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?
 A: I suppose $a<b$.  The intersection points of the two ellipses are (see the figure where $a=2$ and $b=4$) $A,B,C,D =(\pm k,\pm k)$ with
$$k=\frac{ab}{\sqrt{a^2+b^2}}$$ 

and, by symmetry, the searched area is
$$
4\left(k^2+\frac{2b}{a}\int_k^a \sqrt{a^2-x^2}dx\right)
$$
A: 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 $x = y$
$a\cos\theta = b \sin \theta\\
\tan \theta = \frac {a}{b}$
$8\int_0^{\arctan \frac {a}{b}}\int_0^{1} abr\ dr\ d\theta$
$4 ab \arctan\frac {a}{b}$
A: Consider the picture below:

The description of two ellipses in polar coordinates is:
$$r^2={1\over {{\cos^2\theta\over{a^2}}+{\sin^2\theta\over{b^2}}}}$$
$$r^2={1\over {{\cos^2\theta\over{b^2}}+{\sin^2\theta\over{a^2}}}}$$
so the surface we seek to find (assuming $a>b$) is:
$$S=\int_{\pi\over4}^{3\pi\over4}{r^2\over2}d\theta={1\over 2}\int_{\pi\over4}^{3\pi\over4}{1\over {{\cos^2\theta\over{a^2}}+{\sin^2\theta\over{b^2}}}}d\theta=abtan^{-1}{a\over b}$$
Then the area between two ellipses is $\Large 4abtan^{-1}{max(a,b)\over min(a,b)}$
