Answer
The total charge is $q \simeq 10.39 \times {10^{ - 6}}$ C.
Work Step by Step
We have the charge density of $\delta \left( {x,y} \right) = x/y$.
The curve can be parametrized by
${\bf{r}}\left( t \right) = \left( {t,{t^{4/3}}} \right)$, ${\ \ \ \ }$ for $1 \le t \le 8$
So,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {1,\frac{4}{3}{t^{1/3}}} \right)\cdot\left( {1,\frac{4}{3}{t^{1/3}}} \right)} $
$ = \sqrt {1 + \frac{{16}}{9}{t^{2/3}}} $
Total charge: $q = \mathop \smallint \limits_1^8 \delta \left( {x,y} \right){\rm{d}}s$
Using Eq. (4):
$\mathop \smallint \limits_C^{} f\left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$,
where in this case $f\left( {x,y} \right) = \delta \left( {x,y} \right) = \frac{x}{y}$. So,
$q = \mathop \smallint \limits_1^8 \delta \left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_1^8 {t^{ - 1/3}}\sqrt {1 + \frac{{16}}{9}{t^{2/3}}} {\rm{d}}t$
Write $u = 1 + \frac{{16}}{9}{t^{2/3}}$. So, $du = \frac{{32}}{{27}}{t^{ - 1/3}}dt$.
By changing the variable from $t$ to $u$, the integral becomes
$q = \frac{{27}}{{32}}\mathop \smallint \limits_{25/9}^{73/9} {u^{1/2}}{\rm{d}}u$
$ = \frac{{27}}{{32}}\cdot\frac{2}{3}\left( {{u^{3/2}}|_{25/9}^{73/9}} \right)$
$ = \frac{9}{{16}}\left( {{{\left( {\frac{{73}}{9}} \right)}^{3/2}} - {{\left( {\frac{{25}}{9}} \right)}^{3/2}}} \right) \simeq 10.39$
So, the total charge is $q \simeq 10.39 \times {10^{ - 6}}$ C.
