An unexpectedly difficult geometry problem

Question

I've run into a geometry problem that feels like it should have an easy answer. But short of numerical integration, I can't find a way to solve it.

Consider a filled circle on top of a filled ellipse with their origins overlapping, and the major axis of the ellipse on the X axis (minor axis aligned with y). (See picture)

The radius of the circle and major/minor axes of the ellipse are known constants. Find the green shaded area.

If at all possible, I would like a function $A(Green) = f(a,b,R)$. I've tried finding the intersection point of the two shapes in the first quadrant and integrating, but the integrand does not lend itself to a simple analytical integration technique, and Mathematica has been chewing on this problem for about an hour now to no success.

Numerical integration could work for my purposes, and I may need to resort to that. Is there a better way to proceed before I do that?

Answer 1

Draw line segments from the origin to the two intersection points to the right, which are $\displaystyle \biggl( \frac{a \sqrt{R^2-b^2}}{\sqrt{a^2-b^2}},\pm\frac{b\sqrt{a^2-R^2}}{\sqrt{a^2-b^2}} \biggr)$. The shaded green area to the right is the area of a sector of an ellipse minus the area of a sector of a circle.

• The angle each segment makes with the $x$-axis is $\arctan \dfrac{b\sqrt{a^2-R^2}}{a \sqrt{R^2-b^2}}$, and therefore the area of the circle sector is $R^2 \arctan\dfrac{b\sqrt{a^2-R^2}}{a \sqrt{R^2-b^2}}$.
• Now scale the $y$-coordinate by $\frac ab$ while leaving the $x$-coordinate alone; this multiplies all areas by exactly $\frac ab$. The ellipse becomes a circle of radius $b$, the new points of intersection are $\displaystyle \biggl( \frac{a \sqrt{R^2-b^2}}{\sqrt{a^2-b^2}},\pm\frac{a\sqrt{a^2-R^2}}{\sqrt{a^2-b^2}} \biggr)$, the new angle is $\arctan \dfrac{\sqrt{a^2-R^2}}{\sqrt{R^2-b^2}}$, and thus the area of the new circle sector is $b^2 \arctan \dfrac{\sqrt{a^2-R^2}}{\sqrt{R^2-b^2}}$. Therefore the area of the original elliptical sector is $ab \arctan \dfrac{\sqrt{a^2-R^2}}{\sqrt{R^2-b^2}}$.

Subtracting and doubling to take the left green area into account yields that the total green area is $$ 2\biggl( ab \arctan \dfrac{\sqrt{a^2-R^2}}{\sqrt{R^2-b^2}} - R^2 \arctan\dfrac{b\sqrt{a^2-R^2}}{a \sqrt{R^2-b^2}} \biggr). $$ Moral: areas involving ellipses are just as easy as areas involving circles. (Arc lengths, now those are harder.)

Answer 2

In polar coordinates, the circle is $r=R$ and the ellipse is $\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}=\frac1{r^2}$. Their angular intercept $\theta_0$ is given by $\frac{\cos^2\theta_0}{a^2} + \frac{\sin^2\theta_0}{b^2}=\frac1{R^2}$, or, $\tan^2\theta_0= \frac{1-\frac{R^2}{a^2}}{\frac{R^2}{b^2}-1}$, and each of the two green areas is \begin{align} \\ A=&\int_{-\theta_0}^{\theta_0} \frac12\left[r^2(\theta)-R^2\right]d\theta = \int_{0}^{\theta_0} \left(\frac{a^2b^2}{b^2\cos^2\theta +a^2\sin^2\theta}-R^2\right)d\theta\\ =& \ ab \tan^{-1}\left(\frac ab \tan\theta_0\right) -R^2\theta_0= ab \tan^{-1} \sqrt{\frac{\frac{a^2}{R^2}-1}{1-\frac{b^2}{R^2}} } -R^2 \tan^{-1} \sqrt{\frac{1-\frac{R^2}{a^2}}{\frac{R^2}{b^2}-1} } \end{align}