Question

Find the moment of inertia about the z -axis of the solid.

Short answer

The moment of inertia about the z –axis is: \(\frac{{{\rm{\pi kh}}{{\rm{a}}^{\rm{4}}}}}{{\rm{2}}}\).

Step-by-step solution

Concept Introduction

Triple integrals are the three-dimensional equivalents of double integrals. They're a way to add up an unlimited number of minuscule quantities connected with points in a three-dimensional space.

Write cylindrical coordinates value

The coordinates of a cylinder

\({\rm{(r,\theta ,z)}} \in {\rm{E}}\mid {\rm{0}} \le {\rm{r}} \le {\rm{a,0}} \le {\rm{\theta }} \le {\rm{2\pi ,0}} \le {\rm{z}} \le {\rm{h}}\)

Find the moment of inertia about the z –axis

\(\begin{aligned}{{I}_{z}}&=\iint_{E}{\left( {{x}^{2}}+{{y}^{2}} \right)}\rho (x,y,z)dV \\ & =k\int_{0}^{2\pi }{\int_{0}^{h}{\int_{0}^{a}{\left( {{r}^{2}} \right)}}}(r)drdzd\theta \\ & =k\int_{0}^{2\pi }{\int_{0}^{h}{\int_{0}^{a}{{{r}^{3}}}}}drdzd\theta \\ & =k\left( \int_{0}^{2\pi }{d}\theta \right)\left( \int_{0}^{h}{d}z \right)\left( \int_{0}^{a}{{{r}^{3}}}dr \right) \\ & =k(\theta )_{0}^{2\pi }(z)_{0}^{h}\left( \frac{{{r}^{4}}}{4} \right)_{0}^{a} \\ & =k(2\pi )(h)\left( \frac{{{a}^{4}}}{4} \right) \\ & =\frac{\pi kh{{a}^{4}}}{2} \end{aligned}\)

Therefore, the moment of inertia about the z –axis is \(\frac{{{\rm{\pi kh}}{{\rm{a}}^{\rm{4}}}}}{{\rm{2}}}\).