Moments and the centre of mass for point masses and for a rod with varying density, the centroid of a plane region and the factor of one half in the vertical moment, the symmetry principle, and the Theorem of Pappus.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Moments and Centers of Mass
Objectives
Five outcomes. A balance point is a weighted average, and the integrals compute exactly that.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 610-625 — the section these objectives are drawn from
Warm-up
Section 6.5 noticed that a pumping problem can be checked by lifting the whole mass from its centre of mass, without defining what that point is.
Discussion prompt
Where does a seesaw with a 3 kg mass at position 0.6 and a 2 kg mass at position 4.4 balance?
Hint: Weight each position by its mass.
Answer:
The balance point is the average of the positions weighted by the masses: multiply each position by its mass, add, and divide by the total mass.
\[ \bar{x} = \frac{3(0.6)+2(4.4)}{3+2} = \frac{10.6}{5} = 2.12 \]
The numerator is called the moment and the denominator is the total mass. That is the whole definition, and this section replaces the sums by integrals when the mass is spread continuously rather than concentrated at points.
Concept
The centre of mass is the moment divided by the total mass — the average position, weighted by how much mass sits at each place. For a continuous object both are integrals.
moment and centre of mass — The moment about a point is the integral of position times mass density. Dividing by the total mass gives the centre of mass, the point at which the object balances.
\[ \bar{x} = \frac{M}{m} = \frac{\int x\rho(x)\,dx}{\int\rho(x)\,dx} \]
The moment appears in Section 6.5's pumping shortcut, where the total weight lifted from the centre of mass gave the same answer as the full integral. That was this section's result used in advance.
Figure (svg): The balance point of point masses: moment divided by total mass
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 716-726
Section
Section 1
Concept
For discrete masses the moment is the sum of each mass times its position, and the centre of mass is that sum divided by the total mass.
moment — The sum or integral of position times mass. It measures the tendency to rotate about the origin, and dividing it by the total mass gives the balance point.
\[ M = \sum m_{i}x_{i}, \qquad \bar{x} = \frac{M}{\sum m_{i}} \]
The moment is not itself the balance point and has different units — mass times length rather than length. Dividing by the total mass is what converts it into a position.
Figure (svg): The balance point of point masses: moment divided by total mass
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 716-724 — mass and moments
Picture it
The balance point, weighted.
Figure (svg): The balance point of point masses: moment divided by total mass
The fulcrum sits nearer the heavier masses. Doubling every mass would leave the balance point unchanged, since both the moment and the total double.
Worked example
Example 6.33. The weighted average computed.
\[ \text{Masses } 3, 1, 2 \text{ kg sit at } x=0.6, 2.2, 4.4. \text{ Find the centre of mass.} \]
Compute the moment
Why: Each mass times its position.
\[ 1.8 + 2.2 + 8.8 = 12.8 \]
Compute the total mass
Why: Add.
\[ 6 \text{kg} \]
Divide
Why: Moment over mass.
\[ \frac{12.8}{6} \]
Evaluate
Why: About.
\[ 2.13 \]
Check it lies among the masses
Why: Between the extremes.
Figure (svg): The balance point of point masses: moment divided by total mass
\[ \bar{x} = \frac{12.8}{6} \approx 2.13 \]
Verify: check the effect of doubling every mass
Why: Doubling all three masses doubles both the moment and the total, so the quotient is unchanged — which is right, since making everything uniformly heavier does not move a balance point. That invariance is a useful check and it shows the centre of mass depends on the RATIOS of the masses rather than their absolute sizes. Note also the answer lies between the extreme positions, as a weighted average always must.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 719-721
Fill the middle
A moment and a total mass.
Fill in the blanks
\bar6 = \frac______}
Why: The moment alone has units of mass times length. Dividing by the total mass converts it into a position, which is what a balance point is.
Worked example
Checkpoint 6.33. What balancing means.
\[ \text{Show the total moment about the centre of mass is zero.} \]
Write the moment about a general point
Why: Positions measured from it.
Expand
Why: Split the sum.
Set it to zero
Why: The balance condition.
\[ M - c \times\text{ total } = 0 \]
Solve for c
Why: Divide.
\[ c = M /\text{ total} \]
Recognise the result
Why: The centre of mass.
Figure (svg): The solution to Worked example the moment about another point shown as a ladder of expressions, one row per legal move
\[ \sum m_{i}(x_{i}-\bar{x}) = 0 \]
Verify: connect the algebra to the physical picture
Why: A seesaw balances when the turning effects on the two sides cancel, and the turning effect of a mass is its weight times its distance from the pivot — which is exactly the moment about that pivot. So the algebraic condition that the moment vanishes IS the physical condition for balance. That equivalence is what justifies calling the quotient a balance point rather than merely an average.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 721-723
Trap
\[ \bar{x} = 12.8 \]
Report the moment without dividing
Why: The student stops one step early.
The masses lie between 0.6 and 4.4, so a balance point of 12.8 is outside the object entirely.
\[ \bar{x} = \frac{12.8}{6} \approx 2.13 \]
Divide the moment by the total mass
Why: The moment has units of mass times length, not length.
Two checks catch this: the units, and the fact that a weighted average must lie between the extreme positions. Either one takes a second.
Sorting
Check the units and the size.
Sort into buckets
Sort each quantity.
The last is the most useful practical check: a weighted average of positions can never fall outside their range, so an answer that does means the division was skipped or reversed.
Two truths and a lie
All three are about moments.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The moment has units of mass times length and can be far larger than any position in the object; the centre of mass is what remains after dividing by the total mass, and it always lies within the object's extent.
Prediction
Commit before reasoning.
Predict first
Why is the centre of mass the moment divided by the total mass rather than the moment itself?
Correct: Because it is a weighted average.
\[ \text{weighted average} = \frac{\sum w_{i}x_{i}}{\sum w_{i}} \]
Why: Any weighted average divides the weighted sum by the sum of the weights, and here the weights are the masses. The division also fixes the units, but that is a symptom rather than the reason — without it the quantity would scale with the object's total mass, so making an object heavier would appear to move its balance point, which is physically false.
Section
Section 2
Concept
For a continuous rod the total mass is the integral of the density and the moment is the integral of position times density. The centre of mass is their quotient.
centre of mass of a rod — The moment integral divided by the mass integral, both taken along the rod's length with the density as the weighting.
\[ \bar{x} = \frac{\int_{a}^{b}x\rho(x)\,dx}{\int_{a}^{b}\rho(x)\,dx} \]
The mass integral is Section 6.5's; the moment integral is the same one with an extra factor of the position. Nothing new is being computed, only weighted.
Figure (svg): A rod with varying density: the moment weights each piece by its position
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 720-728 — centre of mass of a rod
Picture it
A rod heavier at one end.
Figure (svg): A rod with varying density: the moment weights each piece by its position
The centre of mass sits past the geometric midpoint, toward the heavy end. For a uniform rod the two coincide, which is the case worth checking a formula against.
Worked example
Example 6.35. Two integrals and a quotient.
\[ \text{A rod on } [0,4] \text{ has density } 1+0.8x. \text{ Find its centre of mass.} \]
Compute the mass
Why: From Section 6.5.
\[ 10.4 \text{kg} \]
Write the moment integrand
Why: Position times density.
\[ x(1 + 0.8 x) \]
Integrate
Why: Term by term.
\[ 8 + 0.8(\frac{64}{3}) \]
Evaluate
Why: About.
\[ 25.07 \]
Divide
Why: Moment over mass.
\[ \text{about } 2.41 \]
Figure (svg): A rod with varying density: the moment weights each piece by its position
\[ \bar{x} = \frac{25.07}{10.4} \approx 2.41 \]
Verify: check it sits past the midpoint, and by how much
Why: The rod's midpoint is at 2, and the centre of mass is at 2.41 — displaced toward the heavier end, as expected. The displacement is modest because the density only quadruples over the rod's length; a more extreme density would push it further. As a limiting check, a uniform density would give exactly 2, which is what setting the linear term to zero produces.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 723-725
Fill the middle
A rod's moment about the origin.
Fill in the blanks
M = \int_x^______\,\rho(x)\,dx
Why: The moment weights each piece of mass by its position, so the integrand is the position times the density. Without that factor the integral is just the mass.
Worked example
Checkpoint 6.35. The formula tested on a known case.
\[ \text{What does the formula give for a rod of constant density?} \]
Write the mass
Why: Density times length.
\[ \rho(b - a) \]
Write the moment
Why: Density times the integral of x.
\[ \rho(b ^{2} - a ^{2}) / 2 \]
Divide
Why: The density cancels.
\[ \frac{b ^{2} - a ^{2}}{2(b - a)} \]
Factor the difference of squares
Why: Standard.
\[ \frac{b + a}{2} \]
Recognise it
Why: The midpoint.
Figure (svg): The solution to Worked example checking against a uniform rod shown as a ladder of expressions, one row per legal move
\[ \bar{x} = \frac{a+b}{2} \]
Verify: note what the cancellation tells you
Why: The density cancelled, so a uniform rod's balance point does not depend on how heavy it is — only on its geometry. That is why the centre of mass of a uniform object is called its centroid and is a purely geometric quantity. Recovering the midpoint is also the strongest possible check that the formula has been set up correctly, and it is worth doing once for any new formula in this section.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 725-727
Error analysis
A student computes a rod's centre of mass.
Annotate
On: \( \bar{x} = \frac{\int_{0}^{4}(1+0.8x)\,dx}{\int_{0}^{4}(1+0.8x)\,dx} = 1 \)
The moment differs from the mass by exactly one factor of the position, and that factor is what makes the quotient a weighted average rather than the number one.
Ranking
The centre of mass of a rod.
Put in order
Why: Two integrals differing by one factor, then a division. Step e is worth the second it takes: a balance point outside the object means the division was skipped or the moment was computed about the wrong origin.
Sorting
Toward the heavy end.
Sort into buckets
Sort each rod on [0,4] by where its centre of mass lies.
Predicting the direction before computing is a strong check on the arithmetic, and it costs nothing. An answer on the wrong side of the midpoint means an error in the moment.
Prediction
Commit before reasoning.
Predict first
For a rod of constant density the centre of mass is the midpoint whatever the density is. Why?
Correct: Because it factors out of both integrals.
\[ \frac{\rho\int x\,dx}{\rho\int dx} = \frac{\int x\,dx}{\int dx} \]
Why: A constant density comes outside both the moment and the mass integral, so it appears in both the numerator and the denominator and divides away. The balance point is then determined by geometry alone — which is why a uniform object's centre of mass is called its centroid and is treated as a purely geometric quantity in the next idea.
Section
Section 3
Concept
For a region of uniform density, the horizontal moment weights each strip by its position and the vertical moment by half its height — because a strip's own mass sits at its midpoint.
centroid — The centre of mass of a region of uniform density, determined by geometry alone. Its coordinates are the two moments divided by the area.
\[ \bar{x}=\frac{\int xf\,dx}{\int f\,dx}, \qquad \bar{y}=\frac{\int \tfrac12 f^{2}\,dx}{\int f\,dx} \]
The asymmetry between the two formulas is the section's characteristic difficulty. It exists because a vertical strip is spread out vertically but concentrated horizontally.
Figure (svg): The centroid of a plane region: the moment of a strip about each axis
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 726-736 — centroid of a plane region
Picture it
Where a strip's mass acts.
Figure (svg): Why the vertical moment uses half the height
Horizontally the strip is at a single position; vertically its mass is spread from zero up to the curve, so it acts at half that height. The one half is that fact written down.
Worked example
Example 6.37. Both coordinates.
\[ \text{Find the centroid of the region under } y=x^{2} \text{ on } [0,2]. \]
Compute the area
Why: The mass, with unit density.
\[ \frac{8}{3} \]
Compute the horizontal moment
Why: Position times height.
\[ \int\text{ of } x \times x ^{2} = 4 \]
Divide
Why: For the x-coordinate.
\[ \frac{4}{\frac{8}{3}} = 1.5 \]
Compute the vertical moment
Why: Half the height, times the height.
\[ \int\text{ of } x ^{4} / 2 = \frac{16}{5} \]
Divide
Why: For the y-coordinate.
\[ \frac{\frac{16}{5}}{\frac{8}{3}} = 1.2 \]
Figure (svg): The centroid of a plane region: the moment of a strip about each axis
\[ (\bar{x},\bar{y}) = \left(\tfrac32, \tfrac65\right) \]
Verify: check both coordinates against the region's shape
Why: The region runs from 0 to 2 horizontally and the centroid at 1.5 sits toward the right, which is right because the region is much wider there. Vertically it runs from 0 to 4, and the centroid at 1.2 sits low — right again, since most of the region is near the bottom where the parabola is shallow. Both coordinates lie inside the region's bounding box, which is the minimum any centroid must satisfy.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 729-731
Fill the middle
The vertical moment of a strip.
Fill in the blanks
dM_2 = \frac______}\cdot f(x)\,dx
Why: The strip's mass acts at its own midpoint, halfway up. Omitting the two places the mass at the top and doubles the vertical coordinate.
Worked example
Checkpoint 6.37. The strip's own centre.
\[ \text{Why does the vertical moment use half the height?} \]
Consider one vertical strip
Why: It runs from 0 to f(x).
Ask where its mass acts
Why: Its own centre of mass.
Find that midpoint
Why: Halfway up.
\[ \text{at height } f(x) / 2 \]
Write the strip's moment
Why: That height times its mass.
\[ (\frac{f}{2}) (f \,dx) \]
Simplify
Why: Collect.
\[ (\frac{1}{2}) f ^{2} \,dx \]
Figure (svg): Why the vertical moment uses half the height
\[ dM_{x} = \frac{f(x)}{2}\cdot f(x)\,dx \]
Verify: contrast with the horizontal moment
Why: Horizontally the strip is a single position, so no averaging is needed and its moment is simply x times its mass. Vertically it occupies a range, so its mass must be treated as acting at the range's centre. The asymmetry between the two formulas is therefore not arbitrary: it reflects the fact that a vertical strip is thin in one direction and extended in the other. Choosing horizontal strips instead would reverse which formula carries the half.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 731-733
Trap
\[ \bar{y} = \frac{\int_{0}^{2}x^{4}dx}{8/3} = 2.4 \]
Use the full height as the strip's moment arm
Why: The student mirrors the horizontal formula.
That places the strip's mass at its top rather than its middle, doubling the vertical coordinate.
\[ \bar{y} = \frac{\int_{0}^{2}\tfrac12 x^{4}dx}{8/3} = 1.2 \]
The strip's mass acts at half its height
Why: It is spread vertically, so it must be averaged.
A quick check: the region under a parabola is bottom-heavy, so its centroid should sit well below the halfway height of 2. An answer of 2.4 is above even that, which is visibly wrong.
Matching
The two coordinates.
Match the pairs
Why: The two moment integrands differ in structure, not merely in which letter appears. That asymmetry comes from the strip being thin horizontally and extended vertically.
Sorting
Vertical strips.
Sort into buckets
Sort each integrand.
Reading the last two entries alongside the first two shows the formulas are the same statement twice: mass times moment arm. Only the arm differs, and it differs because of the strip's shape.
Prediction
Commit before reasoning.
Predict first
Using horizontal strips instead, which formula would carry the factor of one half?
Correct: The horizontal one.
\[ \text{half goes with the direction the strip spans} \]
Why: A horizontal strip occupies a range of x values and a single y value, so its mass acts at its own horizontal midpoint while its vertical position is unambiguous. The factor of one half attaches to whichever direction the strip is extended in, which shows it is a fact about the strip rather than about the coordinate. Recognising that makes the formulas reconstructible rather than memorised.
Section
Section 4
Concept
If a region is symmetric about a line, its centroid lies on that line. Two axes of symmetry locate the centroid completely, with no integration at all.
the symmetry principle — The centroid of a region symmetric about a line lies on that line, because the moments of the two mirror halves about it cancel exactly.
\[ \text{symmetric about } x=c \;\Longrightarrow\; \bar{x}=c \]
Checking for symmetry before setting up any integral can halve the work or remove it. A circle's centroid is its centre, established without computing anything.
Figure (svg): The symmetry principle: a centroid lies on every axis of symmetry
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 730-738 — the symmetry principle
Picture it
How much symmetry saves.
Figure (svg): The symmetry principle: a centroid lies on every axis of symmetry
One axis fixes one coordinate and the other must be computed; two fix both. The right-hand shape has neither and needs the full pair of integrals.
Worked example
Example 6.39. One coordinate free.
\[ \text{Find the centroid of the region under } y=4-x^{2} \text{ above the axis.} \]
Check for symmetry
Why: The function is even.
Apply the principle
Why: The centroid lies on it.
\[ \text{x-coordinate is } 0 \]
Compute the area
Why: The remaining work.
\[ \frac{32}{3} \]
Compute the vertical moment
Why: Half the height squared.
\[ \frac{256}{15} \]
Divide
Why: For the y-coordinate.
\[ 1.6 \]
Figure (svg): The symmetry principle: a centroid lies on every axis of symmetry
\[ (\bar{x},\bar{y}) = \left(0, \tfrac85\right) \]
Verify: confirm the symmetry argument and the remaining coordinate
Why: Integrating for the x-coordinate would give an odd integrand over a symmetric interval, which Section 5.4's shortcut sends to zero — so the symmetry principle and the integral agree, as they must. The y-coordinate of 1.6 sits below the halfway height of 2, which is right for a region wider at the bottom. Half the work was avoided by one observation.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 734-736
Sorting
Count the axes of symmetry the region has.
Sort into buckets
Sort each region.
The third is the trap: the parabola is symmetric but the region on that interval is not. Symmetry of the region is what matters, and restricting an interval usually destroys it.
Worked example
Checkpoint 6.39. The moments cancel.
\[ \text{Prove the centroid of a symmetric region lies on its axis of symmetry.} \]
Place the axis at the origin
Why: Without loss of generality.
Note the height function is even
Why: By the symmetry.
\[ f(-x) = f(x) \]
Write the moment integrand
Why: Position times height.
\[ x f(x) \]
Identify its parity
Why: Odd times even.
Apply Section 5.4's shortcut
Why: Over a symmetric interval.
Figure (svg): The solution to Worked example why symmetry works shown as a ladder of expressions, one row per legal move
\[ \int_{-a}^{a}xf(x)\,dx = 0 \;\Longrightarrow\; \bar{x}=0 \]
Verify: see which earlier result did the work
Why: The proof is entirely Section 5.4's even-and-odd shortcut: the position is odd, the height is even by symmetry, and their product is odd — so the integral over a symmetric interval vanishes. The symmetry principle is therefore not a separate fact to remember but a consequence of a result already proved, which is a good reason to trust it and an easy way to reconstruct it.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 736-738
Error analysis
A student skips an integral by appealing to symmetry.
Annotate
On: \( y=x^{2} \text{ on } [0,2]: \; \bar{x}=1 \text{ by symmetry} \)
The symmetry must belong to the region, not merely to the curve bounding it. Restricting an interval usually destroys a symmetry the whole curve had.
Fill the middle
A region symmetric about the y-axis.
Fill in the blanks
\text0 x=0 \;\Longrightarrow\; \bar___ = ___
Why: The centroid lies on every axis of symmetry, so one coordinate is fixed without integration. The moment integrand would be odd over a symmetric interval, which Section 5.4 sends to zero.
Two truths and a lie
All three are about symmetry.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The parabola is symmetric about the y-axis, but the region under it from 0 to 2 is not symmetric about anything — it is far wider at the right. The symmetry must belong to the region being measured.
Prediction
Commit before reasoning.
Predict first
Why must a symmetric region's centroid lie on its axis of symmetry?
Correct: Because the mirror halves' moments cancel.
\[ \text{odd integrand over a symmetric interval} \Rightarrow 0 \]
Why: Every piece of mass on one side has a mirror partner at the same distance on the other, so their moments about the axis are equal in size and opposite in sign. The total moment about the axis is therefore zero, and a zero moment about a line means the balance point lies on it. Algebraically that is exactly the statement that an odd integrand over a symmetric interval vanishes.
Section
Section 5
Concept
Revolving a plane region about an axis it does not cross produces a solid whose volume is the region's area times the distance its centroid travels.
the Theorem of Pappus — The volume of a solid of revolution equals the area of the revolved region multiplied by the circumference of the circle traced by its centroid.
\[ V = 2\pi d A \]
It turns Section 6.2's integrals into a multiplication whenever the area and centroid are known — and it can be run backwards to find a centroid from a known volume.
Figure (svg): The Theorem of Pappus: volume as area times the distance the centroid travels
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 736-742 — the Theorem of Pappus
Picture it
A region swept once around.
Figure (svg): The Theorem of Pappus: volume as area times the distance the centroid travels
Neither the area nor the centroid's distance requires the axis to be near the region. The one condition is that the axis must not pass through the region, or parts would sweep over each other.
Worked example
Example 6.40. No integral at all.
\[ \text{Find the volume of a torus made by revolving a circle of radius } 1 \text{ centred } 3 \text{ from the axis.} \]
Find the region's area
Why: A circle.
Locate its centroid
Why: By symmetry.
Find the distance travelled
Why: One full circle.
\[ 2 \pi \times 3 \]
Multiply
Why: Area times distance.
\[ \pi \times 6 \pi \]
State
Why: The volume.
\[ 6 \pi\text{ squared} \]
Figure (svg): The Theorem of Pappus: volume as area times the distance the centroid travels
\[ V = 2\pi(3)(\pi) = 6\pi^{2} \]
Verify: compare with what the washer method would need
Why: By washers the torus requires two square-root boundaries, a split, and a trigonometric substitution — a substantial computation for the same answer. Pappus needed the circle's area and its centre, both known without work. As a rough check, a torus is roughly a cylinder of radius 1 bent into a circle of circumference 6 pi, giving pi times 6 pi, which is exactly the answer — and here the approximation happens to be exact.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 739-741
Fill the middle
A centroid at distance d from the axis, revolved once.
Fill in the blanks
\text2\pi = ___\,d
Why: The centroid traces a circle of radius d, so it travels that circle's circumference. Multiplying by the region's area gives the volume.
Worked example
Checkpoint 6.40. A centroid from a volume.
\[ \text{Use a hemisphere's volume to find the centroid of a semicircular region.} \]
Revolve a semicircle about its diameter
Why: Gives a sphere.
Revolve a quarter-disk instead
Why: About one straight edge.
Write Pappus
Why: Volume equals area times distance.
\[ 2 \pi d(\pi r ^{2} / 4) \]
Set it equal to the hemisphere
Why: Two thirds pi r cubed.
Solve
Why: The centroid's distance.
\[ 4 r / (3 \pi) \]
Figure (svg): The solution to Worked example Pappus run backwards shown as a ladder of expressions, one row per legal move
\[ d = \frac{4r}{3\pi} \approx 0.424r \]
Verify: sanity-check the position
Why: The quarter-disk extends from the axis out to r, and its centroid at about 0.42r sits somewhat inside the halfway point — which is right, because a quarter-disk has more of its area near the corner than near the arc. Running the theorem backwards like this is a standard way of finding centroids of shapes whose solids of revolution are already known, and it avoids two integrals.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 741-742
Trap
\[ V = 2\pi d A \text{ with the axis through the region} \]
Apply the theorem regardless of the axis's position
Why: The student ignores the hypothesis.
Parts of the region on opposite sides sweep through the same space, so the swept volume is not the product.
\[ \text{the axis must not cross the region} \]
Check the hypothesis before applying the theorem
Why: The region must lie entirely on one side.
The clearest case: a circle centred ON the axis has a centroid at distance zero, so the theorem would predict zero volume — while revolving it plainly gives a sphere.
Sorting
The axis must not cross the region.
Sort into buckets
Sort each situation.
A region touching the axis along an edge is fine, since nothing crosses. It is only genuine straddling that breaks the theorem, and the circle centred on the axis is the clearest failure — predicted volume zero, actual volume a sphere.
Ranking
A volume by Pappus.
Put in order
Why: Step a is the hypothesis and skipping it can give an answer of zero for a solid of obvious size. Step c is where symmetry usually removes the remaining work, which is what makes the theorem so quick for standard shapes.
Prediction
Commit before reasoning.
Predict first
Why should a volume equal an area times a distance?
Correct: Because summing each piece's journey gives the centroid's.
\[ V = 2\pi\int r\,dA = 2\pi\bar{r}A \]
Why: A small piece at distance r sweeps a ring of volume its area times twice pi r, so the total volume is twice pi times the integral of r over the region — which is exactly the moment about the axis. Dividing and multiplying by the area turns that moment into the area times the centroid's distance. So the theorem is the definition of the centroid rearranged, which is why it is exact rather than approximate.
Comparison
Fill the blanks. The asymmetry comes from the strip's shape.
Comparison matrix
| About the y-axis | About the x-axis | |
|---|---|---|
| Moment arm | the strip's position x | half the strip's height |
| Integrand | x times f(x) | half f(x) squared |
| Gives | the x-coordinate | the y-coordinate |
| Why | the strip is at one position horizontally | the strip is spread out vertically |
The last row explains the whole difference. Using horizontal strips instead would move the factor of one half to the other column, which shows it belongs to the strip rather than to a coordinate.
Pattern
Given a region or object whose balance point is wanted.
Step one is worth the few seconds it takes and can halve the work. Step four's check catches the commonest error, which is reporting a moment as though it were a position.
Stewart, Calculus: Early Transcendentals 8e, §8.3 Applications to Physics and Engineering §8.3, pp. 558-568
Check
Point masses.
Check your understanding
Masses 3, 1 and 2 kg sit at 0.6, 2.2 and 4.4. Where is the centre of mass?
Answer: A
Why: The moment is 12.8 and the total mass 6, so the quotient is about 2.13.
Check
Centroids.
Check your understanding
Why does the moment about the x-axis use half the strip's height?
Answer: A
Why: A vertical strip is spread from 0 to f(x), so its mass is centred halfway up.
Check
Pappus.
Check your understanding
A circle of radius 1 centred 3 from an axis is revolved. What is the volume?
Answer: A
Why: The area is pi and the centroid travels 6 pi, so the product is 6 pi squared.
Real world
An aircraft's loadmaster must keep the centre of gravity within a narrow band as passengers, fuel and cargo are loaded. Each item's position along the fuselage and its mass are known, and fuel burns off during flight.
Discussion prompt
Explain how the centre of gravity is computed, why it moves during flight, and what makes the calculation safety-critical.
Hint: The aircraft is a system of point masses.
Answer:
The centre of gravity is the moment divided by the total mass — exactly the point-mass computation, with each passenger, pallet and fuel tank contributing its mass times its distance from a reference station. The reference point is arbitrary, but it must be the same for every item.
\[ \bar{x} = \frac{\sum m_{i}x_{i}}{\sum m_{i}} \]
It moves during flight because fuel is consumed, and fuel tanks are not at the centre of gravity. Burning fuel from a rear tank shifts the balance forward; from a wing tank near the centre it barely shifts at all. Modern aircraft pump fuel between tanks specifically to control this.
The band is narrow because control depends on it. Too far forward and the elevator cannot raise the nose at rotation speed; too far aft and the aircraft becomes unstable in pitch, with the tendency to diverge rather than return to level. Both failures occur at the extremes of a range that is often only a few percent of the fuselage length.
Note that this is the point-mass formula rather than an integral, because the loads genuinely are discrete. The continuous version is used for the airframe itself, whose mass is distributed — and the two are combined by adding moments, since moments add exactly as masses do.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does the moment about the x-axis carry a factor of one half?
Correct: Because the strip's mass acts at its midpoint.
\[ dM_{x} = \frac{f}{2}\cdot f\,dx, \qquad dM_{y} = x\cdot f\,dx \]
Why: The strip runs from the axis up to the curve, so its mass is spread over that range and its effective height is the range's centre. Horizontally the strip sits at a single position, so no such averaging is needed — which is why the two moment formulas look so different. Using horizontal strips instead would move the half to the other formula, showing it belongs to the strip's shape rather than to a particular coordinate.
Explain it
They computed a centroid's y-coordinate without the factor of one half and got twice the right answer.
Discussion prompt
In four sentences or fewer, show them where it comes from.
Hint: Ask where a strip's mass sits.
Answer:
Ask them to draw one vertical strip and mark where its own mass is concentrated: not at the top and not at the bottom, but halfway up. So the strip's moment about the x-axis is that midpoint height times the strip's mass, and the midpoint height is half the function's value.
Without the half they are treating all the strip's mass as sitting at the curve, which is why the answer came out too high. A quick check: the region under a parabola is bottom-heavy, so its centroid must sit well below the middle of its height range.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For point masses, remember to divide by the total. For a rod, the moment is the mass integral with one extra factor of the position. For regions, the half attaches to whichever direction the strip spans. For Pappus, check the axis does not cross the region. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw three masses on a beam with the fulcrum marked, and write the weighted-average formula beside it with the moment and the total mass labelled separately. Below, write the rod's two integrals side by side, marking that they differ by exactly one factor of the position, and check the formula on a uniform rod in three lines. In the middle of the page, draw a region with one vertical strip, mark where the strip's own mass sits, and write both moment integrands with an arrow from the midpoint to the factor of one half. Beside it, draw three shapes with none, one and two axes of symmetry and note how much each saves. In the lower half, draw a region beside an axis with its centroid marked and the circle it traces, and write Pappus with its hypothesis. At the bottom, work the torus in five lines.
If your strip's mass is drawn at the top of the strip, move it — it belongs at the midpoint, and that single mark is where the factor of one half comes from.
Recap
Five things, and every one of them is a weighted average.
| If you see | Then |
|---|---|
| A weighted balance point | Moment divided by total |
| A moment reported as a position | The division was skipped |
| A uniform density | It cancels: the answer is geometric |
| A vertical strip | Its mass acts at half its height |
| An axis of symmetry | One coordinate is free |
| A symmetric curve on an asymmetric interval | The region is not symmetric |
| A known area and centroid | Pappus gives the volume in one line |
Section 6.7 returns to functions, building the natural logarithm from scratch as an integral — which finally proves the properties Chapter 1 assumed and Chapter 3 used.
OpenStax Calculus Volume 1, §6.6 Moments and Centers of Mass §6.6, pp. 610-625 — everything on these slides traces back here
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