This is the region from which I want to calculate the centroid: $\{(x, y) \in \mathbb{R}^2 : 0 < 2x < y < 3-x^2\}$.
I calculated the area for the region and it is $A = \frac{5}{3}$. Now, for the $x$ coordinate of the centroid I'm trying to calculate:
$$\bar{x} = \frac{S_x}{A} = \frac{\int_0^1\int_{2x}^{3-x^2}ydydx}{\int_0^2\int_0^{\frac{1}{2}y}dxdy + \int_2^3\int_0^\sqrt{3-y}dxdy} = \frac{\frac{1}{2}\int_0^1\frac{y^2}{2}\biggr\rvert_{2x}^{3-x^2}dx}{\frac{5}{3}} = \frac{21/6}{5/3} = \frac{21}{15}$$
But the solutions give $\frac{7}{20}$.
I'm not understanding what am I doing wrong. Is it in my logic by constructing the $\bar{x}$ integrals? Or is it my integral calculations? I'm being around this problem for a while now.
Yes, the answer is $\frac7{20}$. You should have computed$$\frac{\displaystyle\int_0^1\int_{2x}^{3-x^2}\color{red}x\,\mathrm dy\,\mathrm dx}{\displaystyle\int_0^1\int_{2x}^{3-x^2}1\,\mathrm dy\,\mathrm dx}.$$What you got was $\overline y$.