Moment of Inertia / Mass Element proof

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SUMMARY

The discussion focuses on calculating the mass and moment of inertia of a thin square plate with a density function p(x,y) = xy. The mass M of the plate is conclusively derived as M = (1/4)a^4 through double integration over the defined area. The moment of inertia I is defined as I = ∫∫ r² p dx dy, where r is the distance from the origin. The challenge lies in expressing r² in terms of x and y to complete the proof that I = Ma².

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Homework Statement



A thin square plate of side a has one corner at the origin and two sides along the positive x and y axes. If the density of the plate is given by p(x,y) = xy show that its mass is M=(1/4)a^4
If the distance of the mass element dM = pdS from the origin is r the moment of inertia of the plate is I= integral [r^2 (pdS)] where S is the surface of the plate. Prove that I=Ma^2

Homework Equations



The Attempt at a Solution



I've proved the first half by showing

M = pdV and taking a small mass element, then integrating

(0->a)int(dx) (0->a)int(dy) xy = (a^2)/2 * (a^2)/2 = (1/4)a^4

However I can't seem to get started with the second part...
 
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Just use the given definition:

[tex]I=\int_0^a \int_0^a r^2\rho dxdy[/tex].

What is r^2 in terms of x and y?
 

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