Circular Sector

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A wedge obtained by taking a portion of a disk with central angle theta<pi radians (180 degrees), illustrated above as the shaded region. A sector of pi radians would be a semicircle. Let R be the radius of the circle, a the chord length, s the arc length, h the sagitta (height of the arced portion), and r the apothem (height of the triangular portion). Then

R=h+r
(1)
s=Rtheta
(2)
r=Rcos(1/2theta)
(3)
=1/2acot(1/2theta)
(4)
=1/2sqrt(4R^2-a^2)
(5)
a=2Rsin(1/2theta)
(6)
=2rtan(1/2theta)
(7)
=2sqrt(R^2-r^2)
(8)
=2sqrt(h(2R-h)).
(9)

The angle theta obeys the relationships

theta=s/R
(10)
=2cos^(-1)(r/R)
(11)
=2tan^(-1)(a/(2r))
(12)
=2sin^(-1)(a/(2R)).
(13)

The area of the sector is

A=1/2Rs
(14)
=1/2R^2theta
(15)

(Beyer 1987). The area can also be found by direct integration as

 int_(-Rsin(theta/2))^(Rsin(theta/2))int_(|x|cot(theta/2))^(sqrt(R^2-x^2))dydx.
(16)

It follows that the weighted mean of the y is

<y>=int_(-Rsin(theta/2))^(Rsin(theta/2))int_(|x|cot(theta/2))^(sqrt(R^2-x^2))ydydx
(17)
=2/3R^3sin(1/2theta),
(18)

so the geometric centroid of the circular sector is

y^_=(<y>)/A
(19)
=(4Rsin(1/2theta))/(3theta)
(20)
=2/3Rsinc(1/2theta)
(21)

(Gearhart and Schulz 1990). Checking shows that this obeys the proper limits y^_=4R/(3pi) for a semicircle (theta=pi) and y^_->2R/3 for an isosceles triangle (theta->0).

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