Expected value of area of triangle Here is the problem:
Let $A$ be the point with coordinates $(1, 0)$ in $\mathbb R ^2$. Another point $B$ is chosen randomly over the unit circle. What is then the expected value of the area of the triangle $OAB$ ? 
What I do is to define a random variable $X$, which is the angle $AOB$ and assume that it is uniformly ditributed between $0$ and $\pi$. Then the random variable $Y = \frac{\sin X}{2}$ is the area of the triangle. Unfortunately I don't know how to calculate it's expected value (or even it's distribution). Can someone help me with this ? Thanks in advance!
 A: If $X$ is uniform in $[0, \pi]$ then its density is just $1/\pi dx$.  Then the formula for calculating the expected value of a function of a random variable of X is:
$$E[f(X)] = \int_0^\pi f(x) \frac{1}{\pi} dx$$
In your case $Y = \frac{\sin X}{2}$ so:
$$E[Y] = \int_0^\pi \frac{\sin x}{2} \frac{1}{\pi} dx$$
A: Another way to calculate the expected area is to realize that the area of the triangle depends only on the $y$-coordinate of the point $B$. This is because the base of the triangle is the distance between $O$ and $A$, namely $1$, and the height of the triangle is $|y|$. Therefore the area is just $\frac {|y|}2$. By symmetry you can just find the expected value of the area for $B$ above the $x$-axis: then area is $\frac y2$.
If $B$ is chosen in the unit disk, just take a weighted average of $\frac y2$ from $0$ to $1$, where the weight is the length of the chord that is the intersection of the horizontal line with the unit circle.
If $B$ is chosen on the unit circle, take the average over the points on the upper semicircle. This ends up being 
$$\frac 1{\pi}\int_0^{\pi} \frac{\sin x}2\,dx$$
which is the same formula that others have given you.
(Thanks to @AndréNicolas for clarifying the possible confusion between disk and circle in the OP.)
