How to Find the Centroid of an Area Using a Double Integral
By Know.it editorial teamPublished Updated 7 min readEnglish
Voice + animated visuals, free — no sign-in needed.
Quick answer
The centroid of an area is its balance point. You find it with two averages: x̄ equals the double integral of x over the region divided by the double integral of 1 over the region, and ȳ is the same with y. The bottom integral is just the area.
In one lineThe centroid of an area is the average position of every point in that region — the single point where the flat shape would balance perfectly.
Here's a strange thought. A flat metal plate can balance on the tip of a single pencil. Somewhere on that plate is one magic point where it sits perfectly still. That point is the centroid.
The lovely part? Finding it is a double integral in disguise. Once you see the pattern, a scary-looking formula turns into something you already understand: an average.
What is the centroid of an area?
The centroid is the average position of all the area in a region. Every tiny piece of the shape counts equally, and the centroid is where all those positions average out.
Think of a football pitch. The centroid isn't necessarily the centre circle. It's the balance point of the whole playing surface — the average of every single point of grass.
For a thin flat plate of uniform material, the centroid is also the centre of mass. So the balance-point picture is literally true, not just a metaphor.
- Centroid = average position of all the area
- For a uniform flat plate, it's also the centre of mass
- It can sit outside the shape for some curved or L-shaped regions
- It always lies on any axis of symmetry the region has
Why does the centroid need two double integrals?
The centroid has two coordinates, x̄ and ȳ. Each one is a separate average, and each average needs its own double integral. That's why the formula comes in a pair.
The bottom integral, the double integral of 1 over the region, is simply the area. It's the total amount of stuff you're averaging over.
The top integral is a weighted sum. It gives more weight to points further from the origin, which is exactly what pulling out an average x-coordinate requires.
The formulas to memorise
x̄ equals the double integral of x dA over the region, divided by the double integral of 1 dA over the region. ȳ is identical, with y in place of x. Examiners love this pair, so write it out until it feels obvious.
How do you set up the double integral for a region?
You set up the integral by describing the region with limits. The inner integral's limits can depend on the outer variable. The outer limits must be plain numbers.
Take the right triangle with vertices at (0,0), (1,0) and (0,1). The slanted edge is the line y = 1 − x. So for a fixed x, y runs from 0 up to 1 − x. Then x sweeps from 0 to 1.
Draw the region first. Always. The picture tells you which variable to put inside and what the limits should be.
- Sketch the region before writing any integral
- Inner limits may contain the outer variable
- Outer limits must be constants
- dA becomes dy dx (or dx dy) in the order you choose
Worked example: centroid of a triangle
Let's do the whole thing for the triangle with vertices (0,0), (1,0) and (0,1). First the area: integrate 1 with y from 0 to 1 − x, then x from 0 to 1. The inner integral gives 1 − x, and integrating that from 0 to 1 gives 1/2.
Now the numerator for x̄. The inner integral of x with respect to y gives x(1 − x). Integrating x − x² from 0 to 1 gives 1/6.
Divide: x̄ equals 1/6 divided by 1/2, which is 1/3. By the same working, ȳ is also 1/3. So the centroid is at (1/3, 1/3).
The beautiful check
For any triangle, the centroid sits at the intersection of the medians — one third of the way along from each base. Our double integral just proved that from scratch. See? Simple!
What is the most common mistake with inner limits?
The classic exam trap is forgetting the order of integration. The inner limits must be written in terms of the outer variable. For our triangle, y goes from 0 to 1 − x, never from 0 to 1.
Here's why it matters. If you integrate y from 0 to 1, you're covering a square, not a triangle. The shape of the region is defined entirely by those inner limits.
A handy memory hook from gaming: think of a two-pass render. The inner loop draws each column of pixels, the outer loop sweeps across all the columns. Get the loops wrong and the whole scene glitches.
- Wrong: y from 0 to 1 (that's a square)
- Right: y from 0 to 1 − x (that's the triangle)
- Check: does your inner limit match the slanted edge?
- Sketch the region and shade it — the limits should match the shading
When can symmetry save you the whole integral?
If a region is symmetric about the y-axis, then x̄ is automatically zero. If it's symmetric about the x-axis, ȳ is zero. You can skip that integral entirely.
Try a semicircle sitting above the x-axis. It's symmetric about the y-axis, so x̄ equals 0, and the centroid lies right on the axis of symmetry. For a semicircle of radius R, ȳ works out to 4R/3π.
This isn't a trick — it's the average doing its job. For every point on one side, a mirror point cancels it out.
- Symmetric about the y-axis → x̄ = 0
- Symmetric about the x-axis → ȳ = 0
- The centroid always lies on an axis of symmetry
- Semicircle of radius R: x̄ = 0, ȳ = 4R/3π
Where do engineers actually use centroids?
Think of a car's suspension. The centroid of the chassis is where all the weight averages out. Engineers compute it with exactly these double integrals so the car doesn't tip over in a sharp corner.
The same maths shows up in ship design, bridge decks and aircraft wings. Anywhere a flat or layered shape has to stay stable, someone is averaging x and y over an area.
And once you're comfortable with this, moments of inertia are the natural next step. Same double integral, but the weights become squared distances from an axis.
Key takeaways
- The centroid is the average position of all the area — its balance point.
- x̄ equals the double integral of x dA divided by the double integral of 1 dA; ȳ is the same with y.
- The bottom integral is just the area of the region.
- Inner limits may depend on the outer variable; outer limits must be constants.
- Symmetry about the y-axis makes x̄ zero instantly, and symmetry about the x-axis makes ȳ zero.
- For the triangle (0,0), (1,0), (0,1), the centroid is (1/3, 1/3) — matching the medians rule.
Key terms
- Centroid
- The average position of all the area in a region — its balance point.
- Double integral
- An integral taken over a two-dimensional region, written with two integral signs and dA.
- dA
- The tiny element of area; in Cartesian coordinates it becomes dy dx or dx dy.
- Inner limits
- The limits on the inner integral; they may depend on the outer variable, like y from 0 to 1 − x.
- Outer limits
- The limits on the outer integral; they must be constants, like x from 0 to 1.
- Axis of symmetry
- A line that splits the region into two mirror halves; the centroid lies on it.
- Centre of mass
- The point where a body's mass balances; for a uniform flat plate it coincides with the centroid.
Practice questions
For the triangle with vertices (0,0), (1,0) and (0,1), what are the correct inner limits for y?
- (a) 0 to 1
- (b) 0 to 1 − x
- (c) 0 to x
- (d) 1 − x to 1
Answer key
(b) 0 to 1 − x — The slanted edge is y = 1 − x, so for a fixed x, y runs from 0 up to that line.
What does the double integral of 1 over a region give?
- (a) The centroid
- (b) The area of the region
- (c) The x-coordinate of the centroid
- (d) The perimeter
Answer key
(b) The area of the region — Adding up 1 over every tiny piece of area just totals the area itself.
A region is symmetric about the y-axis. What can you say about its centroid?
- (a) x̄ = 0
- (b) ȳ = 0
- (c) Both are zero
- (d) Nothing without integrating
Answer key
(a) x̄ = 0 — Mirror points on either side of the y-axis cancel in the average, so x̄ is zero.
For the triangle (0,0), (1,0), (0,1), what is x̄?
- (a) 1/2
- (b) 1/6
- (c) 1/3
- (d) 2/3
Answer key
(c) 1/3 — x̄ equals (1/6) divided by (1/2), which is 1/3 — one third along from the y-axis.
Which statement about limits is correct?
- (a) Inner limits must be constants
- (b) Outer limits may contain the outer variable
- (c) Inner limits may depend on the outer variable
- (d) Both sets of limits must be constants
Answer key
(c) Inner limits may depend on the outer variable — The inner limits describe the shape and can contain the outer variable; the outer limits must be constants.
Frequently asked questions
What is the centroid of an area in simple words?
It's the balance point of a flat shape — the average position of every point in the region. If you cut the shape out of card, it would balance on a pin placed at the centroid.
Why is the denominator of the centroid formula just the area?
The double integral of 1 over the region adds up all the tiny pieces of area, so it equals the total area. In an average, you divide by the total amount you averaged over — here, that's the area.
Do I always have to integrate to find a centroid?
No. If the region is symmetric about the y-axis, x̄ is zero without any work. If it's symmetric about the x-axis, ȳ is zero. Use symmetry first, then integrate only what's left.
What happens if I use y from 0 to 1 for the triangle?
You'd be integrating over a unit square instead of the triangle. The inner limits describe the shape, so y must run from 0 to 1 − x to follow the slanted edge.
What is the centroid of the triangle with vertices (0,0), (1,0) and (0,1)?
It's (1/3, 1/3). The double integral gives x̄ = (1/6) ÷ (1/2) = 1/3, and ȳ comes out the same. This agrees with the rule that a triangle's centroid is where its medians meet.
What comes after centroids in this topic?
Moments of inertia. The setup is the same double integral, but the weights become squared distances from an axis instead of plain x or y.
Quick recap: A double integral finds the centroid by averaging every point of an area: divide the weighted integral of x (or y) by the area itself. Set the inner limits from the region's edge, keep the outer limits constant, and let symmetry zero out a coordinate when you can. For the triangle (0,0), (1,0), (0,1), that gives (1/3, 1/3) — exactly where the medians meet.
Sources and further reading
- Watch the animated lesson: How to Find the Centroid of an Area Using a Double Integral (Know.it)
- Centroid — en.wikipedia.org
- Multiple integral — en.wikipedia.org
- Center of mass — en.wikipedia.org
- Image: Wikimedia Commons
Know.it editorial team
Written by the Know.it editorial team from the script of an animated Know.it lesson, checked against the lesson's own facts and the sources listed below. Know.it turns any question into a short visual lesson with voice and animation.
Reviewed: 11 November 2026 · Editorial standards · AI transparency