How to find the area of a segment of an ellipse 
I need to find the area of the yellow part of the arc 
given the a, b , start and end angle of the sector points
Also the ellipse is centered at the origin
How to find the area of the yellow part?
 A: A calculus-free derivation:
Consider the analagous figure drawn for a unit circle. We find the area (assuming an angle is given as $\theta$) as
$$
A = \frac12(\theta-\sin\theta)
$$
Stretch the graph left-right by a factor of $a$, and stretch it up-down by a factor of $b$. Having stretched the region with the rest of the picture, we can deduce that the new area will be
$$
A = \frac{ab}{2}(\theta-\sin\theta)
$$
Where $\theta$ is still the angle of our squished ellipse. To make this a complete formula, we must find an expression for $\theta$ given an elliptical angle.
In fact, if we are given an elliptical angle $\phi$ from the x-axis, we have
$$
\theta = \arctan\left[\frac{a}b \tan\phi\right] = 
$$
Which gives us
$$
A = \frac{ab}{2}\left(\arctan\left[\frac{a}b \tan\phi\right]-\sin\left(\arctan\left[\frac{a}b \tan\phi\right]\right)\right)
$$
In the case that $\phi$ is not given as an angle from the x-axis, we can break $\phi$ into $\phi = \phi_1+\phi_2$, where $\phi_1$ is the part of the angle going clockwise from the x-axis, and $\phi_2$ is the counterclockwise part from the x-axis.  We then have
$$
\theta = \arctan\left[\frac{a}b\phi_1\right] + \arctan\left[\frac{a}b\phi_2\right]\\
%\tan\theta= ab\frac{\tan\phi_1+\tan\phi_2}{1-a^2\tan\phi_1\tan\phi_2}
$$
Which can be substituted as before.

This was more complicated than I expected it to be.
Please comment, edit, or let me know if there is anything I have left out that makes this answer less understandable.
A: If we are allowed to use calculus,
 the eqaution of the ellipse, $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ and $\alpha,\beta$ be the known angles
From the defintion of  parametric angle,
We get, $\tan\alpha=\frac{b\sin\phi}{a\cos\phi}$
and $\tan\beta=\frac{b\sin\delta}{a\cos\delta}$ where $\phi,\delta$ are the parametric angles
Consequently,the area will be $$\left|\int_{a\cos\phi}^{a\cos\delta}ydx\right|=\left|\int_{a\cos\phi}^{a\cos\delta}b\left(\sqrt{1-\frac{x^2}{a^2}}\right)dx\right|=\frac ba\left|\int_{a\cos\phi}^{a\cos\delta}\sqrt{a^2-x^2}dx\right|$$ 
