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

Ix=112m(a2+b2)Iy=112m(b2+c2)Iz=112m(a2+c2)\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}


Answered 9 months ago
Answered 9 months ago
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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., ρ=1\rho=1.

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

c2xc2-\frac{c}{2}\leq x \leq \frac{c}{2}

a2ya2-\frac{a}{2}\leq y\leq \frac{a}{2}

b2zb2-\frac{b}{2}\leq z\leq \frac{b}{2}


m=Qρ(x,y,z) dV=c/2c/2a/2a/2b/2b/2dz dy dx\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:


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