Find the moment of inertia of a homogeneous rectangle

Find the moment of inertia of a homogeneous rectangle with sides $a,b$ ($\varrho=1$) with respect to the intersection of its diagonals (the center).

Step 1 · Setup

With $D:-\frac a2\le x\le\frac a2,\,-\frac b2\le y\le\frac b2$: $I_0=\iint_D(x^2+y^2)\,dA=I_x+I_y$.

Step 2 · Compute $I_y,I_x$

$$ I_y = \int_{-b/2}^{b/2}dy\int_{-a/2}^{a/2}x^2\,dx = b\cdot\frac{a^3}{12} = \frac{a^3b}{12}, \qquad I_x = \frac{ab^3}{12}\ \text{(by symmetry)}. $$

Step 3 · Combine

$$ I_0 = I_x+I_y = \frac{ab\big(a^2+b^2\big)}{12}. $$

With mass $m=ab$: $I_0=\dfrac{m(a^2+b^2)}{12}$.

Additional Notes

• This derivation matches the source exactly. For a square ($a=b$): $I_0=\dfrac{ma^2}6$, the standard result.
• Verified numerically (e.g. $a=2.1,\,b=1.3$): $I_0\approx1.3878$, matching $\dfrac{ab(a^2+b^2)}{12}$ exactly.