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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.

Year FMEDEXCEL 9FM0 FM2 3.1

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.
  • Use symmetry to reduce the number of integrals needed.

COMMON MISCONCEPTION

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 sits at height y/2. Sum the moments, let the width tend to zero, and you have:

mean x=∫xydx∫ydx,mean y=∫12y2dx∫ydx\text{mean x} = \frac{\text{∫} xy \, dx}{\text{∫} y \, dx}, \qquad \text{mean y} = \frac{\text{∫} \frac{1}{2}y^{2} \, dx}{\text{∫} y \, dx}NOT IN THE BOOKLET — LEARN IT

None of the integration formulae are in the booklet. Only the finished results for the standard bodies are, so learn the method itself.

The two are not symmetric, and the asymmetry is worth pausing on. For x the whole strip acts at distance x, so the integrand is xy. For y the strip acts at its own halfway height, giving y × y/2, hence the half and the square.

The lamina under y = x² from 0 to 2, with its centre of mass at (1.5, 1.2)a thin stripGy = x²mean x = ∫xy dx / ∫y dxmean y = ∫½y² dx / ∫y dx
FIG. 1A lamina under a curve, cut into thin strips, with its centre of mass marked.

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 with its centre on the axis at x, so symmetry puts the centre of mass on the axis and only xG needs finding:

mean x=∫xy2dx∫y2dx\text{mean x} = \frac{\text{∫} x y^{2} \, dx}{\text{∫} y^{2} \, dx}NOT IN THE BOOKLET — LEARN IT

The π cancels, top and bottom. For a non-uniform body the density varies with position, so each element carries a mass ρ dV instead of dV and ρ goes inside both integrals. Nothing else changes; the method stays the same weighted average throughout.

The solid formed by rotating y = x²: thin discs of radius y, with the centre of mass on the axis at 5/3G at 5/3y = x²disc volume πy² δx, all on the axis
FIG. 2The solid of revolution cut into thin discs, with its centre of mass on the axis at five thirds.

GUIDED PRACTICE

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.

ASSESSMENT FOCUS

  • 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 instead of 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

Area = ∫x dx = 4.5. ∫xy dx = ∫x² dx = 9. So xG = 9/4.5 = 2, which is two thirds of the way along as a triangle's centroid should be.

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.

Question workbook

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Question workbook

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Answer book

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7 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

Or read them with their worked answers on the centres of mass by integration questions page.

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.
  • Use symmetry to reduce the number of integrals needed.

Open the full revision checklist to see every objective in the course in one place.