# How to find the center of mass of a plane region?

I have a plane region bounded by

$$y = 2 x^2$$

and $$y = 6 - 4x.$$

How to find the center of mass of that plane?

-
See here. – David Mitra Dec 4 '12 at 13:22

Assume that the density per unit area $\sigma$ of the plane region is constant. Its mass is $M=\sigma A$, where $A$ is the area of region $R$ of the plane bounded by the straight line $y=u(x)=6-4x$ and the parabola $y=v(x)=2x^{2}$ (see plot). Equate $u(x)=v(x)$ and conclude that $u(x),v(x)$ cross each other at the points $(x_{1},y_{1})=(-3,18)$ and $(x_{2},y_{2})=(1,2)$.

The area $A$ is given by the integral $$\begin{eqnarray*} A &=&\iint_{R}dA=\int_{x_{1}}^{x_{2}}\int_{y=v(x)}^{u(x)}dx\ dy= \int_{-3}^{1}u(x)-v(x)\ dx \\ &=&\int_{-3}^{1}6-4x-2x^{2}\ dx.\tag{1} \end{eqnarray*}$$

By definition the coordinates $(h,k)$ of the center of mass satisfy

$$\begin{eqnarray*} Mh &=&\sigma Ah=\iint_{R}x\sigma\ dA=\sigma \int_{x_{1}}^{x_{2}}\int_{y=v(x)}^{u(x)}x\ dx\ dy \\ &=&\sigma \int_{-3}^{1}x\left( u(x)-v(x)\right)\ dx=\sigma \int_{-3}^{1}x\left( 6-4x-2x^{2}\right)\ dx, \\ Mk &=&\sigma Ak=\iint_{R}y\sigma\ dA=\sigma \int_{x_{1}}^{x_{2}}\int_{y=v(x)}^{u(x)}y\ dx\ dy \\ &=&\sigma \int_{x_{1}}^{x_{2}}\left. \frac{y^{2}}{2}\right\vert _{x_{1}}^{x_{2}}dx=\sigma \int_{-3}^{1}\frac{u(x)^{2}-v(x)^{2}}{2}\ dx \\ &=&\frac{\sigma }{2}\int_{-3}^{1}\left( 6-4x\right) ^{2}-\left( 2x^{2}\right) ^{2}\ dx. \end{eqnarray*}$$

Solving for $h,k$ we get \begin{eqnarray*} h &=&\frac{1}{A}\int_{-3}^{1}x\left( 6-4x-2x^{2}\right)\ dx,\tag{2} \\ k &=&\frac{1}{2A}\int_{-3}^{1}\left( 6-4x\right) ^{2}-\left( 2x^{2}\right) ^{2}\ dx.\tag{3} \end{eqnarray*}

Finally evaluate the integrals $(1),(2),(3)$.

ADDED: I computed these integrals and found $A=\frac{64}{3},(h,k)=(-1,\frac{34}{5})$, which agrees with WimC's answer.

-

If the boundary of a region is traced by the curve $\gamma:[0,1] \to \mathbb{R}^2$ in counter clockwise direction then the center of mass is given by

$$\frac{2}{3}\frac{\int_0^1\det(\gamma(t), \gamma'(t))\, \gamma(t) dt} {\int_0^1 \det(\gamma(t), \gamma'(t)) dt}$$

where $\det$ is the determinant of a $2 \times 2$ matrix and the integration in the numerator is per coordinate in $\mathbb{R}^2$. In your case the boundary is naturally composed of two parts and we can do the computation for each part separately. First a linear part

$$\gamma_1: [0,\tfrac{1}{2}] \to \mathbb{R}^2,\, \gamma_1(t) = \left(1-8t,\, 2 + 32t\right)$$

and second a parabolic part

$$\gamma_2: [\tfrac{1}{2},1] \to \mathbb{R}^2,\, \gamma_2(t) = \left(8t-7,\, 2(8t-7)^2 \right).$$

The integral in the numerator becomes

$$\int_0^{\frac{1}{2}} 48 \left(1-8t, 2+32t\right)dt + \int_{\frac{1}{2}}^1 16 (8t-7)^2 \left(8t-7, 2(8t-7)^2 \right)dt =$$ $$(-24,240) + (-40, 195 \! \tfrac{1}{5}) = (-64, 435 \! \tfrac{1}{5})$$ and in the denominator

$$\int_0^{\frac{1}{2}} 48 dt + \int_{\frac{1}{2}}^1 16(8t-7)^2 dt = 24 + 18 \! \tfrac{2}{3} = 42 \! \tfrac{2}{3}.$$

So for the center of mass we get

$$\frac{2}{3} \frac{(-64, 435 \! \tfrac{1}{5})}{42 \! \tfrac{2}{3}} = (-1, 6 \! \tfrac{4}{5}).$$

And if I'm not mistaken this is the same result as Américo obtained.

-
+1, I confirm that the evaluation of the integrals in my answer gives the same result as yours. – Américo Tavares Dec 4 '12 at 21:45