Moment of Inertia Integral

The integral evaluates to:

$I_{xx}=\frac{\rho hwd}{12}(w^2+d^2)$

or equivalently, if $M=\rho \cdot h\cdot w\cdot d$ (total mass),

$I_{xx}=\frac{M}{12}(w^2+d^2)$

Step-by-step evaluation

The integral is:

$I_{xx}={\int }_{-h/2}^{h/2}{\int }_{-w/2}^{w/2}{\int }_{-d/2}^{d/2}(y^2+z^2)\rho dxdydz$

Assuming uniform (constant) density $\rho$.

1. Integrate with respect to $x$ (the integrand does not depend on $x$):

${\int }_{-h/2}^{h/2}dx=h$

So the integral simplifies to:

$I_{xx}=\rho h{\int }_{-w/2}^{w/2}{\int }_{-d/2}^{d/2}(y^2+z^2)dydz$

2. Split the integrand:

$I_{xx}=\rho h(I_1+I_2)$

where

$I_1={\int }_{-w/2}^{w/2}{\int }_{-d/2}^{d/2}{y}^{2}dydz,I_2={\int }_{-w/2}^{w/2}{\int }_{-d/2}^{d/2}{z}^{2}dydz$

3. Evaluate $I_1$:

   $y^2$ is independent of $z$, and ${\int }_{-d/2}^{d/2}dz=d$, so

$I_1=d{\int }_{-w/2}^{w/2}{y}^{2}dy$

Since $y^2$ is even:

${\int }_{-w/2}^{w/2}{y}^{2}dy=2{\int }_0^{w/2}{y}^{2}dy=2{\left[\frac{y^3}{3}\right]}_0^{w/2}=2\cdot \frac{(w/2{)}^3}{3}=2\cdot \frac{w^3}{24}=\frac{w^3}{12}$

Thus, $I_1=d\cdot \frac{w^3}{12}$, and the first term is $\rho h\cdot d\cdot \frac{w^3}{12}=\rho \frac{h{w}^{3}d}{12}$.

4. Evaluate $I_2$:

   $z^2$ is independent of $y$, and ${\int }_{-w/2}^{w/2}dy=w$, so

$I_2=w{\int }_{-d/2}^{d/2}{z}^{2}dz=w\cdot \frac{d^3}{12}$

(by the same calculation as above). The second term is $\rho h\cdot w\cdot \frac{d^3}{12}=\rho \frac{hw{d}^3}{12}$.

5. Combine:

$I_{xx}=\rho h(\frac{w^{3}d}{12}+\frac{w{d}^3}{12})=\frac{\rho hwd}{12}(w^2+d^2)$

This is the standard formula for the moment of inertia of a rectangular prism (box) of uniform density about its central axis parallel to the length $h$.

Notes:

• The symmetry of the limits around zero made the odd-powered terms vanish automatically.
• If density $\rho$ is not constant, you cannot pull it out and would need to keep it inside the integral.
• You can verify this using the parallel axis theorem or known formulas for rectangular laminas.