{(about the }x\text{-axis)},\qquad V=\pi\int_c^dx^2\,\mathrm{d}y\quad\text{(about the }y\text{-axis)}.$ For rotation about the $y$-axis you must write $x^2$ in terms of $y$ and use limits that are $y$-values. Always give exact answers as multiples of $\pi$ unless told to use decimals.

Example: the region under $y=\sqrt x$ from $x=0$ to $x=4$: $V=\pi\int_0^4x\,\mathrm{d}x=\pi\left[\frac{x^2}{2}\right]_0^4=8\pi$. Example: $y=\frac1x$ from $x=1$ to $x=3$: $V=\pi\int_1^3x^{-2}\,\mathrm{d}x=\pi\left[-\frac1x\right]_1^3=\frac{2\pi}{3}$. Example with a trigonometric curve: $y=\sin x$, $0\leq x\leq\pi$. Use $\sin^2x=\frac12(1-\cos2x)$: $V=\frac\pi2\left[x-\frac12\sin2x\right]_0^\pi=\frac{\pi^2}{2}$.

Rearrange to $x$ in terms of $y$ and integrate with respect to $y$. Example: $y=\frac1x$ means $x=\frac1y$, so for $1\leq y\leq5$, $V=\pi\int_1^5y^{-2}\,\mathrm{d}y=\frac{4\pi}{5}$. Example: the region bounded by $y=\sqrt x$, the $x$-axis and $x=4$ rotated about the $y$-axis. For each $y$ from 0 to 2 the solid has an outer radius $4$ and an inner radius $x=y^2$, so $V=\pi\int_0^2\left(16-y^4\right)\mathrm{d}y=\frac{128\pi}{5}$.

If a region lies between an outer curve and an inner curve, subtract the volumes, not the radii: $V=\pi\int_a^b\left(y_{\text{outer}}^2-y_{\text{inner}}^2\right)\mathrm{d}x.$ Example: between $y=2x$ and $y=x^2$: they meet at $x=0$ and $x=2$, so $V=\pi\int_0^2\left(4x^2-x^4\right)\mathrm{d}x=\frac{64\pi}{15}$. About the $y$-axis the outer radius is $\sqrt y$ and the inner radius is $\frac y2$, giving $\pi\int_0^4\left(y-\frac{y^2}{4}\right)\mathrm{d}y=\frac{8\pi}{3}$. Example: between $y=\sin x$ and $y=\frac12$, the limits are where $\sin x=\frac12$: $x=\frac\pi6$ and $\frac{5\pi}{6}$.

The formulae reproduce familiar results.

• Cone (radius $r$, height $h$): rotate $y=\frac rhx$ for $0\leq x\leq h$: $V=\pi\int_0^h\frac{r^2x^2}{h^2}\,\mathrm{d}x=\frac13\pi r^2h$.
• Sphere (radius $r$): rotate $y=\sqrt{r^2-x^2}$ for $-r\leq x\leq r$: $V=\pi\int_{-r}^{r}\left(r^2-x^2\right)\mathrm{d}x=\pi\left[r^2x-\frac{x^3}{3}\right]_{-r}^{r}=\frac43\pi r^3$.

For a curve given by $x=\mathrm{x}(t)$, $y=\mathrm{y}(t)$ rotated about the $x$-axis, change the variable in $V=\pi\int y^2\,\mathrm{d}x$ using $\mathrm{d}x=\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t$ and convert the limits to values of $t$: $V=\pi\int_{t_1}^{t_2}y^2\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t.$ Example: $x=t^2$, $y=2t$ for $0\leq t\leq2$: $\frac{\mathrm{d}x}{\mathrm{d}t}=2t$, so $V=\pi\int_0^2(2t)^2(2t)\,\mathrm{d}t=8\pi\int_0^2t^3\,\mathrm{d}t=32\pi$. For rotation about the $y$-axis use $V=\pi\int x^2\frac{\mathrm{d}y}{\mathrm{d}t}\,\mathrm{d}t$.