The formulae for the centre of mass of a lamina or solid of revolution often appear in textbooks as things to memorise and apply. They are not. Each formula is the natural result of one idea: the centre of mass of a continuous body is a weighted average position, and an integral is the limit of a weighted sum. Seeing where the formula comes from makes it far easier to reconstruct under exam pressure — and to apply correctly to unfamiliar shapes.
The unifying idea: weighted average position
For a system of discrete point masses mi at positions xi, the centre of mass is the average x-position weighted by mass:
For a continuous body, the sum becomes an integral. The formula does not change in concept — only in how the sum is taken.
From discrete sum to integral
A-level Further Maths — Mechanics (optional)Edexcel (Further Mechanics 1)AQA (Mechanics)OCR (Mechanics A)OCR MEI (Mechanics)
Imagine slicing a one-dimensional rod of varying density ρ(x) into n thin pieces, each of width Δx. The piece centred at xi has:
mass ≈ ρ(xi) Δx
position xi
The discrete centre of mass is:
As n → ∞ and Δx → 0, each sum becomes an integral:
This is not a new formula — it is the same weighted average, evaluated exactly rather than approximately. The same argument applies in 2D and 3D by slicing into strips or shells.
Uniform lamina bounded by y = f(x)
A-level Further Maths — Mechanics (optional)Edexcel (Further Mechanics 1)AQA (Mechanics)OCR (Mechanics A)OCR MEI (Mechanics)
For a uniform lamina (constant surface density ρ) bounded above by y = f(x), below by the x-axis, between x = a and x = b, slice it into vertical strips. Each strip at position x has:
width dx, height f(x)
mass dm = ρ f(x) dx
its own centre at x horizontally, f(x)/2 vertically
The weighted sums become:
The denominator ∫f(x) dx is just the area of the lamina — the total “mass” when density is 1. The ½[f(x)]² in the ȳ numerator comes from the centre of each strip being at height f(x)/2, so the moment is (f(x)/2) × f(x) dx = ½[f(x)]² dx.
Example 1 — lamina under y = x² between x = 0 and x = 2
First compute the area (denominator):
Then the x-moment (numerator for x̄):
And the y-moment (numerator for ȳ):
The centre of mass is at (3/2, 6/5). As a sanity check: ȳ = 6/5 = 1.2, which is less than the maximum height of the curve (f(2) = 4), and x̄ = 1.5, which is between 0 and 2 — both plausible.
Try it yourself
Find the centre of mass of the uniform lamina bounded by y = x3, the x-axis, x = 0 and x = 2.
Show answer
Area (denominator):
x̄ numerator:
ȳ numerator:
Centre of mass: (8/5, 16/7).
Region between two curves
When the lamina lies between y = f(x) above and y = g(x) below (with f(x) ≥ g(x)), each vertical strip has height f(x) − g(x) and its centre is at height (f(x) + g(x))/2. The formulae extend naturally:
The ȳ numerator uses [f(x)]² − [g(x)]² because the moment of the strip about the x-axis is the moment of the full strip from 0 to f(x) minus the moment of the removed part from 0 to g(x).
Solid of revolution about the x-axis
A-level Further Maths — Mechanics (optional)Edexcel (Further Mechanics 1)AQA (Mechanics)OCR (Mechanics A)OCR MEI (Mechanics)
When the region under y = f(x) is rotated 360° about the x-axis, the resulting solid can be sliced into thin circular discs. Each disc at position x has:
radius f(x), thickness dx
volume dV = π[f(x)]² dx
centre of mass at x (by symmetry — it is a disc)
The solid is symmetric about the x-axis, so ȳ = z̄ = 0. Only x̄ is non-trivial:
The π cancels from numerator and denominator. The denominator ∫[f(x)]² dx is proportional to the volume of the solid (V = π∫[f(x)]² dx).
Example 2 — hemisphere of radius r
A solid hemisphere of radius r is generated by rotating the semicircle y = √(r² − x²) about the x-axis, for x from 0 to r.
Denominator (proportional to volume):
Numerator (x-moment):
The centre of mass of a solid hemisphere lies at 3r/8 from the flat face — a standard result worth verifying rather than memorising.
Try it yourself
Find x̄ for the solid of revolution formed by rotating y = √x about the x-axis for x from 0 to 4.
Show answer
With f(x) = √x, f(x)² = x. Denominator (proportional to volume):
x-moment (numerator):
Example 3 — solid cone of height h and base radius r
A cone is generated by rotating y = (r/h)x about the x-axis for x from 0 to h.
The centre of mass of a solid cone is at 3h/4 from the apex — or equivalently h/4 from the base. Both are standard results, and again the derivation is straightforward once you see how the disc argument sets up the integrals.
Try it yourself
Find x̄ for the solid of revolution formed by rotating y = 2x about the x-axis for x from 0 to 3.
Show answer
f(x) = 2x, so f(x)² = 4x².
This is a cone of height h = 3, so x̄ = 3h/4 = 9/4 ✓ — consistent with the general formula.
Composite bodies
When a body is made of several standard pieces joined together, return to the discrete weighted average:
Removing a piece is handled by subtracting a negative mass — so a lamina with a circular hole is treated as the full lamina plus a disc of negative mass at the hole's location.
Example 4 — hemisphere glued to a cone
A solid hemisphere of radius r (mass Mh, centre of mass at 3r/8 from its flat face) is glued to a solid cone of base radius r and height h (mass Mc, centre of mass at h/4 from its base). Place the origin at the flat face/base join, with x increasing into the cone.
The hemisphere contributes a negative x position because its centre of mass lies on the hemisphere side of the join. Substitute the masses (Mh = ⅔πr³ρ, Mc = ⅓πr²hρ) and simplify.
Try it yourself
A uniform lamina consists of a 4 cm × 2 cm rectangle joined along its right edge to a 1 cm × 2 cm rectangle. Taking x = 0 at the left edge of the larger rectangle, find x̄ of the composite lamina.
Show answer
Large rectangle: area = 8 cm², centre at x = 2 cm. Small rectangle: area = 2 cm², extends from x = 4 to x = 5, centre at x = 4.5 cm.
