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# 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(3 a^2+L^2\right) \\ & I_y=\frac{1}{2} m a^2 \\ & I_z=\frac{1}{12} m\left(3 a^2+L^2\right) \end{aligned}

Solution

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First 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:

$-a\leq x \leq a$

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

$x^2+z^2=a\Rightarrow z=\pm \sqrt{a^2-x^2}$

$-\sqrt{a^2-x^2}\leq z\leq \sqrt{a^2-x^2}$

Now:

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

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

$\color{#4257b2}m=L\pi a^2$

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