# Marginal density of bivariate density that is a circle

I have the following density function: $f(x,y) = \frac{2}{\pi}$ for $x^2 + y^2 \leq 1$ and $y > x$.

I figured out that this represents half of the unit circle (the upper half when cut along the line $y=x$). I would like to find the marginal density of Y. To do this, I understand that I will have to integrate over x, as the marginal density is usually $f_y(y) = \int{f(x)dx}$, which in this case would be $f_y(y) = \int{\frac{2}{\pi}dx}$. I am having trouble understanding how to find the values over which to integrate. Could someone please clarify this for me? I don't think I have a strong grasp on the intuition behind this.

To find the (marginal) density of $Y$, we "integrate out" $x$. Draw the circle, and the line $y=x$. The geometry is everything. Your post shows that you are aware that the density function is $\frac{2}{\pi}$ in the part of the disk that is "above" the line $y=x$, and $0$ elsewhere.
First deal with negative $y$. The smallest negative value of $y$ is $-\frac{1}{\sqrt{2}}$. In that part of the world, $x$ travels from $-\sqrt{1-y^2}$ to $y$. So for $-\frac{1}{\sqrt{2}}\le y\lt 0$ we have $$f_Y(y)=\int_{x=-\sqrt{1-y^2}}^y \frac{2}{\pi} \,dx=\frac{2}{\pi}\left(y+\sqrt{1-y^2} \right).$$
Now look at $y\ge 0$. For $0\le y\lt \frac{1}{\sqrt{2}}$, the variable $x$ travels from $-\sqrt{1-y^2}$ to $y$: we get exactly the same expression as for negative $y$. So in fact we could have done it in one step for all $y$ with $\frac{1}{\sqrt{2}}\le y\lt \frac{1}{\sqrt{2}}$.
The geometry changes in the interval $\frac{1}{\sqrt{2}}\le y\le 1$. There $x$ travels from one side of the circle to the other. that is, from $-\sqrt{1-y^2}$ to $\sqrt{1-y^2}$. Thus in that interval we have $$f_Y(y)=\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{2}{pi}\,dx=\frac{4}{\pi}\sqrt{1-y^2}.$$