Answer
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{8}{3}$
Work Step by Step
We have the vector field ${\bf{F}}\left( {x,y,z} \right) = \left( {{z^3},yz,x} \right)$.
The circle in the $yz$-plane mentioned in this exercise can be parametrized by
${\bf{r}}\left( t \right) = \left( {0,2\cos t,2\sin t} \right)$, ${\ \ \ }$ for $0 \le t \le \frac{\pi }{2}$
So, we obtain
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {8{{\sin }^3}t,4\cos t\sin t,0} \right)$
$d{\bf{r}} = {\bf{r}}'\left( t \right)dt$
$d{\bf{r}} = \left( {0, - 2\sin t,2\cos t} \right)dt$
Evaluate the dot product ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt$:
${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right)dt = \left( {8{{\sin }^3}t,4\cos t\sin t,0} \right)\cdot\left( {0, - 2\sin t,2\cos t} \right)dt$
$ = - 8\cos t{\sin ^2}tdt$
Evaluate $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$ using Eq. (8):
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_a^b {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - \mathop \smallint \limits_{\pi /2}^0 8\cos t{\sin ^2}t{\rm{d}}t$
$ = - 8\mathop \smallint \limits_{\pi /2}^0 {\sin ^2}t{\rm{d}}\left( {\sin t} \right)$
$ = - \frac{8}{3}\left( {{{\sin }^3}t|_{\pi /2}^0} \right) = \frac{8}{3}$
So, $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \frac{8}{3}$.
