Independence from random vector eith joint density function

Question

If (X,Y) is a random vector with the following joint density function

f(x,y) =\begin{cases} \pi^{-1} \; \; x^2 + y^2\leq 1 \\ 0 \; \text{elsewhere} \end{cases}

Are X and Y independent?

My approach:

I know that they are independent if $f_{X,Y}(x,y) = f_X(x)f_Y(y)$

So

1. integrate wrt to x from $-\infty$ to 1
2. integrate wrt to y from $-\infty$ to 1

Then multiply the two results and see if it equals f(x,y).

Is that the right approach? I'm especially unsure about the bounds for the integrals.

Answer 1

The support of $X,Y$ is $x^2 + y^2 \leq 1$. This is

Now, for $X$ we have: \begin{align*} -1 \leq x \leq 1, \quad -\sqrt{1-x^2} \leq y \leq \sqrt{1-x^2} \end{align*} $$ f_{X}(x) = \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} dy $$ for $Y$ we have \begin{align*} -1 \leq y \leq 1, \quad -\sqrt{1-y^2} \leq x \leq \sqrt{1-y^2} \end{align*} $$ f_{Y}(y) = \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \frac{1}{\pi} dx $$