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enter image description here

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?

Chris Godsil
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andikat dennis
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    If you stretch the figure vertically by a factor of $a/b$ (keeping the $x$-axis fixed), you'll get a circle of radius $a$. The stretched yellow area will be $a/b$-times the original yellow area. The (slightly-)tricky part is determining the starting and ending angles in the stretched figure, but once you have them, the area of a circular segment is straightforward to compute. – Blue Aug 03 '13 at 04:37
  • I got this link: http://keisan.casio.com/exec/system/1343722709 but don't know hat is the r(delta) in the formula – andikat dennis Aug 03 '13 at 05:11

2 Answers2

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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.

Ben Grossmann
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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 $$\left|\int_{a\cos\phi}^{a\cos\delta}ydx\right|=\left|\int_{a\cos\phi}^{a\cos\delta}b\left(\sqrt{1-\frac{x^2}{a^2}}\right)dx\right|=\frac ba\left|\int_{a\cos\phi}^{a\cos\delta}\sqrt{a^2-x^2}dx\right|$$

lab bhattacharjee
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  • @AndréNicolas, not sure if you have seen the edited version? We need $\int_{x_1}^{x_2} ydx$ where $x_1=a\cos\phi$ where $\tan\phi=\frac ab\tan\alpha,$ where $\alpha$ is known – lab bhattacharjee Aug 03 '13 at 04:42