Find f(u,v) = joint pdf of U and V.

A circular dartboard has a radius of 1 foot. When a dart is thrown at the board, where it sticks is uniformly distributed on the face of the board, meaning that probability is proportional to area. (And the probability for the whole board is 1, since throws which miss the board are repeated until the dart hits the board.) Hitting points for different darts are independent.

(a). Throw 2 darts. Let X be the distance to the center for the first dart, in feet. Let Y be the distance to the center for the second dart. Let U = min(X,Y) and V = max(X,Y).

Find f(u,v) = joint pdf of U and V. Remember to say where f(u,v) is positive.

Attached is the solution for this question.

However, I do not understand how we got f(u,v)= 8uv. Would you guys mind explaining? Or is there any simpler way to do this? Something that is more clear.

Thanks

Firstly let's calculate the CDF of $X$: $$F_X(x) = P(X \le x) = \frac{\pi x^2}{\pi r^2} = x^2, 0 \le x \le 1$$ hence the PDF of $X$ is $f_X(x) = 2x, 0 \le x \le 1$ then the joint PDF of $X, Y$ is $$f(x, y) = f_X(x) f_Y(y) = 4xy, \ \ 0 \le x \le 1, 0 \le y \le 1$$
Now, compute the joint CDF of $U, V$:
$$P(min(X, Y) \le u, max(X, Y) \le v) = \\ P(min(X, Y) \le u, max(X, Y) \le v|X>Y) \times P(X>Y) + P(min(X, Y) \le u, max(X, Y) \le v|X \le Y) \times P(X \le Y) = \\ = P(Y \le u, X \le v | X <Y) \times\frac{1}{2} + P(X \le u, Y\le v|X \le Y) \times \frac{1}{2} = \\ = \frac{u^2v^2}{\frac{1}{2}} \times \frac{1}{2} + \frac{v^2u^2}{\frac{1}{2}} \times \frac{1}{2} = 2u^2v^2, 0 \le u \le v \le 1$$
Finally the joint PDF of $U, V$ is $$\frac{\partial^2}{\partial u\partial v} 2u^2v^2 = 8uv, \ 0 \le u \le v \le 1$$