Question

If the solid of part (a) has constant density k, find its moment of inertia about the z-axis.

Short answer

The moment of inertia of the solid about the -axis is\(\frac{{4k\pi }}{{15}}\left( {{a^2} + {b^2}} \right)(abc){\rm{.\;}}\)

Step-by-step solution

Formula

Formula for the Jacobian

\(\frac{{\partial (x,y)}}{{\partial (u,v)}} = \left| {\begin{array}{*{20}{l}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\partial y}&{\partial y}\end{array}} \right|\)

Using the Jacobian theorem and transform the integral

The moment of inertia of the solid about theaxis will be

Now we use the transformation

\(\begin{array}{l}\;y = bv\\z = cw\end{array}\) \(\)

Hence the Jacobian \(\begin{array}{l}J = \left| {\frac{{\partial (x,y,z)}}{{\partial (u,v,w)}}} \right|\\ = abc\end{array}\)
So we now have,

Again substituting

\(\begin{array}{l}v = r\sin \theta sin\phi ,\\w = r\cos {\rm{\backslash }}\phi .\end{array}\)

Hence the Jacobian

\(\begin{array}{l}J = \left| {\frac{{\partial (u,v,w)}}{{\partial (r,\theta ,\phi )}}} \right|\\ = {r^2}\sin \phi \end{array}\)

Simplify the final calculation

Thus, the moment of inertia of the solid about the -axis is\(\frac{{4k\pi }}{{15}}\left( {{a^2} + {b^2}} \right)(abc){\rm{.\;}}\)