Volumes of revolutionEdexcel International A Level Maths: Revision notes

Rotate the region between a curve $y=f(x)$, the $x$-axis and the lines $x=a$ and $x=b$ through $2\pi$ radians about the $x$-axis. The solid is made of thin discs of radius $y$ and thickness $\delta x$, each of volume $\pi y^2\,\delta x$. Adding them as $\delta x\to0$ gives $V=\pi\int_a^b y^2\,dx.$ The formula needs $y^2$ inside the integral. Only $\pi\int y^2\,dx$ is required for the course; $\pi\int x^2\,dy$ (rotation about the $y$-axis) is not.

Step 1: square $y$ fully, remembering that $(2\sqrt{x})^2=4x$ and $(e^{2x})^2=e^{4x}$. Step 2: integrate. Step 3: substitute limits and give an exact answer unless told to use a calculator. Example: $y=2\sqrt x$ between $x=1$ and $x=4$: $V=\pi\int_1^4 4x\,dx=\pi\left[2x^2\right]_1^4=30\pi$. Example: $y=e^{2x}$ from $x=0$ to $x=1$: $V=\pi\int_0^1e^{4x}\,dx=\frac{\pi}{4}(e^4-1)$. Example: $y=\frac{1}{\sqrt{2x+1}}$ from $0$ to $4$: $y^2=\frac1{2x+1}$, so $V=\pi\left[\frac12\ln(2x+1)\right]_0^4=\frac{\pi}{2}\ln9=\pi\ln3$.

If the region meets the $x$-axis, the limits are the roots of $y=0$. If a volume is given, form an equation in the unknown limit: for $y=2\sqrt x$ from $x=1$ to $x=a$, $\pi(2a^2-2)=48\pi$ gives $a^2=25$, so $a=5$ (taking the positive root as $a>1$). For logarithmic integrals, set $\frac\pi2\ln(2k+1)=2\pi$ to get $\ln(2k+1)=4$, so $k=\frac{e^4-1}{2}$. To split a solid at a given $x$, evaluate the integral over each interval separately and compare or add the volumes.

If $x=f(t)$ and $y=g(t)$, then $dx=\frac{dx}{dt}\,dt$ and $V=\pi\int_{t_1}^{t_2} y^2\frac{dx}{dt}\,dt,$ where $t_1$ and $t_2$ are the parameter values at $x=a$ and $x=b$. Replace every $x$ by its expression in $t$, so the integral is entirely in $t$. Example: $x=t^2$, $y=3t(2-t)$, $0\le t\le2$. $\frac{dx}{dt}=2t$, so $V=\pi\int_0^2 9t^2(2-t)^2\cdot2t\,dt=\pi\int_0^2(72t^3-72t^4+18t^5)\,dt=\frac{96\pi}{5}$. For the part with $x\le1$ use $t=0$ to $t=1$, giving $\frac{33\pi}{5}$.

In a modelling question the units are cubic units, such as cm$^3$. To test a claim, work out each volume and compare using a ratio or difference, then state a conclusion. A vase 'holding more than twice as much' means the ratio of volumes exceeds 2: $\frac{63}{33}=1.91<2$, so the claim is false. The standard 3 s.f. is enough for decimal answers; $\frac{33\pi}{5}=20.7$ cm$^3$.