Parallel axis theorem; centroid of area under y = Kx²

Parallel axis theorem

Statement: the moment of inertia of an area about any axis equals its moment of inertia about a parallel axis through its centroid plus the product of the area and the square of the distance between the two axes.

I = I_G + A d²

Proof: let G be the centroid of the area A, and let X′ be the centroidal axis. Take a parallel axis X at distance d from X′. For a small element dA at distance y′ from X′, its distance from X is y = y′ + d.

I_X = ∫ y² dA = ∫ (y′ + d)² dA
    = ∫ y′² dA + 2d ∫ y′ dA + d² ∫ dA

• ∫ y′² dA = I_G, the moment of inertia about the centroidal axis.
• ∫ y′ dA = A·ȳ′ = 0, because the distance of the centroid from the centroidal axis is zero.
• ∫ dA = A.

So I_X = I_G + A d². Similarly, I_Y = I_Gy + A d_x². Since A d² is never negative, the moment of inertia about the centroidal axis is the least of the moments of inertia about all parallel axes.

Centroid of the area under y = Kx²

Reading of the figure: the shaded area is under the curve y = Kx² between x = 0 and x = a, where the height is h, so K = h/a². Use a vertical strip of height y = Kx² and width dx.

Area A = ∫₀ᵃ Kx² dx = K a³/3 = ah/3

x̄ ∫ x dA = ∫₀ᵃ x · Kx² dx = K a⁴/4 = a²h/4 x̄ = (a²h/4)/(ah/3) = 3a/4

ȳ (strip centroid at y/2) ∫ ȳ dA = ∫₀ᵃ ½ (Kx²)(Kx²) dx = K² a⁵/10 = a h²/10 ȳ = (a h²/10)/(ah/3) = 3h/10

Answer: x̄ = 3a/4, ȳ = 3h/10 (from O).