Common area of two rectangles Suppose we have a rectangle at the center of the coordinates. One top point of the rectangle has the coordinates (a, b), the second (-a, b), third (-a, -b) and (a, -b). We rotate this rectangle with the angle $\phi $ counterclockwise to get a second rectangle. 
Question: Find the common area of these two rectangles.
By common area I mean the area of the polygon that is limited by the intersection points of the two rectangles

I'll assume $0<a<b$ and $0\leq\phi\leq\frac\pi2$. All other cases can be reduced to this.
Case 1: $\phi < 2\arctan\left(\frac ab\right)$
In this case you have an octagonal intersection, with an area of
$$2\frac{2ab\sin(\phi) + \bigl(a^2 + b^2\bigr)\bigl(\cos(\phi) - 1\bigr)}
{\sin(\phi)\,\cos(\phi)}$$
Case 2: $\phi \geq 2\arctan\left(\frac ab\right)$
In this case you get a quadrilateral intersection, with an area of
$$\frac{4a^2}{\sin(\phi)}$$
I obtained the above results using symbolic expressions for all coordinates, performing the computation using sage.
