Analytic proof of area probability $X,Y$ are i.i.d. $unif(-1,1)$ random variables.
Prove that
$$P(X^2+Y^2\leq 1)=\frac{π}{4}$$

Geometrically, I understand how that happens. $(X,Y)$ is a random point in square having centre at origin and vertices $(-1,-1),(-1,1),(1,-1),(1,1)$. Probability that a random point in this square lies in the unit circle is the ratio of their areas.

But, is there any analytic proof ?
Let $D = \{(x, y): x^2+y^2 \leq 1\}$. Let $g(x, y)$ be the density of $(X, Y)$. It exists because $X$ and $Y$ are independent, and equals to multple of their densities. (Fubini theorem)
$$P(X^2+Y^2\leq1) = \iint_D g(x, y)dxdy = \frac{1}{4}\iint_D dxdy = \frac{\pi}{4}$$
$$\mathbb P\{X^2+Y^2\leq 1\}=\frac{1}{4}\int_{\{(x,y)\mid x^2+y^2\leq 1\}}\boldsymbol 1_{[-1,1]\times [-1,1]}(u,v)\,\mathrm d u\,\mathrm d v=\frac{1}{4}Area(\{(x,y)\mid x^2+y^2\leq 1\})$$