Maths › Further Pure 2 › Arc length and surface area
Arc length and surface area
Chop a curve into tiny straight pieces and Pythagoras measures each one. Add them with an integral for length, or spin them for the area of a surface of revolution.
Builds on Volumes of revolution and Polar curves.
IN THIS TOPIC
- Apply the arc length formula in cartesian, parametric and polar form.
- Compute the area of a surface of revolution as 2π times the integral of (radius) ds.
- Choose the correct radius when the axis of rotation changes.
- Check answers against circles, cones and spheres.
COMMON MISCONCEPTION
The length of a curve between two points is found by integrating y with respect to x.
Pythagoras on a small scale
Over a tiny step the curve is near enough straight, with horizontal run dx and rise dy, so its length is √(dx² + dy²). Factor out dx and the cartesian formula appears.
Factor out dt instead and the element is √((dx/dt)² + (dy/dt)²) dt, while polar form gives √(r² + (dr/dθ)²) dθ. Integrating y itself measures the area under the curve, never the distance along it. The square root is what turns an area into a length.
WORKED EXAMPLE
A length that comes out exactly
Find the length of y = x3/2 from x = 0 to x = 4.
dy/dx = (3/2)√x, so 1 + (dy/dx)² = 1 + 9x/4.
s = ∫√(1 + 9x/4) dx = (8/27)[(1 + 9x/4)3/2] from 0 to 4.
= (8/27)(103/2 − 1) = 9.073 to 3 decimal places. This curve was chosen so the root simplifies. Most curves do not oblige, and numerical methods sit in the same option paper for exactly that reason.
Spinning the arc
Rotate the arc about the x-axis and each element sweeps a thin band of radius y and width ds, with area 2πy ds. The surface of revolution is therefore 2π∫y ds, where ds carries the same square root as before. Rotate about the y-axis and the radius becomes x, so read the axis before writing anything down.
WORKED EXAMPLE
The surface of a sphere, from scratch
Find the surface area generated by rotating y = √(a² − x²) about the x-axis, from x = −a to a.
dy/dx = −x/y, so 1 + (dy/dx)² = (y² + x²)/y² = a²/y².
So ds = (a/y) dx, and the integrand 2πy ds becomes 2πa dx, with the y cancelling entirely.
Area = 2πa × 2a = 4πa², the familiar formula, derived instead of quoted. That cancellation is also why bands of equal width on a sphere have equal area whatever their latitude.
Sanity checks and the polar case
Every answer here should be checked against something you already know. A circle of radius a in polar form has r = a and dr/dθ = 0, so the integrand is just a and the length over a full turn is 2πa. Rotate a straight line through the origin about the x-axis and the formula must give the curved surface area of a cone, πrl. If your answer fails a check like that, the error is almost always a dropped square root or a radius measured from the wrong axis.
Watch the limits in polar work as well. A curve traced twice over 0 ≤ θ ≤ 2π gives twice the length it should, so look at the sketch and decide the range before integrating.
GUIDED PRACTICE
An arc that hyperbolic functions tidy
Find the length of the catenary y = cosh x from x = 0 to x = 1.
Show the working
dy/dx = sinh x, so 1 + sinh²x = cosh²x by the hyperbolic identity.
The square root is therefore just cosh x, with no surd left.
s = ∫cosh x dx from 0 to 1 = sinh 1 ≈ 1.1752. The catenary is one of the few curves whose arc length integral collapses this cleanly.
ASSESSMENT FOCUS
- Quote the formula for the right coordinate system before differentiating anything.
- Simplify under the square root first, since most exam curves are built so a surd disappears.
- For a surface of revolution, the radius is the distance to the axis of rotation, which is not always y.
- Sense-check against a known solid; circles, cones and spheres are all fair game.
CHECK YOURSELF
A curve is given parametrically by x = 5 cos t, y = 5 sin t. Write down the integrand for its arc length, and hence its total length for 0 ≤ t ≤ 2π.
Show a hint
Differentiate both, square, add, and take the root.
Show the answer
dx/dt = −5 sin t and dy/dt = 5 cos t, so the integrand is √(25 sin²t + 25 cos²t) = 5. The length is 5 × 2π = 10π, the circumference of a circle of radius 5.
Arc length integrates the hypotenuse: √(1 + (dy/dx)²) dx, or the parametric and polar equivalents.
A surface of revolution is 2π∫(radius) ds, with the same ds as the arc length integral.
Check every answer against a circle, a cone or a sphere before moving on.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the arc length and surface area questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Apply the arc length formula in cartesian, parametric and polar form.
- Compute the area of a surface of revolution as 2π times the integral of (radius) ds.
- Choose the correct radius when the axis of rotation changes.
- Check answers against circles, cones and spheres.
Open the full revision checklist to see every objective in the course in one place.