# [SOLVED] How to find using integration, the area of a circular sector with an angle,x

• Jan 1st 2010, 01:06 PM
Aerospank
[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 PM
skeeter
Quote:

Originally Posted by Aerospank
Equation of the circle is x^2 + y^2 = r^2

and the sector's angle = x

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 PM
Aerospank
Quote:

Originally Posted by skeeter
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 ...

$\displaystyle A = \int_0^{\theta} \frac{1}{2} r^2 \, d\theta$

Thanks so much! but can you find the area derived from the equation of the circle?

Thanks.
• Jan 2nd 2010, 04:36 AM
HallsofIvy
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, $\displaystyle \frac{\theta}{2\pi}(\pi r^2)= \frac{\theta r^2}{2}$.