Arc length and area of a surface of revolutionEdexcel A-Level Further Maths: Revision notes

The length of a curve is found by adding up tiny straight segments $ds=\sqrt{dx^2+dy^2}$. For $y=f(x)$ between $x=a$ and $x=b$: $s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$ Example: $y=\frac23x^{\frac32}$ has $\frac{dy}{dx}=x^{\frac12}$, so $s=\int_0^3\sqrt{1+x}\,dx=\frac23\left[(1+x)^{\frac32}\right]_0^3=\frac{14}{3}$. Curves are usually chosen so that $1+\left(\frac{dy}{dx}\right)^2$ is a perfect square or a simple linear expression.

For $x=x(t)$, $y=y(t)$ between $t=t_1$ and $t=t_2$: $s=\int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.$ Example: $x=3t^2$, $y=2t^3$ gives $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=36t^2(1+t^2)$, so for $0\leq t\leq1$, $s=\int_0^16t\sqrt{1+t^2}\,dt=\left[2(1+t^2)^{\frac32}\right]_0^1=4\sqrt2-2$. For a circle $x=a\cos t$, $y=a\sin t$ the integrand is the constant $a$, so a quarter turn has length $\frac{\pi a}{2}$.

For $r=f(\theta)$ between $\theta=\alpha$ and $\theta=\beta$: $s=\int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta.$ Example: the cardioid $r=2(1+\cos\theta)$: $r^2+\left(\frac{dr}{d\theta}\right)^2=8(1+\cos\theta)=16\cos^2\frac{\theta}{2}$, so for $0\leq\theta\leq\pi$, $s=\int_0^{\pi}4\cos\frac{\theta}{2}\,d\theta=8$. The identity $1+\cos\theta=2\cos^2\frac{\theta}{2}$ is the key to simplifying cardioid problems.

Rotating an arc through $2\pi$ about the $x$-axis: each element $ds$ sweeps a thin band of area $2\pi y\,ds$, so $S=2\pi\int y\,ds=2\pi\int_a^by\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx=2\pi\int_{t_1}^{t_2}y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.$ About the $y$-axis, replace $y$ by $x$: $S=2\pi\int x\,ds$. Example: $y=\sqrt{x}$, $0\leq x\leq2$, about the $x$-axis: $y\sqrt{1+\frac{1}{4x}}=\sqrt{x+\frac14}$, so $S=2\pi\left[\frac23\left(x+\frac14\right)^{\frac32}\right]_0^2=\frac{13\pi}{3}$.

For a polar curve rotated about the initial line, the distance from the axis is $y=r\sin\theta$ and $ds=\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta$, so $S=2\pi\int r\sin\theta\,ds$. For the cardioid $r=2(1+\cos\theta)$, $0\leq\theta\leq\pi$: $y=8\cos^3\frac{\theta}{2}\sin\frac{\theta}{2}$ and $ds=4\cos\frac{\theta}{2}\,d\theta$, so $S=64\pi\int_0^{\pi}\cos^4\tfrac{\theta}{2}\sin\tfrac{\theta}{2}\,d\theta=64\pi\times\frac25=\frac{128\pi}{5}.$ Integrals like $\cos^n u\sin u$ are done by the substitution $u=\cos\frac{\theta}{2}$ or by reverse chain rule.

1. Identify the form: Cartesian, parametric or polar, and choose the matching $ds$.
2. Differentiate carefully and simplify the expression under the root; it almost always becomes a perfect square.
3. Write the integral with correct limits in the variable you are integrating with respect to.
4. For surfaces, identify the axis and the correct radius ($y$ for the $x$-axis, $x$ for the $y$-axis).
5. Give exact values unless the question asks for a decimal. Sense check: arc length is at least the straight-line distance between the end points.