Second moment of area of a hollow cylindrical arc

[NancyHadad]

Homework Statement

I need to find the second moment of area for a hollow cylindrical arc. When I searched the web, they had formulas for several shapes but I couldn't find this one The profile of my part is a little unusual as follows:
R: Outer Radius = 35.089"
r: Inner Radius = 35.050"
(ie the material is 1mm thick)
The angle of arc is only 10 degrees (so the arc length is only about 6 inches, ie the width of my part is only about 6 inches). The length of my part is 5'7", so the reason why I need the second moment of area, is because I'm trying to figure out how much deflection this very long, thin part will have in the middle, if it is supported on both ends.

Homework Equations

I know that, the second moment of area for hollow cylindrical cross section is:

(Pie/64) * (D4 - d4)

where D is the Ouside Diameter and d is the Inside Diameter.

The Attempt at a Solution

I found the second moment of area for half a hollow cylindrical cross section (long complicated equation that I don't want to enter in here but only variables in the equation are R and r (Outside and inside radii). I know I have to get arc angle or length in there somehow. Help!

 

[haruspex]

The easiest way is to use the parallel axis theorem "in reverse".
Start by finding the second moment of area about the diameter.
$\int_{-\theta/2}^{\theta/2}\int_r^R(r\cos(\theta))^2r.dr.d\theta=\frac 18(R^4-r^4)(\theta+\sin(\theta))$.
It remains to subtract the area multiplied by the square of the distance from its mass centre to the diameter.