Solve the problem.
-Find the moment of inertia about the $z$-axis of a region of constant density $\delta$ enclosed by the paraboloid $z = 16 - x ^ { 2 } - y ^ { 2 }$ and the $x y$-plane.
A) $\frac { 8 } { 3 } \delta \pi$
B) $\frac { 2048 } { 3 } \delta \pi$
C) $\frac { 128 } { 3 } \delta \pi$
D) $\frac { 32 } { 3 } \delta \pi$

Question

Quizplus · Accepted Answer

We can use the formula for moment of inertia of a solid about the $z$-axis:
$$I_z = \iint_R r^2 dm$$
where $R$ is the region enclosed by the solid and $dm = \delta dV$ is a small mass element. We can express $dV$ in terms of $dxdydz$ and make the change of variables $z = 16 - x^2 - y^2 \Rightarrow dz = -2x dx - 2y dy$.
Then the integral becomes:
$$I_z = \delta \iiint_V (x^2 + y^2) (-2x dx - 2y dy dz) = 2\delta \iint_R (x^2 + y^2) (16 - x^2 - y^2) dxdy$$
Using polar coordinates $x = r\cos\theta$, $y = r\sin\theta$, we have:
\begin{align*}
I_z &= 2\delta \int_0^{2\pi} \int_0^4 r^2(16-r^2) r dr d\theta \
&= 2\delta \int_0^{2\pi} \left(\int_0^4 16r^2 dr - \int_0^4 r^4 dr\right) d\theta \
&= 2\delta \int_0^{2\pi} \left(\frac{64}{3} - \frac{1024}{5}\right) d\theta \
&= \frac{2048}{3} \delta \pi
\end{align*}
Therefore, the answer is $\boxed{\textbf{(B)}}$.

Solve the Problem zzz -Axis of a Region of Constant Density δ\deltaδ

Solve the Problem $z$ -Axis of a Region of Constant Density $\delta$