Finding Area Moment of Inertia for Axis 33 to X-Axis

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To find the area moment of inertia about an axis 33 degrees to the x-axis, the relevant equation is I_φ = 1/2(I_{xx} + I_{yy}) + 1/2(I_{xx} - I_{yy})cos(2φ) - I_{xy}sin(2φ). The user calculated I_{xx} and noted that due to symmetry, I_{xx} equals I_{yy} and I_{xy} equals zero. This leads to the conclusion that I_φ simplifies to I_{xx}. The reasoning appears sound, confirming that the approach to finding the moment of inertia is correct.
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Homework Statement


I have to find the area moment of inertia about an axis 33 to the x-axis
http://img227.imageshack.us/img227/2392/shape.jpg

Homework Equations


I_\phi=\frac{1}{2}(I_{xx}+I_{yy})+\frac{1}{2}(I{xx}-I_{yy})cos2\phi - I_{xy}sin2\phi


The Attempt at a Solution



I found Ixx, but since it is symmetrical, Ixx=Iyy, and since it is symmetrical about the x and y axes, Ixy=0

So the I about any axis is just the 1/2(Ixx+Iyy) = Ixx.

Is my thinking correct, or am I mistaking it?
 
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Looks ok to me.
 

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