Find the marginal densities of $X$ and $Y$ knowing $h(x,y)$ I have the distribution with joint density $h$ (with respect to the Lebesgue measure):
$$h(x,y)=\frac{3}{2}y 1_{A}(x,y), \ \  A=\{(x,y) \in R^2|0<y, x^2+y^2<1\}$$
And then I have to find the marginal densities for X and Y. While we have that $x^2+y^2<1 \Leftrightarrow y^2<1-x^2 \Leftrightarrow y<\sqrt{1-x^2}$ so $0<y<\sqrt{1-x^2}$ therefore I think that:
$$F_X(x)=\int_{0}^{\sqrt{1-x^2}} \frac{3}{2}ydy=[\frac{3}{4}y^2]_{0}^{\sqrt{1-x^2}}=\frac{3}{4}(\sqrt{1-x^2})^2=\frac{3}{4}(1-x^2)$$
Is that a correct method? Somethings says my out from a simulation I have made that this is wrong. Can anyone correct me And how can I find the marginal distribution of $X$? I must have the upper bound $x<\sqrt{1-y^2}$. But how can I find the lower bound?

your joint density is defined over half of the unit disk thus
$$h_X(x)=\int_0^{\sqrt{1-x^2}}\frac{3}{2}y dy=\frac{3}{4}(1-x^2)\times\mathbb{1}_{[-1;1]}(x)$$
and similarly for the other marginal
$$h_Y(y)=\frac{3}{2}y \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}dx=3y\sqrt{1-y^2}\times\mathbb{1}_{[0;1]}(y)$$
