How to calculate ellipse sector area *from a focus*

The area will be $$\int_{\theta_1}^{\theta_2}\frac{1}{2}r^2d\theta,$$ where $r=r(\theta)$ is the equation of the ellipse, with polar origin at the focus.

Imagine an ellipse with semi-major axis $a$ and eccentricity $e$, and with one of the foci at the origin, and the other focus on the half-line $\theta=0$ (so to the "right" of the origin). Then the ellipse has polar equation $$r=\frac{a(1-e^2)}{1-e\cos\theta}.$$

The rest is unpleasant. There is a closed form for the antiderivative. One might ask Wolfram Alpha. If you want to do it yourself, you can use the Weierstrass substitution. But there is undoubtedly a neater way.

Ben, here's a better suggestion. You can stretch a circle to make an ellipse and, if you start with a unit circle, area is magnified by the factor of $ab$, where $a$ and $b$ are the semi-axes, as usual. Take a point at $(-R,0)$ inside the unit circle and consider the sector it subtends to $(1,0)$ and $(\cos t, \sin t)$. You can find the area pretty easily: I get $\frac 12(t+R\sin t)$. Now stretch by the fudge factor and figure out how to match up $R$ with your focus and $t$ with your arbitrary point on the ellipse.

Ted's answer already outlined what I'd do, but left a lot of things for readers to work out or look up. So here I'll give details on these. The main idea is to transform the problem to the circle, where areas are computed more easily.

Transforming coordinates

The polar parametrization of an (axis-aligned) ellipse from its focus is given by

$$\overline{FP}=r(\theta)=\frac{a(1-e^2)}{1+e\cos\theta}$$

where $a$ is the semi-major axis, $e$ is the eccentricity and $\theta$ is the angle, i.e. the true anomaly. Spelled out in coordinates, this is

$$\overrightarrow{FP}= \begin{pmatrix}r(\theta)\cos\theta\\r(\theta)\sin\theta\end{pmatrix}$$

Now stretch the whole things in the direction of its minor axis (i.e. its $y$ direction) by a factor of

$$\frac ab=\frac1{\sqrt{1-e^2}}$$

The result will be a circle of radius $a$. The origin is still the point that used to be the focus. The distance between that point and the center is $\overline{CF}=ae$ (also called linear eccentricity or focal distance). So add that to $x$ coordinates to move the origin into the center. Together you now have

$$ \overrightarrow{CQ}= \begin{pmatrix}r(\theta)\cos\theta+ae\\ \frac ab r(\theta)\sin\theta\end{pmatrix} = \frac{a}{1+e\cos\theta} \begin{pmatrix}(1-e^2)\cos\theta+e+e^2\cos\theta\\ \sqrt{1-e^2}\sin\theta\end{pmatrix} $$

Eccentric anomaly

The angle for that point (against the horizontal axis) is the eccentric anomaly $E$, so it satisfies the following relation:

$$\tan E=\frac{\sqrt{1-e^2}\sin\theta} {(1-e^2)\cos\theta+e+e^2\cos\theta}$$

Using the Weierstrass substitution André mentioned in his answer, you can turn this into

\begin{align*} \frac{2\tan\frac E2}{1-\tan^2\frac E2}&= \frac{\sqrt{1-e^2}2\tan\frac\theta2} {(1-e^2)(1-\tan^2\frac\theta2)+e(1+\tan^2\frac\theta2)+e^2(1-\tan^2\frac\theta2)} \\&= \frac{2\sqrt{(1+e)(1-e)}\tan\frac\theta2}{(1+e)-(1-e)\tan^2\frac\theta2} =\frac{2\sqrt{\frac{1-e}{1+e}}\tan\frac\theta2} {1-\frac{1-e}{1+e}\tan^2\frac\theta2} \end{align*}

By simply comparing both sides you can see

$$\tan\frac E2 = \sqrt{\frac{1-e}{1+e}}\tan\frac\theta2$$

This agrees with the relation Wikipedia mentions. And dealing with the tangent of half the angle has the benefit that you can compute the arctan without worrying about the quadrant. You can simply compute

$$E=2\arctan\left(\sqrt{\frac{1-e}{1+e}}\tan\frac\theta2\right)$$

except for the special case when $\theta=\pm\pi$ which yields an infinite tangent but results in $E=\pm\pi$ as well.

Area of circular sector

Now you know that the area of a circular sector is proportional to the angle, so for a circle of radius $a$ this would be $\frac12a^2E$. That's the yellow and cyan areas taken together, i.e. the sector $ACQ$.

From center back to focus

But that area is subtended by the center, not by the focus. Therefore subtract a triangle of base $ae$ and height $a\sin E$ (the cyan triangle $\triangle FCQ$) and you get the area subtended by the focus ($AFQ$ colored yellow):

$$ \tfrac12a^2E-\tfrac12a^2e\sin E=\tfrac12a^2\left(E - e\sin E\right) $$

Back to the ellipse

This is still an area in the circle. To get back to the original ellipse, you have to undo the scaling, i.e. scale the $y$ direction by $\frac ba$. This scales areas by the same factor. Thus the area you were asking about ($AFP$ tinted red) would be

$$\tfrac12ab\left(E-e\sin E\right)$$

Mean anomaly

As you vary $\theta$ from $0$ to $2\pi$, the above area will vary from $0$ to $ab\pi$. Since equal areas are swept in equal time, this is closely related to the mean anomaly which sweeps from $0$ to $2\pi$ in constant time.

$$M=E-e\sin E$$

This result is again mentioned on Wikipedia, with a reference to Kepler's equation. So in terms of the mean anomaly, you'd get the area as

$$\tfrac12abM$$

just as ben wrote in this comment.