Answer
$\mathop \smallint \limits_C^{} y{\rm{d}}x + z{\rm{d}}y + x{\rm{d}}z = \frac{{41}}{{10}}$
Work Step by Step
Write the vector field ${\bf{F}}\left( {x,y,z} \right) = \left( {y,z,x} \right)$ and the path ${\bf{r}} = \left( {x,y,z} \right)$, such that
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} y{\rm{d}}x + z{\rm{d}}y + x{\rm{d}}z$
The path is given by ${\bf{r}}\left( t \right) = \left( {2 + {t^{ - 1}},{t^3},{t^2}} \right)$ for $0 \le t \le 1$. So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^3},{t^2},2 + {t^{ - 1}}} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( { - {t^{ - 2}},3{t^2},2t} \right)dt$
Using Eq. (8), the vector line integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} y{\rm{d}}x + z{\rm{d}}y + x{\rm{d}}z = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} y{\rm{d}}x + z{\rm{d}}y + x{\rm{d}}z = \mathop \smallint \limits_0^1 \left( {{t^3},{t^2},2 + {t^{ - 1}}} \right)\cdot\left( { - {t^{ - 2}},3{t^2},2t} \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} y{\rm{d}}x + z{\rm{d}}y + x{\rm{d}}z = \mathop \smallint \limits_0^1 \left( { - t + 3{t^4} + 4t + 2} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {3{t^4} + 3t + 2} \right){\rm{d}}t$
$ = \left( {\left( {\frac{3}{5}{t^5} + \frac{3}{2}{t^2} + 2t} \right)|_0^1} \right) = \frac{3}{5} + \frac{3}{2} + 2 = \frac{{41}}{{10}}$
So, $\mathop \smallint \limits_C^{} y{\rm{d}}x + z{\rm{d}}y + x{\rm{d}}z = \frac{{41}}{{10}}$.
