In a circle $k(O,r)$ we have fixed point A, point B lands randomly on the circle. Find $EX$ of the area of the triangle $AOB$
What I have tried : since the area of the triangle will be $ \frac{1}{2}*r^2(sin(\alpha))$ where $\alpha$ is from 0 to 360 degrees then we have to find the integral $\displaystyle\int_0^{2\pi} \frac{1}{2}*\alpha*r^2(sin(\alpha))\, d\alpha$ .Not sure if I am correct or missing something
The area of a triangle is $\displaystyle ab\sin{C}$.
Consider the scenario where $O$ is the origin, and $A$ is fixed to the positive $x$ axis, from $0$ to $r_1$.
Then, in polar coordinates, let $B$ be $(r_2,\theta)$, where $\theta$ ranges from $-\pi$ to $\pi$.
The area of this triangle is $\displaystyle \frac{1}{2}r_1r_2|\theta|$.
So that means the average value of the integral is $\displaystyle \frac{2}{2\pi r^2}\int_0^{\pi}\int_0^r\int_0^r\frac{1}{2}r_1 r_2\sin(\theta)\,dr_1\,dr_2\,d\theta$.
Notice that I made the integral symmetric by noting the area is the same for $\theta$ and $-\theta$, hence the last bound only ranging from $0$ to $\pi$, and a corresponding $2$ in the numerator.
In your case, which I just realized, $a$ is fixed. Eliminate $r_1$ from the integral, and now you just have $\displaystyle \frac{2}{2\pi r}\int_0^{\pi}\int_0^r\frac{1}{2}r_1 r_2\sin(\theta)\,dr_2\,d\theta$.