Say I have a region with four "corners" that are connected by parabolas (like in the picture below). Is there a nice way to compute the enclosed area? To make things simple, say the equations for the parabolas are $y = a_ix^2 + b_i, i=1,2$ and $x = a_iy^2 + b_i, i = 3,4$. Note there are many different configurations that may occur. For example, if the $a_i$ are too large, then there will be more than 4 "sides".
By connecting opposite corners with diagonal lines and use the intersection point, one creates four regions and can then compute 4 straightforward integrals (two $x$ integrals and two $y$ integrals). However, I was wondering if there is a better way, especially a way which does not require one to compute the intersection points. In particular, in Finding the area bounded by 4 parabolas, the case is simple enough (albeit not exactly my setup) such that a simple change of coordinates works. Can anyone spot a change of coordinates that works here? Alternatively, are there any well-known conformal maps which map such a region to a square or disk?
From any two points $P$ and $Q$ on a parabola, construct line segment $PQ$. From the midpoint $M$ of $PQ$ construct a line parallel to the axis of the parabola, intersecting the parabola at $R$. The area enclosed between the segment $PQ$ and the parabola is $\frac43$ times the area of the triangle $\triangle PQR$.
Once you have done this with each parabola using the intersection points as $P$ and $Q$, what is left is a polygon whose area you should be able to compute.