Expected area of a triangle with vertices independently and uniformly at random chosen within unit circle. I wish to find the expected area of a triangle with vertices independently and uniformly at random chosen from a unit circle. I solved this problem, but my answer does not agree with the one found here. Where did I make a mistake?
Attempt: Let $O$ be the center of the circle and let $P_1, P_2, P_3$ be the vertices. By linearity of expectation, symmetry, and law of sines, $\mathbb{E}([P_1P_2P_3]) = \mathbb{E}([OP_1P_2] + [OP_2P_3] + [OP_3P_1]) = 3\mathbb{E}([OP_1P_2]) = \\ \frac{3}{2}\mathbb{E}(OP_1 \cdot OP_2 \cdot \sin(\angle P_1 O P_2)).$
Specifying $OP_1, OP_2$ does not affect the distribution of the angle, so $\mathbb{E}(OP_1 \cdot OP_2 \cdot \sin(\angle P_1 O P_2)) = \mathbb{E}(OP_1 \cdot OP_2)\mathbb{E}(\sin(\angle P_1 O P_2)).$
Every angle in $[0,\pi]$ is equally likely, so $\mathbb{E}(\sin(\angle P_1 O P_2)) = \frac{1}{\pi} \int_0^{\pi} \sin(x) \, dx = \frac{2}{\pi}.$ The distances $OP_1, OP_2$ are independent, so $\mathbb{E}(OP_1 \cdot OP_2) = \mathbb{E}(OP_1)^2.$ The ring of distance $r$ from the center has circumference $2\pi r,$ so the pdf of $OP_1$ is proportional to $r.$ Thus, $\mathbb{E}(OP_1) = \frac{\int_0^1 r^2 \, dr}{\int_0^1 r \, dr} = \frac{2}{3}.$ Combining everything gives us an expected value of $\frac{3}{2} \cdot \frac{2}{\pi} \cdot \left(\frac{2}{3}\right)^2 = \frac{4}{3\pi}$ for the area of the triangle.
I suspect that the first 2 paragraphs are solid and the separation of angle and distance is legal. Maybe the issue is that $OP_1, OP_2$ are not independent, but how could that be?
 A: Not an answer to the question asked
@Bananach has already answered (in the comments) the "where's the problem in my reasoning?" question well. This answer attempts to take that same reasoning and deform it to something that might yield the correct answer with almost equal simplicity.
Here's an alternative approach: Pick your three points; let $s$ be the angle counterclockwise from $P_1$ to $P_2$; let $t$ be the angle counterclockwise from $P_1$ to $P_3$. I claim that $s$ and $t$ are uniformly distributed on $[0, 2\pi]$. I expect you believe this part, so I'll move on.
I'm now going to rotate the picture so that $P_1$ is at the point $U = (1, 0)$, so we have angles $s$ and $t$ measured CCW in the usual way from the positive $x$-axis. Letting $S = (\cos s, \sin s)$ and $T = (\cos t, \sin t)$, we're interested in the area of the triangle $UST$. As in your solution, let $O$ be the origin, and let $H(P, Q, R)$ be the signed area of the triangle $PQR$. Then the signed area of $UST$ is
$$
J = H(O,U,S) + H(O, S, T) + H(O, T, U).
$$
You'd like to compute the expected value of $|J|$, and this absolute value is a pain in the neck.
Our probability space $X$ consists of uniformly equally likely points $(s, t) \in [0, 2\pi]$.  Suppose we split $X$ into two parts, $X_1 = \{(s, t) \in X \mid s \le t \}$, $X_2 = X - X_1$. Then on $X_1$, we know that $J$ is positive, and on $X_2$, we know that $J$ is negative. And in fact the expected values of $J$ on the two parts are opposites (simply swap $s$ and $t$ to see this), so we find that the expected value of $|J|$ on all of $X$ is twice the expected value of $J$ on $X_1$.
I'll bet that you can compute that expected value using linearity of expectation, etc., just as you did before, except that you'll need a double integral where the inner integral's limits depend on the outer integral (because you're integrating over a triangle rather than a square).
A: The first place I doubted what you were saying was this:
"Every angle in $[0,\pi]$ is equally likely"
Can you give a reason you think this is true for the uniform distribution of the points $P_1$ and $P_2$?
