Volumes of revolution and mean valueAQA A-Level Further Maths: Revision notes

When a region under a curve is rotated through $2\pi$ radians about the $x$-axis, it forms a solid of revolution. To find its volume, slice the solid into thin discs of thickness $\delta x$ and radius $y$. Each disc has volume $\approx\pi y^2\,\delta x$. Adding the discs and letting $\delta x\to0$: $V=\pi\int_a^b y^2\,dx.$ Square $y$ first, then integrate; the $\pi$ stays outside.

Check with a cone: rotate $y=2x$ for $0\le x\le3$. $V=\pi\int_0^3 4x^2\,dx=\pi\left[\frac{4x^3}{3}\right]_0^3=36\pi$. A cone of radius 6 and height 3 has volume $\frac13\pi(6)^2(3)=36\pi$.

Rotating about the $y$-axis, slice into discs of thickness $\delta y$ and radius $x$: $V=\pi\int_c^d x^2\,dy.$ You must write $x^2$ in terms of $y$ and use limits for $y$.

Example: the region bounded by $y=x^2$, the $x$-axis and $x=2$ is rotated about the $y$-axis. At height $y$ the region reaches from $x=\sqrt y$ to $x=2$, so each slice is a disc of radius 2 with a hole of radius $\sqrt y$: $V=\pi\int_0^4\left(2^2-y\right)dy=\pi\left[4y-\frac{y^2}{2}\right]_0^4=8\pi.$ Equivalently, a cylinder of volume $16\pi$ minus the solid $\pi\int_0^4y\,dy=8\pi$.

If the region lies between an upper curve $y=\mathrm{f}(x)$ and a lower curve $y=\mathrm{g}(x)$, each slice about the $x$-axis is a disc with a hole (a washer): $V=\pi\int_a^b\left[\mathrm{f}(x)^2-\mathrm{g}(x)^2\right]dx.$ Find $a$ and $b$ by solving $\mathrm{f}(x)=\mathrm{g}(x)$.

Example: $y=\sqrt x$ and $y=x^2$ meet where $x=0$ and $x=1$. About the $x$-axis, $V=\pi\int_0^1(x-x^4)\,dx=\pi\left(\frac12-\frac15\right)=\frac{3\pi}{10}$.

The mean value of $\mathrm{f}$ over $a\le x\le b$ is $\bar{y}=\frac{1}{b-a}\int_a^b\mathrm{f}(x)\,dx.$ It is the height of the rectangle on the same interval that has the same area as the region under the curve. Equivalently, $\int_a^b\mathrm{f}(x)\,dx=(b-a)\bar y$.

Example: $\mathrm{f}(x)=3x^2-2x$ on $[0,2]$. $\int_0^2(3x^2-2x)\,dx=\left[x^3-x^2\right]_0^2=4$, so the mean value is $\frac42=2$.

Questions often ask for the value of $x$ at which $\mathrm{f}(x)$ equals the mean value. Solve $\mathrm{f}(x)=\bar y$ and keep only solutions in the interval.

Example: $3x^2-2x=2$ gives $x=\frac{1\pm\sqrt7}{3}$; only $\frac{1+\sqrt7}{3}\approx1.22$ lies in $[0,2]$.

The mean value is not the average of the end values and not the value at the midpoint, unless the function is linear. For a model of a physical quantity, give units: the mean value has the units of $\mathrm{f}$.

• State the formula you use: $V=\pi\int y^2\,dx$ or $V=\pi\int x^2\,dy$.
• Check which axis you rotate about and use the matching variable and limits.
• Leave $\pi$ and logarithms exact unless a decimal is asked for, and give volume units such as cm$^3$.
• Where the integrand is $\frac{1}{ax+b}$, the integral gives $\frac1a\ln(ax+b)$; for example $\pi\int_0^4\frac{1}{2x+1}\,dx=\frac{\pi}{2}\ln9=\pi\ln3$.
• Use a known solid (cone, cylinder, sphere) to sanity-check an answer.