### Centers of mass (centroids)

For many (but certainly not all!) purposes in physics and mechanics,
it is necessary or useful to be able to consider a physical object as
being a mass concentrated at a single point, its *geometric
center*, also called its **centroid**. *The centroid is essentially
the ‘average’ of all the points in the object.* For simplicity, we will
just consider the two-dimensional version of this, looking only at
regions in the plane.

The simplest case is that of a rectangle: it is pretty clear that the centroid is the ‘center’ of the rectangle. That is, if the corners are $(0,0), (u,0), (0,v)$ and $(u,v)$, then the centroid is $$\biggl({u\over 2},{v\over 2}\biggr)$$ The formulas below are obtained by ‘integrating up’ this simple idea:

For the center of mass (centroid) of the plane region described by $f(x)\le y\le g(x)$ and $a\le x\le b$, we have \begin{align*} \hbox{ $x$-coordinate of the centroid }&= \hbox{ average $x$-coordinate}\\ &={\int_a^b x[g(x)-f(x)]\;dx\over \int_a^b [g(x)-f(x)]\;dx}\\ &={ \int_{\textit{ left}}^{\textit{ right}} x[\textit{ upper}-\textit{ lower}]\;dx \over \int_{\textit{ left }}^{\textit{ right }} [\textit{ upper}-\textit{ lower}]\;dx } \\ &={ \int_{\textit{ left }}^{\textit{ right }} x[\textit{ upper}-\textit{ lower}]\;dx \over \hbox{ area of the region } } \end{align*}

And also \begin{align*} \hbox{ $y$-coordinate of the centroid }&= \hbox{ average $y$-coordinate}\\ &={\int_a^b {1\over 2}[g(x)^2-f(x)^2]\;dx \over \int_a^b [g(x)-f(x)]\;dx}\\ &={\int_{\textit{ left }}^{\textit{ right }} {1\over 2}[\textit{ upper}^2-\textit{ lower}^2]\;dx \over \int_{\textit{ left }}^{\textit{ right }} [\textit{ upper}-\textit{ lower}]\;dx}\\ &={\int_{\textit{ left }}^{\textit{ right }} {1\over 2}[\textit{ upper}^2-\textit{ lower}^2]\;dx \over \hbox{ area of the region }} \end{align*}

*Heuristic:* For the $x$-coordinate:
there is an amount
$(g(x)-f(x))dx$ of the region at distance $x$ from the $y$-axis. This is
integrated, and then *averaged* dividing by the *total*, that is, dividing by
the *area* of the entire region.

For the $y$-coordinate: in each vertical band of width $dx$ there is amount $dx\;dy$ of the region at distance $y$ from the $x$-axis. This is integrated up and then averaged by dividing by the total area.

For example, let's find the centroid of the region bounded by $x=0$, $x=1$, $y=x^2$, and $y=0$. \begin{align*} \hbox{ $x$-coordinate of the centroid }&= {\int_0^1 x[x^2-0]\;dx\over \int_0^1 [x^2-0]\;dx}\\ &={[x^4/4]_0^1 \over [x^3/3]_0^1} = {1/4-0 \over 1/3-0}={3 \over 4} \end{align*} And \begin{align*}\hbox{ $y$-coordinate of the centroid }&= {\int_0^1 {1\over 2}[(x^2)^2-0]\;dx \over \int_0^1 [x^2-0]\;dx}\\ &={{1\over 2}[x^5/5]_0^1 \over [x^3/3]_0^1}={{1\over 2}(1/5-0) \over 1/3-0}={3 \over 10} \end{align*}

#### Exercises

- Find the center of mass (centroid) of the region $0\leq x\leq 1$ and $0\leq y\leq x^2$.
- Find the center of mass (centroid) of the region defined by $0\leq x\leq 1, 0\leq y\leq 1$ and $x+y\leq 1$.
- Find the center of mass (centroid) of a homogeneous plate in the shape of an equilateral triangle.

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##### Calculus Refresher

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