$$\text { Find the required quantity given the wire that lies along the curve } r \text { and has density } \delta .$$
-Moment of inertia $\mathrm { I } _ { \mathrm { Z } }$ about the $\mathrm { z }$-axis, where $r ( t ) = ( 4 \sin 3 t ) i + ( 4 \cos 3 t ) j + e ^ { 2 t } k , 0 \leq t \leq 1 ; \delta ( x , y , z ) = z ^ { 2 }$
A) $I _ { Z } = \frac { 2 } { 3 } \left[ \left( 144 + 4 e ^ { 4 } \right) ^ { 3 / 2 } - ( 144 + 4 ) ^ { 3 / 2 } \right]$
B) $\mathrm { I } _ { \mathrm { Z } } = \frac { 8 } { 3 } \left( \mathrm { e } ^ { 4 } - 1 \right)$
C) $\mathrm { I } _ { \mathrm { Z } } = \frac { 4 } { 3 }$
D) $\mathrm { I } _ { \mathrm { Z } } = \frac { 2 } { 3 }$

Question

Quizplus · Accepted Answer

The correct answer is A because the moment of inertia about the z-axis is given by the integral of the square of the distance from the z-axis to the wire, multiplied by the density of the wire. The distance from the z-axis to the wire is given by the magnitude of the vector r(t), which is $\sqrt{16\sin^2 3t + 16\cos^2 3t + e^{4t}} = \sqrt{16 + e^{4t}}$. The density of the wire is given by $\delta(x, y, z) = z^2 = e^{4t}$. So the moment of inertia is given by the integral of $\left(\sqrt{16 + e^{4t}}\right)^2 e^{4t}$ from $t = 0$ to $t = 1$. This integral evaluates to $\frac{2}{3}\left[\left(144 + 4e^4\right)^{3/2} - (144 + 4)^{3/2}\right]$.

Find the required quantity given the wire that lies along the curve r and has density δ.\text { Find the required quantity given the wire that lies along the curve } r \text { and has density } \delta . Find the required quantity given the wire that lies along the curve r and has density δ.

$\text { Find the required quantity given the wire that lies along the curve } r \text { and has density } \delta .$