Find the centroid of the boomarang shaped region for the parabolas $y^2=-4(x-1)$ and $y^2=-2(x-2)$ I know the formulas, I only need assistance setting up the initial integral.
So my order of integration must be $\mathrm{d}x$ $\mathrm{d}y$. Then if we solve the parabola for $x$ the new integral we get is:
$\int_{-2}^2 \int_{(-y^2/{2})+2}^{(y^2/{2})+2}\mathrm{d}x$ $\mathrm{d}y$.
I'm pretty confident this is correct but some reassurance would be helpful! I'm graphing the parabolas on my calculator but its still a bit confusing to see.. 
 A: Assume unit density. Let $B$ be the region covered by the boomerang. For the mass, we want to integrate $dx\,dy$ over $B$.  I would automatically take advantage of symmetry, find the mass of the top half and double. The mass is
$$2\int_{y=0}^2 \left(\int_{x=1-y^2/4}^{2-y^2/2} dx\right)\,dy.$$
For the inner curve of the boomerang is $y^2=-4(x-1)$, that is, $x=1-y^2/4$, while the outer curve is $x=2-y^2/2$. The curves meet at $x=0$, $y=\pm 2$. 
By symmetry the $y$-coordinate of the centroid is $0$. For the moment about the $x$-axis, we need 
$$2\int_{y=0}^2 \left(\int_{x=1-y^2/4}^{2-y^2/2} x\,dx\right)\,dy.$$
A: The centroid $(\bar{x},\bar{y})$ of a region bounded by the graphs of continuous functions $f$ and $g$ such that $f(y)\ge g(y)$ on the interval $[a,b]$, $a\le y\le b$, is given by
$$\begin{cases}
\bar{x}=\frac{1}{A}\int_{a}^{b}\left[\frac{f(y)+g(y)}{2}\right]\left[f(y)-g(y)\right]\mathrm{d}y\\
\bar{y}=\frac{1}{A}\int_{a}^{b}y\left[f(y)-g(y)\right]\mathrm{d}y,
\end{cases}$$
where $A$ is the area of the region (given by $A=\int_{a}^{b}\left[f(y)-g(y)\right]\mathrm{d}y$). This is in complete analogy to the formulas for the coordinates of the centroid of a region bounded by graphs of functions of $x$ as opposed to $y$, described here.
For the sake of this problem, $f(y)=2-\frac{y^2}{2}$, $g(y)=1-\frac{y^2}{4}$, $a=-2$, and $b=2$. The area is then:
$$\begin{align}
A
&=\int_{-2}^{2}\left[f(y)-g(y)\right]\mathrm{d}y\\
&=\int_{-2}^{2}\left[\left(2-\frac{y^2}{2}\right)-\left(1-\frac{y^2}{4}\right)\right]\mathrm{d}y\\
&=\int_{-2}^{2}\left(1-\frac{y^2}{4}\right)\mathrm{d}y\\
&=\frac83.
\end{align}$$
By symmetry the $y$-coordinate of the centroid is $\bar{y}=0$. The $x$-coordinate is:
$$\begin{align}
\bar{x}
&=\frac{1}{A}\int_{a}^{b}\left[\frac{f(y)+g(y)}{2}\right]\left[f(y)-g(y)\right]\mathrm{d}y\\
&=\frac38\int_{-2}^{2}\left[\frac{3-\frac{3y^2}{4}}{2}\right]\left(1-\frac{y^2}{4}\right)\mathrm{d}y\\
&=\frac{9}{16}\int_{-2}^{2}\left(1-\frac{y^2}{4}\right)^2\mathrm{d}y\\
&=\frac65.
\end{align}$$
