problem with sector area

[jalcace]

Hi guys!
I have recently run into a geometry problem that I find myself unable to solve it. I need to find the area of a circular segment in terms of c and D (see the file).



I think that I need to use the cosine theoreme to find the subtended angle θ but to do that I need to find the s distance (see picture) and that's where I get stuck



Thank you so much!

 

[skeeter]

[MATH]\theta = 2\arccos\left(\dfrac{D-2c}{D}\right)[/MATH]
[MATH]A = \dfrac{\theta}{2\pi} \cdot \pi\left(\dfrac{D}{2}\right)^2 - \dfrac{1}{2}\left(\dfrac{D}{2}\right)^2 \sin{\theta}[/MATH]

 

[pka]

This link is very useful for circular segments.

 

[HallsofIvy]

Setting up a coordinate system with origin at the center of the circle and y axis perpendicular to the bottom of the shaded region we can write the circle as \(\displaystyle x^2+ y^2= \frac{D^2}{4}\). Further the base of the shaded region is y= D/2-C. That line and circle intersect where \(\displaystyle x^2+(D/2- C)^2= x^2+ D^2/4- DC+ C^2= D^2/4\) so \(\displaystyle x^2= DC- C^2\), \(\displaystyle x=\pm\sqrt{DC- C^2}\).

The area of the shaded region is \(\displaystyle \int_{-\sqrt{DC- C^2}}^{\sqrt{DC- C^2}} \sqrt{D^2/4- x^2}- \frac{D^2}{4} dx\)\(\displaystyle = \int_{-\sqrt{DC- C^2}}^{\sqrt{DC- C^2}} \sqrt{D^2/4- x^2}dx- \int_{-\sqrt{DC- C^2}}^{\sqrt{DC- C^2}} -\frac{D^2}{4}dx\).

Use a trig substitution for the first integral. The second is just \(\displaystyle -\frac{D^2}{2}\sqrt{DC- C^2}\).

 

[jonel]

You were almost there! The sector generated by the arc is easily calculated, but this needs subtracting the area that isosceles triangle and the height is (D/2 - c).

Jonel