Maths › Further Mechanics 2 › Centres of mass by integration
Centres of mass by integration
When the shape has no straight edges to split along, cut it into strips instead. Each strip is a rectangle whose centre you know, and the sum becomes an integral.
Builds on Centres of mass of plane figures and frameworks and Areas and the limit of a sum.
IN THIS TOPIC
- Find the centre of mass of a lamina bounded by a curve, by integration.
- Find the centre of mass of a solid of revolution.
- Handle a non-uniform body whose density varies with position.
WHAT YOU PROBABLY THINK
For a lamina under a curve, the y coordinate of the centre of mass is found by integrating y against the area, just as x is.
Strips for a lamina
Cut the region into vertical strips of width δx. Each strip is a rectangle to within a vanishing error of height y, so its area is y δx and its own centre is at height y/2. Summing the moments and letting the width tend to zero gives:
The two are not symmetric, and that is the point the opening claim misses. The x moment uses the whole strip at distance x, so the integrand is xy. The y moment uses the strip at its own halfway height, giving y × y/2, hence the half and the square.
WORKED EXAMPLE
Under a parabola
Find the centre of mass of the uniform lamina bounded by y = x², the x-axis and x = 2.
Area = ∫x² dx from 0 to 2 = 8/3.
∫xy dx = ∫x³ dx = 4, so xG = 4 ÷ 8/3 = 1.5.
∫½y² dx = ½∫x⁴ dx = ½(32/5) = 3.2, so yG = 3.2 ÷ 8/3 = 1.2.
Both lie inside the region, and xG is well right of centre because the area is concentrated there.
Solids of revolution, and varying density
For a solid formed by rotating a curve about the x-axis, cut it into discs of radius y and thickness δx. Each disc has volume πy² δx and its centre on the axis at x, so by symmetry the centre of mass lies on the axis and only xG needs finding:
The π cancels. For a non-uniform body the density varies with position, so each element carries a mass ρ dV rather than dV, and ρ goes inside both integrals. That is the only change: the method is the same weighted average throughout.
YOUR TURN
A solid of revolution
The region under y = x² from x = 0 to x = 2 is rotated about the x-axis. Find the centre of mass of the solid formed.
Show the working
Volume ∝ ∫y² dx = ∫x⁴ dx = 32/5.
Moment ∝ ∫xy² dx = ∫x⁵ dx = 64/6 = 32/3.
xG = (32/3) ÷ (32/5) = 5/3 ≈ 1.67, on the axis by symmetry.
That is further right than the lamina's 1.5, because squaring the radius weights the wide end more heavily still.
THE EXAM BIT
- Write down the element you are using, its mass and its own centre, before setting up any integral.
- Remember the half and the square in the y integral for a lamina; the two coordinates are not symmetric.
- For a solid of revolution, say that symmetry puts the centre on the axis rather than integrating for it.
- Quote standard results from the formulae book where the question allows, and say that you are doing so.
CHECK YOURSELF
A uniform lamina lies under y = x from x = 0 to x = 3. Find xG.
Show a hint
Two integrals, then divide.
Show the answer
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Cut into strips: xG is ∫xy dx over ∫y dx, but yG is ∫½y² dx over ∫y dx, because each strip acts at its own halfway height.
For a solid of revolution use y² in place of y and take the centre on the axis by symmetry; for a non-uniform body put ρ inside both integrals.
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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- Find the centre of mass of a lamina bounded by a curve, by integration.
- Find the centre of mass of a solid of revolution.
- Handle a non-uniform body whose density varies with position.
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