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MOI of common area between y = K√x and y = mx

Short answer

Common area = ab/6. Using strip method, moment of inertia about x-axis is ab³/20, about y-axis is a³b/28.

Determine the moment of inertia of the common area as shown in figure below about x and y axis. [Figure: area between the curve y2 = K√x and the straight line y1 = mx from the origin O to x = a, y = b.]

Setting up: the common area lies between the curve y₂ = K√x (upper) and the straight line y₁ = mx (lower), as labelled in the figure. Both pass through O and through the point (a, b), so K = b/√a, m = b/a.

Area

A = ∫₀ᵃ (K√x − mx) dx = (2/3) K a^{3/2} − m a²/2 = (2/3)ab − ab/2 = ab/6

Moment of inertia about the x axis

Strip of width dx between the two curves: dI_x = (y₂³ − y₁³) dx / 3

I_x = (1/3) ∫0^a (K³ x^(3/2) − m³ x³) dx
    = (1/3) [ K³ (2/5) a^(5/2) − m³ a⁴/4 ]
    = (1/3) [ (2/5) a b³ − (1/4) a b³ ]
    = (1/3)(3/20) a b³ = a b³ / 20

Moment of inertia about the y axis

dI_y = x² (y₂ − y₁) dx

I_y = ∫0^a (K x^(5/2) − m x³) dx
    = K (2/7) a^(7/2) − m a⁴/4
    = (2/7) a³ b − (1/4) a³ b = a³ b / 28

(Radii of gyration: k_x = √(I_x/A) = √(3/10) b = 0.548 b, k_y = √(I_y/A) = √(3/14) a = 0.463 a.)

Answer: I_x = ab³/20 and I_y = a³b/28.

✓ Answer checked by Claude Opus 5.5 (high effort)

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