Answer
(a) We show that $||{\bf{r}}'\left( \theta \right)|| = \sqrt {f{{\left( \theta \right)}^2} + f'{{\left( \theta \right)}^2}} $
(b) $\mathop \smallint \limits_C^{} {\left( {x - y} \right)^2}{\rm{d}}s = 2\pi - 4$
Work Step by Step
(a) We have ${\bf{r}}\left( \theta \right) = \left( {f\left( \theta \right)\cos \theta ,f\left( \theta \right)\sin \theta } \right)$.
Taking the derivative of ${\bf{r}}\left( \theta \right)$ with respect to $\theta $ gives
${\bf{r}}'\left( \theta \right) = \left( {f'\left( \theta \right)\cos \theta - f\left( \theta \right)\sin \theta ,f'\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta } \right)$
The dot product is
${\bf{r}}'\left( \theta \right)\cdot{\bf{r}}'\left( \theta \right) = {\left( {f'\left( \theta \right)\cos \theta - f\left( \theta \right)\sin \theta } \right)^2} + {\left( {f'\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta } \right)^2}$
${\bf{r}}'\left( \theta \right)\cdot{\bf{r}}'\left( \theta \right) = f'{\left( \theta \right)^2}{\cos ^2}\theta - 2f'\left( \theta \right)f\left( \theta \right)\cos \theta \sin \theta + f{\left( \theta \right)^2}{\sin ^2}\theta $
${\ \ \ \ \ }$ $ + f'{\left( \theta \right)^2}{\sin ^2}\theta + 2f'\left( \theta \right)f\left( \theta \right)\sin \theta \cos \theta + f{\left( \theta \right)^2}{\cos ^2}\theta $
${\bf{r}}'\left( \theta \right)\cdot{\bf{r}}'\left( \theta \right) = f'{\left( \theta \right)^2} + f{\left( \theta \right)^2}$
Thus, $||{\bf{r}}'\left( \theta \right)|| = \sqrt {{\bf{r}}'\left( \theta \right)\cdot{\bf{r}}'\left( \theta \right)} = \sqrt {f{{\left( \theta \right)}^2} + f'{{\left( \theta \right)}^2}} $.
(b) In this case, we have $r = f\left( \theta \right) = 2\cos \theta $. So, using the result in part (a) we obtain
$||{\bf{r}}'\left( \theta \right)|| = \sqrt {f{{\left( \theta \right)}^2} + f'{{\left( \theta \right)}^2}} = \sqrt {4{{\cos }^2}\theta + 4{{\sin }^2}\theta } = 2$
Evaluate $\mathop \smallint \limits_C^{} {\left( {x - y} \right)^2}{\rm{d}}s$.
Using Eq. (4):
$\mathop \smallint \limits_C^{} g\left( {x,y} \right){\rm{d}}s = \mathop \smallint \limits_a^b g\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$,
where in this case $g\left( {x,y} \right) = {\left( {x - y} \right)^2}$, we get
$\mathop \smallint \limits_C^{} {\left( {x - y} \right)^2}{\rm{d}}s = \mathop \smallint \limits_0^{\pi /2} g\left( {{\bf{r}}\left( \theta \right)} \right)||{\bf{r}}'\left( \theta \right)||{\rm{d}}\theta $
$\mathop \smallint \limits_C^{} {\left( {x - y} \right)^2}{\rm{d}}s = 2\mathop \smallint \limits_0^{\pi /2} {\left( {2{{\cos }^2}\theta - 2\cos \theta \sin \theta } \right)^2}{\rm{d}}\theta $
$ = 8\mathop \smallint \limits_0^{\pi /2} {\cos ^2}\theta {\left( {\cos \theta - \sin \theta } \right)^2}{\rm{d}}\theta $
$ = 8\mathop \smallint \limits_0^{\pi /2} {\cos ^2}\theta \left( {{{\cos }^2}\theta - 2\cos \theta \sin \theta + {{\sin }^2}\theta } \right){\rm{d}}\theta $
$ = 8\mathop \smallint \limits_0^{\pi /2} {\cos ^2}\theta \left( {1 - 2\cos \theta \sin \theta } \right){\rm{d}}\theta $
$ = 8\left( {\mathop \smallint \limits_0^{\pi /2} {{\cos }^2}\theta {\rm{d}}\theta - 2\mathop \smallint \limits_0^{\pi /2} {{\cos }^3}\theta \sin \theta {\rm{d}}\theta } \right)$
Using the double-angle formula (Section 1.4):
${\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)$
we get
$\mathop \smallint \limits_C^{} {\left( {x - y} \right)^2}{\rm{d}}s = 8\left( {\frac{1}{2}\mathop \smallint \limits_0^{\pi /2} \left( {1 + \cos 2\theta } \right){\rm{d}}\theta - 2\mathop \smallint \limits_0^{\pi /2} {{\cos }^3}\theta \sin \theta {\rm{d}}\theta } \right)$
$ = 8\left( {\frac{1}{2}\mathop \smallint \limits_0^{\pi /2} {\rm{d}}\theta + \frac{1}{2}\mathop \smallint \limits_0^{\pi /2} \cos 2\theta {\rm{d}}\theta + 2\mathop \smallint \limits_0^{\pi /2} {{\cos }^3}\theta {\rm{d}}\left( {\cos \theta } \right)} \right)$
$ = 8\left( {\frac{1}{2}\theta |_0^{\pi /2} + \frac{1}{4}\sin 2\theta |_0^{\pi /2} + \frac{1}{2}{{\cos }^4}\theta |_0^{\pi /2}} \right)$
$ = 8\left( {\frac{\pi }{4} + \frac{1}{2}\left( { - 1} \right)} \right) = 2\pi - 4$
So, $\mathop \smallint \limits_C^{} {\left( {x - y} \right)^2}{\rm{d}}s = 2\pi - 4$.