Answer
$\mathop \smallint \limits_C^{} {x^{ - 1}}yz{\rm{d}}s \simeq 339.5587$
Work Step by Step
We have $f\left( {x,y,z} \right) = {x^{ - 1}}yz$ and the curve $C$ parametrized by ${\bf{r}}\left( t \right) = \left( {\ln t,t,{t^2}} \right)$ for $2 \le t \le 4$.
So,
$f\left( {{\bf{r}}\left( t \right)} \right) = \frac{1}{{\ln t}}\cdot t\cdot{t^2} = \frac{{{t^3}}}{{\ln t}}$
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {\frac{1}{t},1,2t} \right)\cdot\left( {\frac{1}{t},1,2t} \right)} = \sqrt {\frac{1}{{{t^2}}} + 1 + 4{t^2}} $
By Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$
$\mathop \smallint \limits_C^{} {x^{ - 1}}yz{\rm{d}}s = \mathop \smallint \limits_2^4 \frac{{{t^3}}}{{\ln t}}\sqrt {\frac{1}{{{t^2}}} + 1 + 4{t^2}} {\rm{d}}t$
Using a computer algebra system, we compute the integral and the result is
$\mathop \smallint \limits_2^4 \frac{{{t^3}}}{{\ln t}}\sqrt {\frac{1}{{{t^2}}} + 1 + 4{t^2}} {\rm{d}}t \simeq 339.5587$
So, $\mathop \smallint \limits_C^{} {x^{ - 1}}yz{\rm{d}}s \simeq 339.5587$.
