Joint and Marginal p.d.f. for a point randomly chosen from a circle

Question

My textbook has a question that I'm unsure of my answer on, specifically part (a) and (d):

Suppose that a point $(X,Y)$ is chosen at random from the circle $S$ defined as follows: $$S=\{(x,y):x^2+y^2\le 1\}$$

a) Determine the joint p.d.f. of $X$ and $Y$

My Approach: My guess is that we're basically looking for the probability that we're in $S$? Is that correct? So I think it should be:

$ f(x,y)= \begin{cases} 1&,\ (x,y)\in S\\ 0&,\text{otherwise}\\ \end{cases} $

b) Determine the marginal p.d.f. of $X$

For this, I'll integrate our joint p.d.f. over $y$: $$f_1(x)=\int_{-\infty}^{\infty}f(x,y)dy=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}1dy=2\sqrt{1-x^2}$$

c) Determine the marginal p.d.f. of $Y$

Same as above, $$f_2(y)=\int_{-\infty}^{\infty}f(x,y)dx=\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}1dx=2\sqrt{1-y^2}$$

d) Are $X$ and $Y$ independent?

Here, I believe we're seeing whether $f(x,y)=f_1(x)\cdot f_2(y)$ and $2\sqrt{1-x^2} \cdot 2\sqrt{1-y^2}\ne 1$ so $X$ and $Y$ are not independent.

I think that the joint pdf should be equal to $\frac{1}{\pi}$ inside the circle — framago, Mar 12 '24 at 20:47
It is an old problem; for example, you may see https://math.stackexchange.com/questions/927347/uniform-distribution-over-disk, https://math.stackexchange.com/questions/1092669/uniform-distribution-on-unit-disk?rq=1, and https://math.stackexchange.com/questions/1599869/uniform-distribution-over-the-unit-disk?rq=1 — Amir, Mar 13 '24 at 08:22
Because $\pi$ is the area of the circle and we need $\int_S f(x,y),dxdy=1$ — framago, Mar 14 '24 at 19:38
@framago That kind of makes sense. But doesn't $f(x,y)=P{X=x \text{ and } Y=y}$? I don't understand what $\frac{1}{\pi}$ actually means — Cotton Headed Ninnymuggins, Mar 16 '24 at 01:52
@framago and would that mean $f_1(x)=\frac{2}{\pi}\sqrt{1-x^2}$? — Cotton Headed Ninnymuggins, Mar 16 '24 at 02:07

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