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

A small piece of a curve has length $ds=\sqrt{dx^2+dy^2}$. For a curve $y=f(x)$ between $x=a$ and $x=b$, the arc length is $s=\int_a^b\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$ Example: $y=\frac23x^{\frac32}$ has $\frac{dy}{dx}=x^{\frac12}$, so $1+\left(\frac{dy}{dx}\right)^2=1+x$ and the length from $x=0$ to $x=3$ is $\int_0^3\sqrt{1+x}\,dx=\frac23(8-1)=\frac{14}{3}$. Square roots like this only integrate neatly when $1+\left(\frac{dy}{dx}\right)^2$ is a perfect square or the integrand has a convenient substitution; exam curves are chosen so they do.

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 $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=36t^2+36t^4=36t^2(1+t^2)$. For $t\ge0$ the integrand is $6t\sqrt{1+t^2}$, so from $t=0$ to $t=1$ the length is $\left[2(1+t^2)^{\frac32}\right]_0^1=4\sqrt2-2$. Factorise inside the root to pull out a square, and use the sign of $t$ when you take its square root. Equations in polar form will not be set.

Rotating a small arc $ds$ at height $y$ about the $x$-axis sweeps out a thin band of area $2\pi y\,ds$. So $S=2\pi\int y\,ds=2\pi\int_a^by\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\quad\text{or}\quad S=2\pi\int_{t_1}^{t_2}y\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.$ Example: $y=x^3$, $0\le x\le1$. Then $S=2\pi\int_0^1x^3\sqrt{1+9x^4}\,dx$. With $u=1+9x^4$, $du=36x^3\,dx$, the integral is $\frac{2\pi}{54}\left[(1+9x^4)^{\frac32}\right]_0^1=\frac{\pi}{27}(10\sqrt{10}-1)$.

Rotating about the $y$-axis, the distance of a point from the axis is $x$, so $S=2\pi\int x\,ds=2\pi\int_a^bx\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\quad\text{or}\quad S=2\pi\int_{t_1}^{t_2}x\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.$ Example: $x=t^2$, $y=\frac{t^3}{3}$, $0\le t\le\sqrt5$. Then $ds=t\sqrt{4+t^2}\,dt$ and $S=2\pi\int_0^{\sqrt5}t^3\sqrt{4+t^2}\,dt$. With $u=4+t^2$, $t^2=u-4$ and $t\,dt=\frac12du$: $S=\pi\int_4^9(u-4)u^{\frac12}\,du=\frac{506\pi}{15}$. The same curve gives arc length $\int_0^{\sqrt5}t\sqrt{4+t^2}\,dt=\frac{19}{3}$.

1. Decide which axis the curve is rotated about; that sets the factor $2\pi y$ or $2\pi x$.
2. Differentiate, then simplify $1+\left(\frac{dy}{dx}\right)^2$ or $\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2$ to a perfect square times something simple.
3. Substitute and integrate, changing the limits with the variable.
4. Check against geometry: $y=2x$ for $0\le x\le3$ rotated about the $x$-axis gives a cone with lateral area $\pi rl=\pi(6)(3\sqrt5)=18\sqrt5\,\pi$, and the formula $2\pi\int_0^32x\sqrt5\,dx$ agrees.