Deriving the Area of a Sector of an Ellipse

A sector $P_1OP_2$ of an ellipse is given by angles $\theta_1$ and $\theta_2$.

Could you please explain me how to find the area of a sector of an ellipse?

Solution 1:

Scale the entire figure along the $y$ direction by a factor of $a/b$. The ellipse becomes a circle of radius $a$, and the two angles become $\tan^{-1}(\frac ab\tan\theta_1)$ and $\tan^{-1}(\frac ab\tan \theta_2)$. The area of the original elliptical sector is $b/a$ times the area of the circular sector between these two angles, which is straightforward to find.

Solution 2:

As I show in Evaluating $\int_a^b \frac12 r^2\ \mathrm d\theta$ to find the area of an ellipse the area of an ellipse given its central angle is:

$A(\theta) = \frac{1}{2} a b \tan ^{-1}\left(\frac{a \tan (\theta )}{b}\right)$

for $0\leq\theta\leq2\pi$, so the area of the sector in question is:

$A\left(\theta _2\right) - A\left(\theta _1\right)$

or:

$\frac{1}{2} a b \left(\tan ^{-1}\left(\frac{a \tan \left(\theta _2\right)}{b}\right)-\tan ^{-1}\left(\frac{a \tan \left(\theta _1\right)}{b}\right)\right)$

which sadly does not appear to simplify any further.