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The area of a sector of a circle is the area of the triangle plus an additional portion which is $\int_{r cos\theta}^r \sqrt{r^2 - x^2} dx$

In order to integrate this, a trig substitution is used, $x =rsin\theta, dx = rcos\theta$. But by making that substitution the integrating limits would change from $\frac{\pi}{4}$ to $\frac{\pi}{2}$ since $r = rsin\theta$ and $sin^{-1}(1) = \frac{\pi}{2}$ and for the lower limit we would have $cos\theta = sin\theta$, which $\theta = \frac{\pi}{4}$

But that doesn't make it any easier to solve for the area formula. What is the proper derivation of the area of a sector using calculus?

theta
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  • I don't know if this is at all what you're looking for, but you might perhaps be interested in this question of mine, especially the answer of André Nicolas. – Chris Sep 09 '13 at 07:37
  • At second glance, it seems my reference is not immediately pertinent to your question.... My bad. – Chris Sep 09 '13 at 07:40

2 Answers2

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The angle $\theta$ is fixed, it is given to you.

When you are integrating $\sqrt{r^2-x^2}$ using a trig substitution, you must not use $\theta$, that's taken.

There are plenty of letters left, Greek if you like, let $x=\sin \phi$. Or maybe use $x=\sin t$. Then everything will work nicely.

Remark: This is a very time consuming way to find the area of a sector with angle $\theta$. For the area of the sector, if $\theta$ is given in radians, is$\dfrac{\theta}{2\pi}$ times the area of the circle.

That gives area $\dfrac{\theta}{2}r^2$.

André Nicolas
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  • Here is a derivation: http://www.vias.org/calculus/07_trigonometric_functions_09_01.html. But it doesn't show how the integral becomes that equation – theta Sep 09 '13 at 06:32
  • Do you mean how the integration is carried out? If you have trouble with that, I can add to the post. – André Nicolas Sep 09 '13 at 06:35
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Isn't it simpler to use polar coordinates? The area is then $\int_{\theta_{min}}^{\theta_{max}}\int_0^{r_{max}} J(r,\theta) \, dr d\theta$, where $J(r,\theta)$ is the Jacobian corresponding to a change from Cartesian coordinates $(x,y)$ to polar coordinates.

Kyzariel
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