Finding the centroid

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In mathematics, the centroid is the point at which the region/volume is equally balanced.

1 Finding the center of mass

[edit] Finding the center of mass

[edit] Prerequisites

[edit] Moments

First, picture an unbalanced see-saw. The see-saw will only balance if the mass of the first person times their distance from the center is equal to the mass of the other person times his/her distance from the center. This is a concept from physics stated in Archimedes' law of the lever. Assuming this we can find the center of complicated figures. Let's call the quantity $m 1 x 1$ the moment of the mass $m 1$ with respect to the origin. Clearly if one is to place multiple people on the see-saw the sum of their moments must be equal, so let's try to use this fact to find the center of mass of an object. Let's use the number line, if one object is at k (to the left) and the other is at r with masses $m 1$ and $m 2$ respectively it is easy to see that

m 1 (origin - k) = m 2 (r - origin).

Solving for origin we have

m 1 (origin - k) = m 2 (r - origin) = m 1 origin + m 2 origin = m 1 k + m 2 r

which takes us further to say that

$\frac{m_1 x + m_2 r }{m_1 + m_2} = \mathrm{origin}.$

We see that the above equation is very easy to generalize so we have

$\frac{m_1 x + m_2 r }{m_1 + m_2} = \mathrm{origin} = \frac{m_1 x_1 + m_2 x_2 + \cdots}{m_1 + m_2 + \cdots} = \frac{ \sum_{k=1}^n m_k x_k }{ \sum_{k=1}^n m_k }.$

Using this we can find what is called the moment about the origin.

[edit] Discrete moments in two dimensions

This is easily extendable to two dimensions, just think of the lamina or plane on which the masses sit as a two dimensional see-saw, so we only have to find the moment about the y-axis. To find both the x and y coordinates of the center we will use the perpendicular distance to the x-axis and y-axis respectively. Make sure that you use the signed distances to the x or y axis, i.e. the point (1, −2) should be considered to have a y-coordinate or −2. Commonly the notation for the center of a region is denoted $\left ( M_x, M_y, M_z , \ldots, \right )$

[edit] The center of a planar lamina

Let's say that we have some function f(x) and we are trying to find its center of mass. Assuming that it has uniform density this is not terribly difficult. Assume that f(x) is bounded so we are only considering $x | x \in [a, b]$ . To find the moment in the x we divide the [a,b] into an infinite number of rectangles. The center of the k^th rectangle is $\mathrm{density}\ f(x_k) \triangle$ but we want to find the moment of the k^th rectangle so we simply multiply by $x k$ . The final result gives us the limit of a sum to infinity, this yields an integral, so we have that

$\mathrm{mass}\ M_y = \int_{a}^{b} \mathrm{density}\ x f(x)\, dx$ .

The mass of the points in the x-axis is the same as that of the y-axis only that the distance is not x, but rather $\frac{f(x)}{2}$ integrating this from a to b we have that

$\mathrm{mass}\ M_x = \frac{1}{2} \int_{a}^{b} \mathrm{density} \left( f(x) \right)^2.$

Because the density is constant the mass of the region will simply be the area times the density so because density is on both sides of both equations they will simplify and the final result is that

$\mathrm{center}_x = \frac{M_x}{\int_{a}^{b} f(x)\, dx}$

and

$\mathrm{center}_y = \frac{M_y}{\int_{a}^{b} f(x)\, dx}.$

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[edit] The center of a lamina with variable density

Although finding the center of a planar lamina with variable density seems significantly more difficult it is actually of the same order of difficulty as finding the center of mass of an object of constant density. All we have to do is change the bounds from [a, b] to $[a 1, b 1] x [a 2, b 2]$ , change the density from a constant to some function $d e n s i t y (x, y)$ and as a result the mass is changed from the area to a double integral of the density. We have that making these substutions

$M_y = \int_{a_2}^{b_2} \int_{a_1}^{b_1} y\, \operatorname{density}(x,y) \, dy\, dx$

and

$M_x = \int_{a_2}^{b_2} \int_{a_1}^{b_1} x\, \operatorname{density}(x,y) \, dy\, dx.$

We must however not forget to divide my the total mass of the planar lamina so the final answer is

$\mathrm{center}_x = \frac{M_x}{ \mathrm{total\ mass} }= \frac{\int_{a_2}^{b_2} \int_{a_1}^{b_1} x\; \mathrm{density}(x,y) \, dy\, dx}{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \, \operatorname{density}(x,y) \, dy\, dx}$

and

$\mathrm{center}_y = \frac{M_y}{\mathrm{total\ mass}} = \frac{\int_{a_2}^{b_2} \int_{a_1}^{b_1} y\; \mathrm{density}(x,y) \, dy\, dx}{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \mathrm{density}(x,y) \, dy\, dx}.$

Note the integrals can be done in any order as that is guaranteed by Fubini's Theorem.

[edit] The center of a 3-dimensional object with variable density

Instead of finding the perpendicular distance to a line we find it to a plane, specifically the xy-, yz-, and xz-planes. The derivation is the same as for a Lamina with variable density with only an extra dimension.

$\mathrm{center}_x = \frac{M_x}{\mathrm{total\ mass}} = \frac{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \int_{a_3}^{b_3} x \,\mathrm{density}(x,y,z) dz\, dy\, dx}{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \int_{a_3}^{b_3} \mathrm{density}(x,y,z)\, dz\, dy\, dx}$

and

$\mathrm{center}_y = \frac{M_y}{\mathrm{total\ mass}} = \frac{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \int_{a_3}^{b_3} y \,\mathrm{density}(x,y,z)\, dz\, dy\, dx}{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \int_{a_3}^{b_3} \mathrm{density}(x,y,z)\, dz\, dy\, dx}$

$\mathrm{center}_z = \frac{M_z}{\mathrm{total\ mass}} = \frac{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \int_{a_3}^{b_3} z \,\mathrm{density}(x,y,z)\, dz\, dy\, dx}{\int_{a_2}^{b_2} \int_{a_1}^{b_1} \int_{a_3}^{b_3} \mathrm{density}(x,y,z)\, dz\, dy\, dx}$