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

Over a tiny step the curve is almost straight, so its length is $\delta s\approx\sqrt{\delta x^2+\delta y^2}$. Summing and taking limits gives the arc length of $y=f(x)$ from $x=a$ to $x=b$: $s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$ If the curve is given as $x=g(y)$, swap the roles: $s=\int_c^d\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy$. Example: $y=\frac23x^{3/2}$, $0\le x\le3$. Then $\frac{dy}{dx}=x^{1/2}$, so $s=\int_0^3\sqrt{1+x}\,dx=\left[\frac23(1+x)^{3/2}\right]_0^3=\frac23(8-1)=\frac{14}{3}$.

For $x=x(t)$, $y=y(t)$ with $t$ from $t_1$ to $t_2$: $s=\int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.$ Example: $x=3t^2$, $y=2t^3$, $0\le t\le1$. Then $\frac{dx}{dt}=6t$ and $\frac{dy}{dt}=6t^2$, so $\sqrt{36t^2+36t^4}=6t\sqrt{1+t^2}$ (as $t\ge0$). With $u=1+t^2$: $s=\int_0^1 6t\sqrt{1+t^2}\,dt=\left[2(1+t^2)^{3/2}\right]_0^1=4\sqrt2-2$.

Rotating a small piece of curve of length $\delta s$ at height $y$ about the $x$-axis sweeps out a thin band of radius $y$, so its area is about $2\pi y\,\delta s$. Therefore $S=2\pi\int y\,ds,$ that is $S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$ (Cartesian) or $S=2\pi\int_{t_1}^{t_2} y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$ (parametric). Example: $y=\sqrt x$, $0\le x\le2$. $\frac{dy}{dx}=\frac{1}{2\sqrt x}$, so $y\sqrt{1+\frac{1}{4x}}=\sqrt{x+\frac14}$ and $S=2\pi\left[\frac23\left(x+\frac14\right)^{3/2}\right]_0^2=\frac{13\pi}{3}$.

About the $y$-axis the radius of each band is $x$, so $S=2\pi\int x\,ds.$ For a Cartesian curve $y=f(x)$ this is $2\pi\int_a^b x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$; for a curve written as $x=g(y)$ it is $2\pi\int_c^d x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy$; and in parametric form it is $2\pi\int x\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$. Always ask: how far is each point of the curve from the axis of rotation? That distance goes in front of $ds$.

Exam curves are chosen so that the root disappears. Example: $y=\frac{x^3}{6}+\frac{1}{2x}$ gives $\frac{dy}{dx}=\frac{x^2}{2}-\frac{1}{2x^2}$, and $1+\left(\frac{dy}{dx}\right)^2=\frac{x^4}{4}+\frac12+\frac{1}{4x^4}=\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)^2.$ So $s=\int_1^2\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)dx=\frac{17}{12}$, and $S=2\pi\int_1^2 y\left(\frac{x^2}{2}+\frac{1}{2x^2}\right)dx=\frac{47\pi}{16}$ about the $x$-axis. Check each answer is positive, give exact values (with $\pi$ and surds) unless told otherwise, and include units: cm for length, cm$^2$ for area.