Answer
The total charge is $85.51 \times {10^{ - 6}}$ $C$.
Work Step by Step
In this exercise, we have $r = f\left( \theta \right) = \theta $. Using the result of Exercise 58(a), we obtain
$||{\bf{r}}'\left( \theta \right)|| = \sqrt {f{{\left( \theta \right)}^2} + f'{{\left( \theta \right)}^2}} = \sqrt {1 + {\theta ^2}} $
Compute the total charge $\mathop \smallint \limits_C^{} \delta \left( {x,y} \right){\rm{d}}s$.
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( {r,\theta } \right) = \delta \left( {r,\theta } \right) = r = \theta $, we get
$\mathop \smallint \limits_C^{} \delta \left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_0^{2\pi } \theta \sqrt {1 + {\theta ^2}} {\rm{d}}\theta $
Write $u = 1 + {\theta ^2}$. So, $du = 2\theta d\theta $.
By changing the variable from $\theta $ to $u$, the integral becomes
$\mathop \smallint \limits_C^{} \delta \left( {x,y} \right){\rm{d}}s = \frac{1}{2}\mathop \smallint \limits_1^{1 + 4{\pi ^2}} {u^{1/2}}{\rm{d}}u$
$ = \frac{1}{3}\left( {{u^{3/2}}|_1^{1 + 4{\pi ^2}}} \right)$
$ = \frac{1}{3}\left( {{{\left( {1 + 4{\pi ^2}} \right)}^{3/2}} - 1} \right) \simeq 85.51$
So, the total charge is $85.51 \times {10^{ - 6}}$ $C$.
