Maths › Further Mechanics 2 › Centres of mass by integration
Centres of mass by integration
When a shape has no straight edges to split along, it is divided into strips instead. Each strip is a rectangle with a known centre, and the sum over the strips 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.
- Use symmetry to reduce the number of integrals needed.
- Find the centre of mass of a composite solid from standard results.
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.
The two coordinates are not symmetric. A vertical strip of height y acts at distance x horizontally but at its own halfway height y/2 vertically, so the moment integrands are xy and y2/2, each divided by ∫y dx. Under y = x2 from 0 to 2 that gives xG = 1.5 and yG = 1.2.
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:
Pearson Edexcel 9FM0 formulae booklet: these integration formulae are not included; only the finished results for the standard bodies are, so the method itself has to be reproduced.
The two are not symmetric. 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.
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:
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.
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.
Composite solids from standard results
A solid built from pieces whose centres of mass the booklet lists needs no integration at all. Quote the standard result for each piece, weight it by that piece's volume, since a uniform solid's mass is proportional to its volume, and take moments about a convenient plane. The flat face where two pieces join is usually that plane, because the two centres then lie at known distances on either side of it.
WORKED EXAMPLE
A cone on a hemisphere
A uniform solid is a cone of base radius 2 and height 6 joined at its base to the flat face of a hemisphere of radius 2. Find its centre of mass.
Both centres are in the Pearson Edexcel 9FM0 formulae booklet: h/4 above the base for a solid cone, which is 1.5, and 3r/8 from the flat face for a solid hemisphere, which is 0.75.
The solid is uniform, so the masses are in proportion to the volumes: cone ⅓π(2²)(6) = 8π, hemisphere ⅔π(2³) = 16π/3, a ratio of 3 : 2.
Take moments about the common flat face, measuring towards the vertex, so the hemisphere's centre sits 0.75 on the other side: xG = (3 × 1.5 − 2 × 0.75)/5 = 3/5 = 0.6.
Symmetry puts the centre of mass on the axis, and 0.6 above the flat face puts it inside the cone, which is where the greater volume is.
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.
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.
- Find the centre of mass of a composite solid from standard results.
Open the full revision checklist to see every objective in the course in one place.