How can I describe the area between two ellipses? Given two ellipses that take up regions $E_1$ and $E_2$ in $\mathbb{R^2}$, with the following properties:


*

*Centers defined in the Cartesian coordinate system $(c_1, 0)$ for $E_1$ and $(c_2, 0)$ for $E_2$ such that $c_2>c_1$

*Semi-diameters $x_1$ & $y_1$ for $E_1$ and $x_2$ & $y_2$ for $E_2$ such that the two ellipses intersect at exactly two points.


Let the surface $\Sigma=E_1\cap E_2$
Describe $\Sigma$ with two parameters, $u$ and $v$, in the form $a\le u\le b$ and $f(u)\le v \le g(u)$ for $\left\{a,b:a,b\in\mathbb{R}\land a<b\right\}$, $f:\mathbb{R}\rightarrow\mathbb{R}$, $g:\mathbb{R}\rightarrow\mathbb{R}$, and $g(u)>f(u)$ over $[a,b]$.  In other words, please find the Lebesgue measure of the set of points that satisfy both of the aforementioned inequalities.
-My work so far-
The two equations for the ellipses are easy to find given the conditions.
$$
\frac{y^2}{y_1^2}+\frac{(x-c_1)^2}{x_1^2}=1
$$
$$
\frac{y^2}{y_2^2}+\frac{(x-c_2)^2}{x_2^2}=1
$$
After multiplying by a constant and subtracting one from another, I arrive at two solutions for $x$:
$$
x=\frac{b\pm \sqrt{b^2-4ac}}{2a}
$$
Where $a=\frac{y_1^2}{x_1^2}-\frac{y_2^2}{x_2^2}$, $b=\frac{y_2^2c_2}{x_2^2}-\frac{y_1^2c_1}{x_1^2}$, and $c=\frac{y_1^2c_1^2}{x_1^2}+\frac{y_2^2c_2^2}{x_2^2}$.  However, from the fact that $c_2>c_1$ and that the ellipses only intersect at two points, I suppose that the only value of $x$ must be the larger one.  I evaluated and checked the determinant with Mathematica, and it is not equal to 0.  How can I be sure to pick the right value of $x$?
Even assuming I found the right value of $x$ and therefore have the intersection points $(x_0, y_0)$ and $(x_0, -y_0)$, with $y_0=\sqrt{y_2^2(1-\frac{(x_0-c_2)^2}{x_1^2})}$, I still have the problem of defining the intersecting area.

If I have $u=y$ and $v=x(y)$, then I'm assuming I have a type II region, whereas I really have a type II region combined with two type I regions.  Pursuing that piecewise definition of $\Sigma$ is not optimal, since I would have to find what parts of the two curves break the horizontal line test, set up three different integrals, etc.  A potential workaround I see is converting to polar, but I don't know how to approach that.
So, to summarize:


*

*If you could find a way to solve the original, that would be much appreciated.

*If you could tell me which $x$ value in the quadratic equation is $x_0$, please let me know.

*If you could elaborate a polar approach, I'd love to see it.


EDIT
As pointed out in the comments section below, the value $x_0$ must be only one of the following solutions $x=\frac{b\pm \sqrt{b^2-4ac}}{2a}$, since the $x$ value farther from $c_2$ will yield an imaginary $y_0$ value.  However, this means I cannot find $x$ symbolically.  If someone has a different approach to finding the intersection, please let me know.
 A: After a translation we may assume
$$E_1:=\left\{(x,y)\Biggm|{x^2\over a_1^2}+{y^2\over b_1^2}\leq 1\right\},\qquad E_2:=\left\{(x,y)\Biggm|{(x-c)^2\over a_2^2}+{y^2\over b_2^2}\leq 1\right\}\ , \quad c>0\ .$$
As assumed the two ellipses intersect in two points $(p,\pm q)$ such that $-a_1<c-a_2<p<a_1$. The value $p$ can be computed from the given data. For $x\leq p$ the boundary $\partial S$ of $S:=E_1 \cap E_2$ is an arc of $\partial E_2$, and for $x\geq p$ the boundary $\partial S$ is an arc of $E_1$. Therefore the set $S$ can be described in the form
$$S=\bigl\{(x,y)\,\bigm|\, c-a_2\leq x\leq a_1, \ -f(x)\leq y\leq f(x)\bigr\}\ ,$$
where the bounding function $f$ is given by
$$f(x):=\cases{b_2\sqrt{1-(x-c)^2/a_2^2} &$(c-a_2\leq x \leq p)$ \cr
b_1\sqrt{1-x^2/a_1^2} &$(p\leq x\leq r)$\cr}\ .$$
A: Is this an assigned problem for a course? The strategy that I would use for this is quite different, and dependent only on single-variable calculus methods. Once I found the two intersection points, I would draw the vertical line between them, and calculate the area of (in the case of your picture) the smaller ellipse that lies to the left of the line, and the larger ellipse that lies to the right of that line. Both of these are calculated by a one-variable integral of the shape $\int\sqrt{a^2-x^2}\,dx$, and this can be done by a trigonometric substitution. Or even by appeal to Euclidean geometry, if you’re feeling antagonistic to the Calculus tonight!
