What is the center of mass of the region bounded by $y=x^2$ and $y=4$? 
English: That the centroid of the area bounded by $y=x^2$ and $y=4$? The answer is $(0,\frac{12}{5}$) I would like to understand the calculations!

$$A=\int_{-2}^{2}|4-x^2|dx=\frac{32}{3}$$

Em Português: Qual o centróide da região limitada por $y=x^2$ e $y=4$? A resposta é $(0,\frac{12}{5}$) Gostaria de entender os cálculos!

 A: The coordinates of the center of mass $(\hat{x},\hat{y})$ is
$$(\hat{x},\hat{y})=\left( \frac{1}{A}\int_{a}^{b}x(f(x)-g(x))\ dx,\:\frac{1}{2A}\int_{a}^{b}(f^2(x)-g^2(x))\ dx \right) $$ 
where $$A=\int_{a}^{b}f(x)-g(x)\ dx$$
Take $f(x)=4$ and $g(x)=x^2$ 
$$\hat{x}=\frac{3}{32}\int_{-2}^{2}x(4-x^2)\ dx=0$$
$$\hat{y}=\frac{3}{64}\int_{-2}^{2}(16-x^4)\ dx=\frac{12}{5}$$
so 
$$(\hat{x},\hat{y})=\left( 0,\frac{12}{5}\right) $$ 
A: Center of mass is the "average point".  For example, if I just gave you 2 points, $p_1$ and $p_2$, and asked for the center of mass, the answer would be $\frac {p_1 + p_2} 2$.
I'm assuming that your object has uniform density; in other words, it's not made of steel on one side and cotton on the other.
Center of mass then becomes the "average point" of your object: add up the points, then divide by the area.
$\begin {align}\frac {\int_{\omega} {\begin{bmatrix} x \\ f(x) \end{bmatrix}}}
       {\int_{\omega} {1}} \end{align}$
where $\omega$ is the area you are integrating over (note, this is assuming a finite surface and uniform density).
The numerator is the sum of all the points in your area:
$\begin {align}\int_{x=-2}^{x=2} \int^{y=4}_{y=x^2} {{\begin{bmatrix} x \\ y \end{bmatrix}\ dy}\ dx} \end{align}$
$x$ is a constant when you are integrating across $y$.
$\begin {align}\int_{x=-2}^{x=2} {\left. \begin{bmatrix} xy \\ \frac 1  2 y^2 \end{bmatrix}\right |_{y=x^2}^{y=4}\ dx} \end{align}$
$\begin {align}\int_{x=-2}^{x=2} { \begin{bmatrix} 4x - x^3 \\ 8 - \frac 1 2  x^4 \end{bmatrix}\ dx} \end{align}$
$\begin {align} \left. \begin{bmatrix} 2x^2 - \frac 1 4 x^4 \\ 8x - \frac 1 {10}  x^5 \end{bmatrix} \right |_{x=-2}^{x=2}\end{align}$
$\begin {align} \begin{bmatrix} 0 \\ \frac {128} 5 \end{bmatrix} \end{align}$
The denominator is the area:
$\begin {align}\int_{x=-2}^{x=2} \int^{y=4}_{y=x^2} {{1\ dy}\ dx} \end{align}$
$\begin {align}\int_{x=-2}^{x=2} {4 - x^2 \ dx} \end{align}$
$\frac {32} 3 $
So numerator over denominator is:
$\begin{bmatrix} 0 \\ \frac {128} 5 \end{bmatrix} \cdot \frac 3 {32} = \begin{bmatrix} 0 \\ \frac{12} 5\end{bmatrix}$
A: $$
{\displaystyle{\int_{0}^{4}{\rm d}y\,y\int_{-\sqrt{y\,}}^{\sqrt{y\,}}{\rm d}x}
 \over
 \displaystyle{\int_{0}^{4}{\rm d}y\int_{-\sqrt{y\,}}^{\sqrt{y\,}}{\rm d}x}}
=
{\displaystyle{\int_{0}^{4}{\rm d}y\,y^{3/2}}
 \over
 \displaystyle{\int_{0}^{4}{\rm d}y\,y^{1/2}}}
=
{4^{5/2}/\left(5/2\right) \over 4^{3/2}/\left(3/2\right)}
=
{3 \over 5}\,4 = \color{#ff0000}{\Large{12 \over 5}}
$$
