Is this question correct because if I find the PDF in terms of X and Y wouldn't that be the marginal PDF f XY(x,y)?

Question

Let S = { (,,) ∈ R³ | (² + ² + ²) ≤ 1, ≥ 0 }

for any lebesgue measurable subset A of S, let P(A) = (3÷(2π)) × (volume of A)

let X (,,) = and Y(,,) = , ∀(,,) ∈ S. find the joint pdf of X and Y.

Answer 1

Nothing is wrong with the question.

If $\lambda$ denotes the Lebesgue measure on $\mathbb{R}^{3}$ and $T:=\left\{ \left(u,v\right)\in\mathbb{R}^{2}\mid u^{2}+v^{2}\leq1\right\} $ then we can write:

$$P\left(X\leq x,Y\leq y\right)=\frac{3}{2\pi}\lambda\left(\left\{ \left(u,v,w\right)\in S\mid u\leq x,v\leq y\right\} \right)=$$$$\frac{3}{2\pi}\int_{-\infty}^{x}\int_{-\infty}^{y}\int_{-\infty}^{\infty}1_{S}\left(u,v,w\right)dwdvdu=\int_{-\infty}^{x}\int_{-\infty}^{y}\frac{3}{2\pi}\left(1-u^{2}-v^{2}\right)1_{T}\left(u,v\right)dvdu$$

This tells us that the function prescribed by: $$(x,y)\mapsto\frac{3}{2\pi}\left(1-x^{2}-y^{2}\right)1_{T}\left(x,y\right)$$ serves as PDF of $\left(X,Y\right)$.