# 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:

1. Centers defined in the Cartesian coordinate system $(c_1, 0)$ for $E_1$ and $(c_2, 0)$ for $E_2$ such that $c_2>c_1$
2. 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:

1. If you could find a way to solve the original, that would be much appreciated.
2. If you could tell me which $x$ value in the quadratic equation is $x_0$, please let me know.
3. 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.

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You're getting results that don't make sense to you, which indicates that you've made a mistake. Try constructing a specific case in which you know the intersection occurs at only one x. Crank through the calculation and figure out whether you're really getting two x's as solutions. If so, then your mistake happened before you set up the quadratic. If not, then it happened after that. (And consider all three possible cases: discriminant less than zero, zero, or greater than zero.) –  Ben Crowell Aug 27 '12 at 2:40
In agreement with @Ben, I recommend strongly looking at one concrete example and seeing what happens. In disagreement with him, however, I suspect that both your values of $x$ are “correct”, but that only one is in the closed interval of the $x$-axis that’s inside both ellipses. The point outside this interval will give $y$-coordinates for the intersection that are conjugate complex. This is what ordinarily happens when two conics intersect in only two real points (except that two circles of different radii have two common complex points on the line at infinity!). –  Lubin Aug 27 '12 at 3:07
@BenCrowell I agree with Lubin - most likely, the term $(1-\frac{(x_0-c_2)^2}{x_1^2})$ in $y_0=\sqrt{y_2^2(1-\frac{(x_0-c_2)^2}{x_1^2})}$ becomes negative, causing the smaller $x$ values to become imaginary. Therefore, the $x$ that we're looking at has to be the larger one. I guess that answers that. I don't think we need to go through a numerical case if my explanation sounds convincing enough. That still only gets me down to where I ended in my question, though. –  VF1 Aug 27 '12 at 3:21
FYI - The image I posted used (1) $\frac{(x-5)^2}{3^2}+\frac{y^2}{6^2}=1$ and (2) $\frac{(x-7)^2}{12}+\frac{y^2}{3^2}$. Should I put that in there? –  VF1 Aug 27 '12 at 3:22
Actually - something I just noticed - for the term that I mentioned to become imaginary, only $(x_0-c_2)^2>x_1^2$. Thus, only the value that is closer to $c_2$ can be real. That really summarizes the scenario, though: since I didn't bring in any information about exactly two intersections, the statement holds true for ellipses that intersect once, twice, or even four times (three is not applicable in this scenario). As the distance from $c_2$ is more of a concern, we can't tell which $x$-values to choose until we have a set of numbers to work with - either the larger or smaller x could –  VF1 Aug 27 '12 at 14:50
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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 $$f(x):=\cases{b_2\sqrt{1-(x-c)^2/a_2^2} &(c-a_2\leq x \leq p) \cr b_1\sqrt{1-x^2/a_1^2} &(p\leq x\leq r)\cr}\ .$$

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Why can we assume that $E_1$ is a circle, not an ellipse? –  VF1 Aug 27 '12 at 22:14
Also, you're assuming that the intersection $S$ is a type I region. I can draw a circle and ellipse intersecting in a way such that they are not a type I region –  VF1 Aug 27 '12 at 22:19
Any one ellipse can be stretched or squashed $(x,y)\mapsto(\lambda x,y)$ to become a circle, and the area is just multiplied thereby by the stretch-factor $\lambda$. The other ellipse will not ordinarily become a circle, of course, but its area will be multiplied by the same factor. –  Lubin Aug 27 '12 at 23:25
Okay, @Lubin. That makes sense now. And an intersection between a circle and ellipse is guaranteed to be a simple type II region. Thanks, Christian. –  VF1 Aug 27 '12 at 23:27
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!
No. This is a problem I made up. I understand the approach that you have, but, unfortunately, it is not finding the area of the region $\Sigma$ that I'm interested in, it's more about defining it. I am more interested in finding a definition of $\Sigma$ with $u$ and $v$. –  VF1 Aug 27 '12 at 3:34
@HenningMakholm No, I am not looking for the explicit area of the intersection, but a definition with inequalities (which I can use in a surface integral's limits). Perhaps an example will get my point across better. The area of a rectangle $R$ is $ab$, but to define its area with inequalities would be to use $0\le x\le a$ and $0 \le y\le b$. The former I can't use to find $\int\int_R f(\vec r) dS$ for some $f$ and position vector $\vec r$, whereas the latter would be perfect. I didn't want to complicate the question with unnecessary information about surface integration. –  VF1 Aug 27 '12 at 22:12