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

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    [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
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  2. #2
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    Quote Originally Posted by Aerospank View Post
    Equation of the circle is x^2 + y^2 = r^2

    and the sector's angle = x
    you're mixing variables ...

    x^2+y^2=r^2 is a circle of radius 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 \theta is

    A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{r^2 \theta}{2}<br />

    as an integral ...

    A = \int_0^{\theta} \frac{1}{2} r^2 \, d\theta
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    Quote Originally Posted by skeeter View Post
    you're mixing variables ...

    x^2+y^2=r^2 is a circle of radius 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 \theta is

    A = \frac{\theta}{2\pi} \cdot \pi r^2 = \frac{r^2 \theta}{2}<br />

    as an integral ...

    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.
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    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 y= tan(\theta)x from x=0 to x= r cos(\theta). Part of the area, then, is given by \int_0^{r cos(\theta)} tan(\theta)x dx which is easy since r and \theta are constants. But then you have to find the area of the portion under the circular arc, y= \sqrt{r^2- x^2}. That is \int_{r cos(\theta)}^r \sqrt{r^2- x^2}dx. The total area of the sector is the sum of those integrals:
    \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 \theta cuts \frac{\theta}{2\pi} of the entire circle, its area is that same fraction of the entire area of the circle, \frac{\theta}{2\pi}(\pi r^2)= \frac{\theta r^2}{2}.
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