$$\text { Find the required quantity given the wire that lies along the curve } r \text { and has density } \delta .$$
-Moment of inertia $I _ { x }$ about the $x$-axis, where $\mathbf { r } ( t ) = \frac { 10 } { 3 } \sqrt { 2 } t 3 / 2 \mathbf { i } + ( 5 t \cos t ) j + ( 5 t \sin t ) \mathbf { k } , 0 \leq t \leq 1$; $\delta ( x , y , z ) = \frac { 9 } { y ^ { 2 } + z ^ { 2 } }$
A) $\mathrm { I } _ { \mathrm { X } } = 45$
B) $\mathrm { I } _ { \mathrm { X } } = 0$
C) $I _ { X } = 90$
D) $\mathrm { I } _ { \mathrm { x } } = \frac { 135 } { 2 }$

Question

Quizplus · Accepted Answer

The moment of inertia about the x-axis is given by the formula,
$$I_x = \int_{c} \delta(x,y,z) \big(y^2+z^2\big)~ds$$
where $c$ is the curve in question, $\delta$ is the density, and $ds$ is the element of arc length along $c$. 
The curve $c$ is given by the vector-valued function $\mathbf{r}(t)$, $0\leq t\leq 1$, so we need to write the integral in terms of $t$ using the parameterization $\mathbf{r}(t)$. Thus, 
$$I_x = \int_{0}^{1} \delta\Big(\frac{10}{3}\sqrt{2}t^{3/2}, 5t\cos t, 5t\sin t\Big) \big(25t^2\big)~dt$$
Next, we substitute $\delta(x,y,z)=\frac{9}{y^2+z^2}$ and simplify, 
\begin{align*}
I_x &= \int_{0}^{1} \frac{9}{25t^2\cos^2 t + 25 t ^2\sin^2 t} (25t^2)~dt \
&= \int_{0}^{1} \frac{9}{25t^2}~dt \
&= 9\int_{0}^{1} \frac{dt}{t^2} \
&= 9\left[\frac{-1}{t}ight]_{0}^{1} \
&= \boxed{\frac{135}{2}}
\end{align*}

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.

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