All subjects · Further Mathematics · WFM03 Further Pure Mathematics 3

Area of a surface of revolution

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

When a curve is rotated about an axis it sweeps out a surface. Its area is found by adding the areas of thin bands, each a circle's circumference times a small piece of arc length.

Practise this in the app: instant marking and a fix for every mistake. Free, no email needed to start.

Practise this lesson

The formulae

Rotating through 2π about the x-axis: S = 2π ∫ y ds.

Cartesian: S = 2π ∫ from a to b of y sqrt(1 + (dy/dx)²) dx
Parametric: S = 2π ∫ from t1 to t2 of y sqrt((dx/dt)² + (dy/dt)²) dt

About the y-axis, replace y by x: S = 2π ∫ x ds. (y must be non-negative over the range; split the range if needed.)

Why it works

A band of width δs at height y is nearly a strip of a cylinder (strictly, a slice of a cone) with circumference 2πy, so its area is about 2πy δs. Adding the bands and letting δs tend to 0 gives the integral. Using δx instead of δs would give the wrong answer, because the band is slanted.

Checking with known shapes

Sphere: y = sqrt(r² - x²) from -r to r gives y ds = r dx, so S = 2πr × 2r = 4πr².
Cone: rotating y = (r/h)x from 0 to h gives πr sqrt(r² + h²), the curved surface area πrl.

Worked example

Show that rotating y = sqrt(r² - x²), -r ≤ x ≤ r, about the x-axis gives a surface area of 4πr².

dy/dx = -x/sqrt(r² - x²), so 1 + (dy/dx)² = r²/(r² - x²)
y sqrt(1 + (dy/dx)²) = sqrt(r² - x²) × r/sqrt(r² - x²) = r
S = 2π ∫ from -r to r of r dx = 2πr(2r)

Answer: 4πr²

Key idea: S = 2π ∫ y ds about the x-axis, 2π ∫ x ds about the y-axis, with ds = sqrt(1 + (dy/dx)²) dx or sqrt((dx/dt)² + (dy/dt)²) dt.

Check yourself

The curve y = cosh x, 0 ≤ x ≤ 1, is rotated through 2π about the x-axis. Find the surface area, to 3 decimal places.
Show answer
8.839. S = 2π ∫ from 0 to 1 of cosh² x dx = π[x + sinh 2x/2] from 0 to 1 = π(1 + sinh 2/2) = 8.839.
The line y = 2x, 0 ≤ x ≤ 3, is rotated through 2π about the x-axis. Find the surface area, to 1 decimal place.
Show answer
126.4. ds = sqrt(5) dx, so S = 2π ∫ from 0 to 3 of 2x sqrt(5) dx = 18sqrt(5)π = 126.4. (Check: πrl = π × 6 × sqrt(45).)
The curve x = t², y = 2t, 0 ≤ t ≤ 1, is rotated through 2π about the x-axis. Find the surface area, to 3 decimal places.
Show answer
15.318. ds/dt = sqrt(4t² + 4) = 2sqrt(t² + 1). S = 2π ∫ from 0 to 1 of 2t × 2sqrt(t² + 1) dt = (8π/3)(2sqrt(2) - 1) = 15.318.
Which integral gives the area of the surface formed by rotating a curve through 2π about the y-axis?
- 2π ∫ y ds
- π ∫ x² dy
- 2π ∫ x ds
- 2π ∫ x dy
Show answer
2π ∫ x ds. Each band has radius x (its distance from the y-axis), so its area is 2πx δs.
The curve y = x³, 0 ≤ x ≤ 1, is rotated through 2π about the x-axis. Find the surface area, to 3 decimal places.
Show answer
3.563. S = 2π ∫ from 0 to 1 of x³ sqrt(1 + 9x⁴) dx = (π/27)(10sqrt(10) - 1) = 3.563.
The curve x = cos³ t, y = sin³ t, 0 ≤ t ≤ π/2, is rotated through 2π about the x-axis. Find the surface area, to 3 decimal places.
Show answer
3.770. ds/dt = 3 sin t cos t, so S = 2π ∫ sin³ t × 3 sin t cos t dt = 6π[sin⁵ t/5] = 6π/5 = 3.770.
Why is S = 2π ∫ y dx wrong for the surface area?
- The bands are slanted, so their width is δs, not δx
- It gives the volume, not the area
- It should be π ∫ y dx
- y must be squared
Show answer
The bands are slanted, so their width is δs, not δx. The width of each band is measured along the curve. Only for a horizontal line (a cylinder) is δs = δx.

Practise with instant feedback

Sources

Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP3

Last updated 8 November 2026. Spotted a mistake? Tell us.