Rewriting the domain of my integral using a function for calculating the center of mass $\Omega:=\{(x,y)\in[-1,1]\times [0,1] | y \leq x^2 \}$
Calculate the median point, with $\rho(x,y)=1$
Also the mass is: $M=2/3$ 
We first rewrite Omega.
I know that there is a nicer way to rewrite Omega in a way that we can easily integrate it, but I also though I could do it like this:
We consider the function $f(x)=x^2$ on the set $B:=\{(x,y)\in\mathbb R^2 | -1\leq x \leq 1, \quad 0\leq y \leq 1\}$
We only calculate the y-component:
$y_s=\frac{1}{M}\int_\Omega y\rho(x,y)d\mu(x,y)=\frac{3}{2}\int_B f(x)yd\mu(x,y)=\frac{3}{2}\int_{-1}^1\int_0^1yx^2 dydx=\frac{3}{2}\int_{-1}^1\frac{1}{2}x^2dx$
$=\frac{3}{2}\frac{1}{2}[\frac{1}{3}x^3]_{-1}^1=\frac{3}{2}\frac{1}{6}(1-(-1))=\frac{3}{2}\frac{2}{6}=1/2$
Question: The correct result would be 1/10. How come, my idea doesn't work?
 A: We want to find the following integral over $\Omega$ :
$$ y_\mathrm{s} = \frac{3}{2}\int \limits_\Omega y \, \mathrm{d} \mu (x,y) \, .$$
We can also integrate over $B$ instead if we find a suitable function $f$ :
$$ y_\mathrm{s} = \frac{3}{2} \int \limits_B y f(x,y) \, \mathrm{d} \mu (x,y) \, .$$
We need to make sure, however, that we only integrate over the part of $B$ on which $y \leq x^2$ holds. We cannot do this by choosing $f(x,y) = x^2$ , since this would lead to a totally different integral which has nothing to do with the centre of mass. 
Instead we can use the Heaviside step function $H$ and let $f(x,y) = H(x^2 - y)$ . Then $f(x,y) = 1$ holds for $(x,y) \in \Omega$ (except possibly for a set of measure zero) and $f$ vanishes on $B \setminus \Omega$ , so the two integrals are indeed equal. This is the correct way to incorporate the condition $y \leq x^2$ into the integration over $B$.
We then end up with the integral
$$ y_\mathrm{s} = \frac{3}{2} \int \limits_{-1}^1 \int \limits_0^1 y H(x^2-y) \, \mathrm{d} y \, \mathrm{d} x = \frac{3}{2} \int \limits_{-1}^1 \int \limits_0^{x^2} y \, \mathrm{d} y \, \mathrm{d} x \, . $$ 
Note that the final result is $y_\mathrm{s} = \frac{3}{10}$ and not $y_\mathrm{s} = \frac{1}{10}$ though.
