Suppose $x,y$ are sampled from a uniform distribution on the interval $[0,1]$.
Find the probability of $z= (2x-1)^2 + (2y-1)^2 \leq 1$, that is, $P(z\leq1)$.
What I've tried:
$P(z\leq 1) = F_z(1)=\int_0^1 z dz$, how can I further simplify it, do I plug $z$ with $x,y$? If so, $z$ is a non-linear function on $x,y$, how would I deal with $dz$ as well as the domains for the integration.
Observing the function form of $z$ is actually an ellipse, can I use the the property of ellipse calculating the area proportion of $z=1$ to the upper bound area $z=2$ so that to obtain the probability $P(z \leq 1)$? Feels like doing so we need some extra assumptions, for example, each ellipse is drawing with the same probability.
Firstly, let us write capital letters for r.v.'s and small letters for real numbers.
$$F_Z(1)=\int_0^1 f_Z(z)\,dz, $$ and to use this formula, you need to find probability density function $f_Z(z)$ first, which is much more complicated than solving the original problem.
From the other side, $$ \mathbb P(Z\leq 1)=\mathbb P((2X-1)^2 + (2Y-1)^2 \leq 1)=\iint\limits_{(2x-1)^2 + (2y-1)^2 \leq 1} f_X(x)f_Y(y)\, dx\,dy. $$ Note that $f_X(x)f_Y(y)=1$ for $(x,y)\in[0,1]\times[0,1]$ only. And to find the integral, one need to find the area of intersection of the filled circle $(2x-1)^2 + (2y-1)^2 \leq 1$ and unit square. Find that the circle is inscribed in a square and the desired probability equals to its area ($\pi/4$).