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?

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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.

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I think you want to multiply $y$ by a factor $\frac ab$ to make a circle, so the fractions inside the $\arctan$s should be inverted. – Ross Millikan Apr 24 '13 at 20:15

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

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

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.

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Using polar elliptical coordinates:$$x=a\rho cos(\theta)$$ $$y=b \rho sin(\theta)$$ the part of the plane enclosed from the ellipse is $$\{(\rho,\theta):0\le \rho \le 1,\theta\in(0,2\pi)\}$$ the Jacobian of the inverse transform is: $$J=\begin{bmatrix}a cos(\theta) & -a \rho sin(\theta) \\ b sin(\theta) & b \rho cos(\theta)\end{bmatrix}$$ so the area of a sector is: $$\int_{\theta_1}^{\theta_2}d\theta \int_0^1 d\rho a b\rho$$

Because the integral in $d\rho$ is $\frac{1}{2}ab$ the result is $$\frac{1}{2}ab(\theta_2-\theta_1)$$

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I don't think this is correct. In your coordinate system, the line joining the origin to the point $(x,y) = (a\rho\cos\theta, b\rho\sin\theta)$ does not subtend an angle of $\theta$ with the $x$-axis, so your limits of integration won't be the $\theta_1$ and $\theta_2$ given in the figure. – Rahul Feb 28 '12 at 10:37

protected by Community♦Oct 11 '14 at 21:37

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