Finding the area between the two ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ I'm trying to find the area inside the intersection of the region bounded by the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ using vector calculus methods. 
I understand how to calculate area of regions by taking the line integral around the boundary and using Green's Theorem. However what I'm stuck with is finding a paramterization of the boundary! Can anyone point me in the right direction?

The figure will be symmetric across the line $x = y$ (along with $x = 0$ and $y = 0$)
We can analyze $\frac 18$ the figure, and then use this symmetry to our advantage.
Suppose $a<b$
$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$ will be on the "inside" from $0$ to the point of intersection.
Parameterize the area inside the curve:
$x = ar\cos\theta\\
y = br\sin\theta\\
dx\ dy = ab r\ dr\ d\theta$  
The last line is the Jacobean for this coordinate system.
And the point of intersection is where $x = y$
$a\cos\theta = b \sin \theta\\
\tan \theta = \frac {a}{b}$
$8\int_0^{\arctan \frac {a}{b}}\int_0^{1} abr\ dr\ d\theta$
$4 ab \arctan\frac {a}{b}$
