Answer
$\mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}x + \left( {y - z} \right){\rm{d}}y + z{\rm{d}}z = \frac{{13}}{2}$
Work Step by Step
Write the vector field ${\bf{F}}\left( {x,y,z} \right) = \left( {x - y,y - z,z} \right)$ and the path ${\bf{r}} = \left( {x,y,z} \right)$, such that
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}x + \left( {y - z} \right){\rm{d}}y + z{\rm{d}}z$
The line segment from $\left( {0,0,0} \right)$ to $\left( {1,4,4} \right)$ can be parametrized by
${\bf{r}}\left( t \right) = \left( {t,4t,4t} \right)$, ${\ \ \ }$ for $0 \le t \le 1$
So,
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( { - 3t,0,4t} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {1,4,4} \right)dt$
Using Eq. (8), the vector line integral becomes
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}x + \left( {y - z} \right){\rm{d}}y + z{\rm{d}}z = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}x + \left( {y - z} \right){\rm{d}}y + z{\rm{d}}z = \mathop \smallint \limits_0^1 \left( { - 3t,0,4t} \right)\cdot\left( {1,4,4} \right){\rm{d}}t$
$ = 13\mathop \smallint \limits_0^1 t{\rm{d}}t = \frac{{13}}{2}{t^2}|_0^1 = \frac{{13}}{2}$
So, $\mathop \smallint \limits_C^{} \left( {x - y} \right){\rm{d}}x + \left( {y - z} \right){\rm{d}}y + z{\rm{d}}z = \frac{{13}}{2}$.
