Find the center of mass of a thin plate of constant density Find the center of mass of a thin plate of constant density and covering the region bounded by the parabola $y = 3x^2$ and the line $y = 12$.
The answer is $\overline{x} = 0$ and $\overline{y} = 36/5$.
Does anyone know how to do this problem?
 A: Start with the definition of the center of mass
$$x_c = \frac{\int x \,{\rm d}m}{\int\,{\rm d}m},~~~
y_c = \frac{\int y \,{\rm d}m}{\int\,{\rm d}m}$$
where ${\rm d}m = \rho {\rm d}x {\rm d}y$ is the mass of a tiny area element and the integral is over the whole region of interest. $\rho$ is the density (mass per area) of the plate. Note that this is the integral form of the usual sum formulas $x_c = \frac{\sum m_i x_i}{\sum m_i}$. 
We can use a trick to avoid calculating the $x_c$ integral. Since the region is symmetric around $x=0$ we automatically get $x_c = 0$.
For $y_c$ we must do the integral. To integrate over the region we first go over the $y$ direction where we go from $y=3x^2$ to $y=12$. We then go over the $x$ direction and the integration limits for $x$ are found by solving $y=3x^2 =12 \to x=\pm 2$. Make a sketch. 
This is the setup. Now it remains to just solve the integrals and we find
$$y_c = \frac{\int_{x=-2}^{x=2}\int_{y=3x^2}^{y=12} y {\rm d}y {\rm d}x}{\int_{x=-2}^{x=2}\int_{y=3x^2}^{y=12} {\rm d}y {\rm d}x} = \frac{1}{2}\frac{\int_{x=-2}^{x=2} 12^2-9x^4 {\rm d}x}{\int_{x=-2}^{x=2} 12-3x^2 {\rm d}x} = \frac{1}{2}\frac{[12^2x-9x^5/5]_{x=-2}^{x=2}}{[12x-x^3]_{x=-2}^{x=2}} = \frac{36}{5}$$
A: Let $D$ denote the domain of the plate, bounded by the parabola and the segment, and let $A$ be its area (compute it).
Then we will have
$$\bar x=\frac1A\cdot\int_Dx\ dxdy \quad \text{ and }\quad \bar y=\frac1A\cdot\int_Dy\ dxdy\,$$
as the 'average' of all $x$, resp. $y$-values over $D$.
