## Related questions with answers

Verify the moments of inertia for the solid of uniform density. Use a graphing utility to evaluate the triple integrals.

$\begin{aligned} & I_x=\frac{1}{12} m\left(a^2+b^2\right) \\ & I_y=\frac{1}{12} m\left(b^2+c^2\right) \\ & I_z=\frac{1}{12} m\left(a^2+c^2\right)\end{aligned}$

Solution

VerifiedFirst we will set up the integral for the mass of the solid. It is assumed that the solid has uniform mass density, i.e., $\rho=1$.

From the picture we can see that the limits for integration are the following:

$-\frac{c}{2}\leq x \leq \frac{c}{2}$

$-\frac{a}{2}\leq y\leq \frac{a}{2}$

$-\frac{b}{2}\leq z\leq \frac{b}{2}$

Now:

$\begin{align*} m&=\int \kern -5pt \int \kern -5pt \int _{Q} \rho(x,y,z) \ dV=\int_{-c/2}^{c/2} \kern -3pt \int_{-a/2}^{a/2} \kern -3pt \int_{-b/2}^{b/2} dz\ dy\ dx \end{align*}$

Using computer algebra system to evaluate the triple integral above we obtain:

$\color{#4257b2}m=abc$

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