MOI and radius of gyration between two parabolas

Setting up: the area lies between y₁ = K₁x² (lower) and y₂ = K₂√x (upper). Both curves pass through the origin and through the point (a, b), so K₁ = b/a², K₂ = b/√a. The figure of the second paper gives a = 80 cm and b = 30 cm. I derive the general formulae and then put in these numbers.

Area

A = ∫₀ᵃ (y₂ − y₁) dx = ∫₀ᵃ (K₂ x^½ − K₁ x²) dx = (2/3) K₂ a^{3/2} − K₁ a³/3 = (2/3)ab − (1/3)ab = ab/3

Moment of inertia about the x axis

Strip of width dx, between y₁ and y₂: dI_x = (y₂³ − y₁³) dx / 3

I_x = (1/3) ∫ (K2³ x^(3/2) − K1³ x^6) dx
    = (1/3) [ K2³ (2/5) a^(5/2) − K1³ a^7 / 7 ]
    = (1/3) [ (2/5) a b³ − (1/7) a b³ ]
    = (1/3)(9/35) a b³ = 3 a b³ / 35

Moment of inertia about the y axis

dI_y = x² (y₂ − y₁) dx

I_y = ∫ (K2 x^(5/2) − K1 x^4) dx
    = K2 (2/7) a^(7/2) − K1 a^5 / 5
    = (2/7) a³ b − (1/5) a³ b = 3 a³ b / 35

Radii of gyration

k_x = √(I_x/A) = √( (3ab³/35) / (ab/3) ) = √(9/35) · b = 0.507 b k_y = √(I_y/A) = √(9/35) · a = 0.507 a

Numbers (a = 80 cm, b = 30 cm)

A   = 80 × 30 / 3                 = 800 cm²
I_x = 3 × 80 × 30³ / 35           = 185 143 cm⁴  (1.851 × 10⁵ cm⁴)
I_y = 3 × 80³ × 30 / 35           = 1 316 571 cm⁴ (1.317 × 10⁶ cm⁴)
k_x = 0.5071 × 30                 = 15.21 cm
k_y = 0.5071 × 80                 = 40.57 cm

Answer: I_x = 3ab³/35 = 1.85 × 10⁵ cm⁴, I_y = 3a³b/35 = 1.32 × 10⁶ cm⁴, k_x = 15.21 cm, k_y = 40.57 cm.