Moment of inertia of thin plate inside QI and x^2+4y^2=4

[chengeto]

Consider a thin plate of constant density which occupies the region in the first quadrant inside the curve:

\(\displaystyle x^2+4y^2=4\)

Find moment of inertia about line y=2

Attempt to solution:

\(\displaystyle y=\frac{\sqrt{4-x^2}}{2}\)

\(\displaystyle I(y=2)=\frac{\rho}{3}\int_0^2(\frac{\sqrt{4-x^2}}{2}-2)^3-(0-2)^3\)

\(\displaystyle I(y=2)=\frac{\rho}{3}\int_0^2(\frac{\sqrt{4-x^2}}{2}-2)^3+\frac{\rho}{3}\int_0^28dx\)

\(\displaystyle I(y=2)=\frac{\rho}{3}\int_0^2(\frac{\sqrt{4-x^2}}{2}-2)^3\)

I am stuck here how do l integrate this this thing ?

 

[BigGlenntheHeavy]

Let x = 2sin(theta), then expand the cubic and then use Wallis's formula for the above integral.