Question

(b) Find the moment of inertia of the solid in part (a) about a diameter of its base.

Short answer

The moment of inertia is \( = \frac{{4\pi {a^5}K}}{{15}}\)

Step-by-step solution

Define spherical coordinates

Three coordinates in three-dimensional space that define the location of a point The length of its radius vector r, the angle between the vertical plane containing this vector and the x-axis, and the angle between this vector and the horizontal x–y plane are the three parameters.

Find the moment of inertia.

The moments of inertia about the three coordinate axes are

\(\begin{aligned} {I_x} &= \iiint_E {\left( {{y^2} + {z^2}} \right)}\rho (x,y,z)dV \hfill \\ {I_y} &= \iiint_E {\left( {{x^2} + {z^2}} \right)}\rho (x,y,z)dV \hfill \\ {I_z} &= \iiint_E {\left( {{x^2} + {y^2}} \right)}\rho (x,y,z)dV \hfill \\ \end{aligned}\)

To find \({I_Z}\)

Here we use spherical coordinates i.e.,

\(\begin{array}{l}x = r\sin \phi \cos \theta \\y = r\sin \phi \sin \theta \\z = r\cos \phi \end{array}\)

The description of the solid E in spherical coordinates is

\(E = \left\{ {(r,\theta ,\phi ):0 \le \theta \le 2\pi ,0 \le \phi \le \frac{\pi }{2},0 \le r \le a} \right\}\)

Find the density.

When\(\rho (x,y,z) = K,v\)

\(\begin{aligned}{I_z} &= \iiint_E {\left( {{x^2} + {y^2}} \right)}KdV \hfill \\&= K\int_0^{2\pi } {\int_0^{\frac{\pi }{2}} {\int_0^a {\left( {{x^2} + {y^2}} \right)} } } {r^2}\sin \phi drd\phi d\theta \hfill \\&= K\int_0^{2\pi } {\int_0^{\frac{\pi }{2}} {\int_0^a {\left( {{{(r\sin \phi \cos \theta )}^2} + {{(r\sin \phi \sin \theta )}^2}} \right)} } } {r^2}\sin \phi drd\phi d\theta \hfill \\ &= K\int_0^{2\pi } {\int_0^{\frac{\pi }{2}} {\int_0^a {{r^4}} } } {\sin ^3}\phi drd\phi d\theta \hfill \\ &= K\int_0^{2\pi } d \theta \int_0^a {{r^4}} dr\int_0^{\frac{4}{2}} {{{\sin }^3}} \phi d\phi \hfill \\ &= K * 2\pi * \frac{{{a^5}}}{5} * \int_0^{\frac{\pi }{2}} {\sin } \phi \left( {1 - {{\cos }^2}\phi } \right)d\phi \hfill \\ &= K * 2\pi * \frac{{{a^5}}}{5} * \int_0^1 {\left( {1 - {u^2}} \right)} du \hfill \\&= K * 2\pi * \frac{{{a^5}}}{5} * \left( {u - \frac{{{u^3}}}{3}} \right)_0^1 \hfill \\&= K * 2\pi * \frac{{{a^5}}}{5} * \left( {1 - \frac{1}{3}} \right) \hfill \\ &= \frac{{4\pi {a^5}K}}{{15}} \hfill \\ \end{aligned} \)

Thus, the moment of inertia is \( = \frac{{4\pi {a^5}K}}{{15}}\)