Answer
The work done by the field along the given path is $\frac{{27}}{{28}}$.
Work Step by Step
We obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {{t^3},{t^5},{t^4}} \right)$
${\bf{r}}'\left( t \right) = \left( {1,2t,3{t^2}} \right)$
The work done by the field:
$W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^1 {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$W = \mathop \smallint \limits_0^1 \left( {{t^3},{t^5},{t^4}} \right)\cdot\left( {1,2t,3{t^2}} \right){\rm{d}}t$
$ = \mathop \smallint \limits_0^1 \left( {{t^3} + 5{t^6}} \right){\rm{d}}t$
$ = \left( {\frac{1}{4}{t^4} + \frac{5}{7}{t^7}} \right)|_0^1 = \frac{1}{4} + \frac{5}{7} = \frac{{27}}{{28}}$
So, the work done by the field along the given path is $\frac{{27}}{{28}}$.
