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Found in: Page 730

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the moments of inertia for a cube of constant density K and side length L if one vertex is located at the origin and three edges lie along the coordinate axes.

The moments of inertia for a cube is $${{\rm{I}}_{\rm{x}}}{\rm{ = }}{{\rm{I}}_{\rm{y}}}{\rm{ = }}{{\rm{I}}_{\rm{z}}}{\rm{ = }}\frac{{{\rm{2k}}{{\rm{L}}^{\rm{5}}}}}{{\rm{3}}}$$.

See the step by step solution

## Step 1: 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.

## Step 2: Find the moments of inertia for a cube

Let us simplify,

\begin{aligned}{I_x} &= \iiint_E {\left( {{y^2} + {z^2}} \right)}\rho (x,y,z)dV \\ &= k\int_0^L {\int_0^L {\int_0^L {\left( {{y^2} + {z^2}} \right)} } } dxdydz \\ &= k\int_0^L {\int_0^L {\int_0^L {\left( {{y^2} + {z^2}} \right)} } } dxdydz \\ &= k\int_0^L {\int_0^L {\left( {{y^2}x + {z^2}x} \right)_0^L} } dydz \\ &= kL\int_0^L {\int_0^L {{y^2}} } + {z^2}dydz \\ &= kL\int_0^L {\left( {\frac{{{y^3}}}{3} + {z^2}y} \right)_0^L} dz \\ &= kL\int_0^L {\frac{{{L^3}}}{3}} + {z^2}Ldz \\ &= kL\left( {\frac{{z{L^3}}}{3} + \frac{{{z^3}L}}{3}}\right)_0^L \\ &= \frac{{2k{L^5}}}{3} \\ \end{aligned}

The required graph is:

Apply symmetry because we're dealing with a cube. As a result, we reach the following conclusion:

${{\rm{I}}_{\rm{x}}}{\rm{ = }}{{\rm{I}}_{\rm{y}}}{\rm{ = }}{{\rm{I}}_{\rm{z}}}{\rm{ = }}\frac{{{\rm{2k}}{{\rm{L}}^{\rm{5}}}}}{{\rm{3}}}$

Therefore, the moments of inertia for a cube is ${{\rm{I}}_{\rm{x}}}{\rm{ = }}{{\rm{I}}_{\rm{y}}}{\rm{ = }}{{\rm{I}}_{\rm{z}}}{\rm{ = }}\frac{{{\rm{2k}}{{\rm{L}}^{\rm{5}}}}}{{\rm{3}}}$.