# Why doesn't a simple mean give the position of a centroid in a polygon?

I was having a look at this question on SO.

From what I know, the centroid is the center of mass of an object. so, by definition its position is given by a simple mean of the positions of all the points in the object.

For a polygon, it only has mass at the vertices. So, the centroid should be given by the arithmetic mean of the coordinates of the vertices.

But Wikipedia says centroid is given by

where A is

Why doesn't a simple arithmetic mean work?

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The centroid of a polygon is indeed its center of mass -- but the mass of a polygon is uniformly distributed over its surface, not only at the vertices. You're right that if the mass were split evenly among the vertices only, the centroid would be the arithmetic mean of the coordinates of the vertices.

It just so happens that both definitions are equivalent (mass evenly distributed over the surface vs mass at the vertices only) for simple shapes like triangles and rectangles.

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+1 for mentioning the coincidence. –  Aryabhata Aug 24 '10 at 6:10
thanks @Laurent, somehow I made the wrong assumption that the polygon was only its vertices - I drew it on paper and all I had were vertices :) –  Lazer Aug 24 '10 at 11:52

Laurent is right, a polygon is not just the vertices, but the whole region. The arithmetic mean of the vertices would give the centroid if the (equal) masses were concentrated at the vertices. While that answers your question, perhaps for the future, the below might be useful.

In case of a mass distributed over a region (not just a polygon), Green's Theorem might be helpful in calculating the area and the centroid:

If $\displaystyle D$ is the region of the polygon, then the x-coordinate of the center of mass is given by the area of the polygon times

$\displaystyle \iint_{D} x dxdy$ which by Green's theorem is same as

$\displaystyle \oint_{C} \frac{x^2}{2} dy$, where the line integral is taken over the perimeter of the polygon.

For centroid we choose $\displaystyle M(x,y) = \frac{x^2}{2}$ and $\displaystyle L(x,y) = 0$

The area of the polygon is given by

$\displaystyle \iint_{D} 1 dxdy$ and can be written as a line integral.

For area we choose $\displaystyle M(x,y) = \frac{x}{2}$ and $\displaystyle L(x,y) = \frac{-y}{2}$

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thanks @Moron. btw, how do you format maths symbols? –  Lazer Aug 24 '10 at 11:52
@Lazer: Using the dollar symbol: Like this: $\pi=3.14...$ gives $\pi=3.14...$. –  Aryabhata Aug 24 '10 at 13:12