Ix = ∫ y2dA
dA = ydx
y = k/x
To get rid of k, use the fact that y(a) = a
So,
a = k/a ---------> k = a2
Substitute this a2 for k and k/x for y to get
dA = (a2/x)dx
Also, y2 = (a2/x)2
Therefore, the equation for the moment of inertia with respect to the x axis is:
Ix = ∫ (a2/x)2(a2/x)dx integrated from a to 2a
= ∫ (a2/x)3dx integrated from a to 2a
= ∫ (a6/x3)dx integrated from a to 2a
= a6/x2(1/-2) evaluated from a to 2a
= (a6/4a2)(1/-2) - (a6/a2)(1/-2)
= (-1/8)(a4) + (1/2)(a4)
= (a4)(4/8 - 1/8)
= (3/8)a4