Moment of Inertia of Plate: Integration Proves 1/3ma^2+b^2

[astr0]

Show by integration that the moment of inertia of the plate about an axis that is perpendicular to the plate and passes through one corner is \frac{1}{3}m(a^{2}+b^{2})

I'm not sure at all how to approach this problem. I know that the moment of inertia is \int r^{2}dm but how do I use that in this instance?

Any help is appreciated.

 

[ehild]

Write up both r² and dm in terms of x,y and integrate with respect to x and y.

ehild

 

[astr0]

Should I use two separate integrals, then add them together?

 

[ehild]

It is a double integral of ρr²=ρ(x²+y²).
First integrate both terms with respect to x from 0 to a, taking y as constant. Then integrate the result with respect to y from 0 to b.

ehild