Use spherical coordinates to find the moment of inertia of the solid homogeneous hemisphere of radius $5$ and density 1 about a diameter of its base.
A) 205.13
B) $2616.67$
C) 195.22
D) 213.5
E) 198.08

Question

Quizplus · Accepted Answer

Let's denote by $R$ the radius of the hemisphere, $r$ the distance from the axis of rotation to any point inside the hemisphere, $ho$ the density of the hemisphere and $d$ the distance between the center of mass of the hemisphere and the plane that contains the base.

We can use the general formula for the moment of inertia of a solid with spherical coordinates 
$$I=\iiint_V(r^2\sin^3\varphiho)\, dV,$$
where $dV=r^2\sin\varphi \, dr\, d\varphi\, d	heta$ is the volume element in spherical coordinates.

Since the hemisphere has uniform density, we have $ho=1$. To compute the limits of integration for $r$, $	heta$, and $\varphi$, we use the representation of the hemisphere in spherical coordinates:
\begin{align*}
0&\leq r\leq R,\
0&\leq 	heta \leq 2\pi,\
0&\leq \varphi \leq \frac{\pi}{2}.
\end{align*}

We choose the z-axis to be along the axis of rotation and the xy-plane to contain the diameter of the base. 
The perpendicular distance from any point in the hemisphere to the xy-plane is given by $d=R\cos\varphi$. Therefore, the distance from the center of mass of the hemisphere to the xy-plane is
$$\bar{d}=\dfrac{1}{\frac{1}{V}\iiint_V ho \cdot z \, dV} \iiint_V ho \cdot z \, dV=\dfrac{3R}{8}.$$

Thus, the axis of rotation is displaced $\bar{d}$ from the plane through the center of the hemisphere containing the diameter. 
By the perpendicular axis theorem, the moment of inertia of the hemisphere about the desired diameter is 
$$I=\frac{2}{3}MR^2+\Delta I,$$
where $\Delta I$ is the moment of inertia of the hemisphere about an axis through its center of mass and parallel to the desired diameter. Since the hemisphere is symmetric with respect to any axis through its center of mass, we have $\Delta I = \frac{2}{5}MR^2$. Therefore, 
$$I=\frac{2}{3}MR^2+\frac{2}{5}MR^2=\frac{16}{15}MR^2.$$ 
Here $M$ is the mass of the hemisphere. We have 
$$M=ho V = ho\cdot \frac{2}{3}\pi R^3 = \frac{2}{3}\pi R^3.$$ 
Substituting this in the expression for $I$, we get 
$$I=\frac{16}{15}\cdot\frac{2}{3}\pi R^3\cdot R^2 = \frac{32}{45}\pi R^5.$$ 
Finally, substituting $R=5$, we get 
$$I=\frac{32}{45}\pi(5)^5 \approx \boxed{2616.67}.$$

Use Spherical Coordinates to Find the Moment of Inertia of the Solid