Statical moment of area, Q, for Semi-Circle

[lizzyb]

Question Exactly from Text
Determine the location e of the shear center, pint O, for the thin-walled member having the cross section shown.

http://img268.imageshack.us/img268/8732/scannedimage058.jpg

Work Done So Far
I'd like to find the statistical moment of area, Q, for a semi-circle in general; using this I should be able to generate Q = Q(R_outer) - Q(R_inner); I've been provided with such a Q as a hint but have failed to produce this value on my own ( 1/3 * (R_outer^3 - R_inner^3)) sin (theta) ).

I begin by determining the area A' by noting that its:
(area of circle)/4 - (area of circle in proportion to theta) - (triangle underneath shaded area)
http://img268.imageshack.us/img268/34/scannedimage059.jpg
I took its derivative and use C for theta:
http://img268.imageshack.us/img268/2466/scannedimage060.jpg

I used a calculator for integration.

What am I doing wrong?

 

[nvn]

lizzyb: Your integral looks correct. But I didn't understand yet why you are integrating from zero to theta. Maybe I'm missing something, but shouldn't that be integrated from theta to 0.5*pi?

I am currently getting Q(theta) = 0.3333(r_outer^3 - r_inner^3)*cos(theta)^3. I currently have no idea how they obtained that hint value for Q; so far, it does not look correct. Are you relatively certain the hint value is correct?

 

[lizzyb]

Its correct in that I dutifully copied the hint value but perhaps someone somewhere messed up; I don't see how they got it.

The hints assign r_i = r - t/2 and r_o as r + t/2.

 

[nvn]

lizzyb: Did you uncover any other information regarding why they claim Q is that hint value you listed in post 1?

 

[lizzyb]

This is from the solutions:
http://img195.imageshack.us/img195/9210/scannedimage061.jpg
http://img188.imageshack.us/img188/1713/scannedimage062.jpg