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 y ds.
- Check answers against circles, cones and spheres.
WHAT YOU PROBABLY THINK
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). Factoring out dx gives the cartesian formula, and factoring out dt or dθ gives the others:
Parametrically the element is √((dx/dt)² + (dy/dt)²) dt, and in polar form √(r² + (dr/dθ)²) dθ. Integrating y itself measures area under the curve, not length along it, which is where the opening claim goes astray: the square root is what turns area into distance.
WORKED EXAMPLE
A length that comes out exactly
Find the length of y = x^(3/2) 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)(10^(3/2) − 1) = 9.073 to 3 decimal places. A curve chosen so the root simplifies; most do not, which is why numerical methods sit in the same option paper.
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. Rotating about the y-axis swaps the radius to x.
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 = 2πa dx: the y cancels entirely.
Area = 2πa × 2a = 4πa², the familiar formula, derived rather than quoted. The cancellation is why bands of equal width on a sphere have equal area, whatever their latitude.
YOUR TURN
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.
THE EXAM BIT
- Quote the formula in the right coordinate system before differentiating anything.
- Simplify under the square root first; most exam curves are built so a surd disappears.
- For a surface of revolution, the radius is the distance to the axis of rotation, 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
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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.
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.
CHECK YOUR PROGRESS
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- 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 y ds.
- Check answers against circles, cones and spheres.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.