Answer
The total charge is $q \simeq 0.0698 \times {10^{ - 6}}$ C.
Work Step by Step
We have the charge density of $\delta \left( {x,y,z} \right) = xy\left( {{y^2} - z} \right)$ and the parametrization of the curve:
${\bf{r}}\left( t \right) = \left( {\sin t,\cos t,{{\sin }^2}t} \right)$, ${\ \ \ }$ for $0 \le t \le \frac{\pi }{8}$
So,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {\cos t, - \sin t,2\sin t\cos t} \right)\cdot\left( {\cos t, - \sin t,2\sin t\cos t} \right)} $
$ = \sqrt {{{\cos }^2} + {{\sin }^2}t + {{\left( {\sin 2t} \right)}^2}} $
$ = \sqrt {1 + {{\sin }^2}2t} $
Total charge: $q = \mathop \smallint \limits_0^{\pi /8} \delta \left( {x,y,z} \right){\rm{d}}s$
Using Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$,
where in this case $f\left( {x,y,z} \right) = \delta \left( {x,y,z} \right) = xy\left( {{y^2} - z} \right)$. So,
$q = \mathop \smallint \limits_0^{\pi /8} \delta \left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_0^{\pi /8} \sin t\cos t\left( {{{\cos }^2}t - {{\sin }^2}t} \right)\sqrt {1 + {{\sin }^2}2t} {\rm{d}}t$
$ = \frac{1}{2}\mathop \smallint \limits_0^{\pi /8} \sin 2t\cos 2t\sqrt {1 + {{\sin }^2}2t} {\rm{d}}t$
Write $u = 1 + {\sin ^2}2t$. So, $du = 4\sin 2t\cos 2tdt$.
By changing the variable from $t$ to $u$, the integral becomes
$q = \frac{1}{8}\mathop \smallint \limits_1^{3/2} {u^{1/2}}{\rm{d}}u$
$ = \frac{1}{8}\cdot\frac{2}{3}\left( {{u^{3/2}}|_1^{3/2}} \right)$
$ = \frac{1}{{12}}\left( {{{\left( {\frac{3}{2}} \right)}^{3/2}} - 1} \right) \simeq 0.0698$
So, the total charge is $q \simeq 0.0698 \times {10^{ - 6}}$ C.
