Arc length and surface area of revolution
Master arc length and surface area of revolution in Year 12 VCE Specialist Mathematics. These applications of integration measure the distance along a curve with \(L=\int\sqrt{1+(dy/dx)^2}\,dx\) and the area of the surface swept out when that curve is rotated about an axis. It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Integral calculus topic of Unit 4.
You will learn to set up the Cartesian and parametric arc-length integrals, build the surface-area integral \(S=\int 2\pi y\sqrt{1+(dy/dx)^2}\,dx\), and spot the perfect-square integrand that makes each one exact by hand — extending integral calculus to length and area on curved shapes.
Theory
Arc length and the surface area of a solid of revolution are applications of integration. In Year 12 Specialist Mathematics you set up \(L=\displaystyle\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\) for a curve \(y=f(x)\), \(L=\displaystyle\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\) for a parametric curve, and \(S=\displaystyle\int_a^b 2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\) for a surface of revolution about the \(x\)-axis — each chosen so the integrand is a perfect square you can integrate by hand.
The arc length of a curve is the distance measured along the curve between two points. Chop the curve into tiny elements: each element \(ds\) is the hypotenuse of a right triangle with legs \(dx\) and \(dy\), so \(ds=\sqrt{(dx)^2+(dy)^2}\). Summing (integrating) these elements gives the total length.
For a curve \(y=f(x)\), factor \(dx\) out of the element to get \(ds=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\). For a curve given parametrically by \(x=x(t)\), \(y=y(t)\), factor \(dt\) out instead to get the speed \(ds=\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\).
A surface of revolution is generated by rotating a curve about an axis. Each element \(ds\) sweeps a thin band; about the \(x\)-axis the band has radius \(y\), so its area is \(dS=2\pi y\,ds\). Rotating about the \(y\)-axis instead uses radius \(x\), giving \(dS=2\pi x\,ds\).
The key skill is spotting the perfect square. These curves are set up so that \(1+\left(\dfrac{dy}{dx}\right)^2\) (or the parametric sum of squares) simplifies to something like \((x^2+1)^2\) or \(\left(\dfrac{x}{2}+\dfrac{1}{2x}\right)^2\). Its square root has no radical left, so the integral becomes elementary — exact by hand, no calculator.
Arc length of a Cartesian curve \(y=f(x)\) from \(x=a\) to \(x=b\):
Arc length of a parametric curve \(x=x(t)\), \(y=y(t)\) for \(t\) from \(a\) to \(b\) (the integrand is the speed):
Surface area of revolution about the \(x\)-axis (band radius \(y\)):
About the \(y\)-axis the band radius is \(x\), not \(y\):
How to find an arc length or a surface area
- Differentiate. Find \(\dfrac{dy}{dx}\) for a Cartesian curve, or \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\) for a parametric curve.
- Form the radicand. Build \(1+\left(\dfrac{dy}{dx}\right)^2\) (Cartesian) or \(\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2\) (parametric).
- Spot the perfect square. Expand and rewrite the radicand as a square \((a+b)^2\); its square root is \(a+b\), with no surd left.
- Integrate over the interval. For an arc length, integrate the square root. For a surface, first multiply by the band factor \(2\pi y\) (about the \(x\)-axis) or \(2\pi x\) (about the \(y\)-axis), then integrate.
Differentiate and square:
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac13\cdot\dfrac32\left(x^2+2\right)^{1/2}\cdot 2x\) |
| \(=\) | \(x\sqrt{x^2+2}\) | |
| \(\left(\dfrac{dy}{dx}\right)^2\) | \(=\) | \(x^2\left(x^2+2\right)=x^4+2x^2\) |
Spot the perfect square:
| \(1+\left(\dfrac{dy}{dx}\right)^2\) | \(=\) | \(x^4+2x^2+1\) |
| \(=\) | \(\left(x^2+1\right)^2\) | |
| \(\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\) | \(=\) | \(x^2+1\) |
Integrate over \([0,3]\):
| \(L\) | \(=\) | \(\int_0^3\left(x^2+1\right)dx\) |
| \(=\) | \(\left[\dfrac{x^3}{3}+x\right]_0^3\) | |
| \(=\) | \(9+3=12\) |
The arc length is \(L=12\).
Differentiate:
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{x^2}{2}-\dfrac{1}{2x^2}\) |
The \(-\tfrac12\) cross term completes a perfect square:
| \(1+\left(\dfrac{dy}{dx}\right)^2\) | \(=\) | \(1+\dfrac{x^4}{4}-\dfrac12+\dfrac{1}{4x^4}\) |
| \(=\) | \(\dfrac{x^4}{4}+\dfrac12+\dfrac{1}{4x^4}\) | |
| \(=\) | \(\left(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\right)^2\) |
Integrate over \([1,2]\):
| \(L\) | \(=\) | \(\int_1^2\left(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\right)dx\) |
| \(=\) | \(\left[\dfrac{x^3}{6}-\dfrac{1}{2x}\right]_1^2\) | |
| \(=\) | \(\left(\dfrac{8}{6}-\dfrac14\right)-\left(\dfrac16-\dfrac12\right)\) | |
| \(=\) | \(\dfrac{17}{12}\) |
The arc length is \(L=\dfrac{17}{12}\).
Differentiate the parametric equations:
| \(\dfrac{dx}{dt}\) | \(=\) | \(1-t^2\) |
| \(\dfrac{dy}{dt}\) | \(=\) | \(2t\) |
Add the squares — a perfect square appears:
| \(\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2\) | \(=\) | \(\left(1-t^2\right)^2+\left(2t\right)^2\) |
| \(=\) | \(1-2t^2+t^4+4t^2\) | |
| \(=\) | \(\left(1+t^2\right)^2\) |
Integrate the speed over \([0,3]\):
| \(L\) | \(=\) | \(\int_0^3\left(1+t^2\right)dt\) |
| \(=\) | \(\left[t+\dfrac{t^3}{3}\right]_0^3\) | |
| \(=\) | \(3+9=12\) |
The arc length is \(L=12\).
Differentiate and build \(1+(dy/dx)^2\):
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{1}{2\sqrt{x}}\) |
| \(1+\left(\dfrac{dy}{dx}\right)^2\) | \(=\) | \(1+\dfrac{1}{4x}=\dfrac{4x+1}{4x}\) |
The two square roots cancel in the surface integrand:
| \(2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\) | \(=\) | \(2\pi\sqrt{x}\cdot\dfrac{\sqrt{4x+1}}{2\sqrt{x}}\) |
| \(=\) | \(\pi\sqrt{4x+1}\) |
Integrate over \([0,2]\):
| \(S\) | \(=\) | \(\int_0^2 \pi\sqrt{4x+1}\,dx\) |
| \(=\) | \(\pi\left[\dfrac{1}{6}\left(4x+1\right)^{3/2}\right]_0^2\) | |
| \(=\) | \(\dfrac{\pi}{6}\left(9^{3/2}-1\right)\) | |
| \(=\) | \(\dfrac{\pi}{6}\left(27-1\right)=\dfrac{13\pi}{3}\) |
The surface area is \(S=\dfrac{13\pi}{3}\).
Common pitfalls
Frequently asked questions
What is the formula for arc length?
For a curve \(y=f(x)\) from \(x=a\) to \(x=b\), the arc length is \(L=\displaystyle\int_a^b\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\). It comes from summing elements \(ds=\sqrt{(dx)^2+(dy)^2}\) along the curve.
How do you find the arc length of a parametric curve?
Use \(L=\displaystyle\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt\). Differentiate \(x(t)\) and \(y(t)\), add the squares of the derivatives (this is the speed squared), take the square root, then integrate over the parameter interval.
What is the surface area of a solid of revolution?
Rotating \(y=f(x)\) about the \(x\)-axis gives \(S=\displaystyle\int_a^b 2\pi y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx\). Each element \(ds\) sweeps a band of radius \(y\) and area \(2\pi y\,ds\). About the \(y\)-axis, replace \(2\pi y\) with \(2\pi x\).
Why do these integrals use a perfect square?
The square root in \(\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\) is usually impossible to integrate by hand. The curves in this topic are chosen so the radicand is a perfect square, e.g. \(\left(x^2+1\right)^2\); its square root has no surd, so the integral becomes elementary and exact.
What is the difference between rotating about the x-axis and the y-axis?
The band radius changes. About the \(x\)-axis the radius of each band is \(y\), so the factor is \(2\pi y\). About the \(y\)-axis the radius is \(x\), so the factor is \(2\pi x\). The slant factor \(\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\) is the same in both.
How do I spot which integrand simplifies nicely?
After forming \(1+\left(\dfrac{dy}{dx}\right)^2\), expand it and check whether the middle term makes a perfect square \(a^2+2ab+b^2\). Curves with a term and its reciprocal (like \(\dfrac{x^2}{2}+\dfrac{1}{2x^2}\)) or of the form \(\left(x^2+c\right)^{3/2}\) are the usual signals.