Equation of the circle is x^2 + y^2 = r^2

and the sector's angle = x

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- Jan 1st 2010, 01:06 PMAerospank[SOLVED] How to find using integration, the area of a circular sector with an angle,x
Equation of the circle is x^2 + y^2 = r^2

and the sector's angle = x - Jan 1st 2010, 02:19 PMskeeter
you're mixing variables ...

$\displaystyle x^2+y^2=r^2$ is a circle of radius $\displaystyle r$ centered at the origin.

x represents the x-coordinate of any point on the circle, so it cannot represent a central angle's measure.

area of a circular sector with central angle $\displaystyle \theta$ is

$\displaystyle A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{r^2 \theta}{2}

$

as an integral ...

$\displaystyle A = \int_0^{\theta} \frac{1}{2} r^2 \, d\theta$ - Jan 1st 2010, 07:53 PMAerospank
- Jan 2nd 2010, 04:36 AMHallsofIvy
That

**is**derived from the equation of the circle! But if you want to do it in the**hard**way, using Cartesian coordinates, you can without loss of generality, assume one edge of the sector lies on x axis, from x= 0 to x= r. The other then lies on the line $\displaystyle y= tan(\theta)x$ from x=0 to $\displaystyle x= r cos(\theta)$. Part of the area, then, is given by $\displaystyle \int_0^{r cos(\theta)} tan(\theta)x dx$ which is easy since r and $\displaystyle \theta$ are constants. But then you have to find the area of the portion under the circular arc, $\displaystyle y= \sqrt{r^2- x^2}$. That is $\displaystyle \int_{r cos(\theta)}^r \sqrt{r^2- x^2}dx$. The total area of the sector is the sum of those integrals:

$\displaystyle \int_0^{r cos(\theta)} tan(\theta)x dx+ \int_{r cos(\theta)}^r \sqrt{r^2- x^2}dx$.

I think using polar coordinates is a much better idea. Of course, as a check, since a central angle of $\displaystyle \theta$ cuts $\displaystyle \frac{\theta}{2\pi}$ of the entire circle, its area is that same fraction of the entire area of the circle, $\displaystyle \frac{\theta}{2\pi}(\pi r^2)= \frac{\theta r^2}{2}$.