(adsbygoogle = window.adsbygoogle || []).push({}); Area moment of inertia--circular cross section

From the bending beam calculation, the moment of inertia of the cross section with regard to a coplanor axis of rotation is used. If we have a circular "beam", the area moment of inertia of a circular disk of radius a about a diameter is [tex] I_d = \frac{\pi a^4}{4}[/tex] according to two separate references. I believe the integral involved can be generally stated as [tex]I_d = \int y^2 dA [/tex] if y is the distance to the diameter d perpendicular to y which diameter (as all diameters of a uniformly mass distributed disk) passes through the centroid of the disk.

Now my stab at actually evaluating this is to do a double integral in polar coordinates and long story short the only way I can come up with the agreed upon answer is [tex] \int_0^\pi\int_0^a r^2\; r\; \mathrm{dr}\;\mathrm{d\theta}[/tex] where [tex] r\;\mathrm{dr}\;\mathrm{d\theta}=dA[/tex] and [tex]r^2[/tex] is the distance to the origin (centroid)

If I were in a creative writing glass I might get a passing grade for this fudge but I really would like to understand what I am doing better than backing into an answer like this. Any help would truly be appreciated.

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# Area moment of inertia-circular cross section

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