Answer
The total mass is $64\pi $ gram.
Work Step by Step
The wire of radius $4$ cm centered at the origin can be parametrized by
${\bf{r}}\left( t \right) = \left( {4\cos t,4\sin t} \right)$, ${\ \ \ }$ for $0 \le t \le 2\pi $
So,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( { - 4\sin t,4\cos t} \right)\cdot\left( { - 4\sin t,4\cos t} \right)} $
$ = \sqrt {16{{\sin }^2}t + 16{{\cos }^2}t} = \sqrt {16} = 4$
Total mass: $m = \mathop \smallint \limits_0^{2\pi } \delta \left( {x,y} \right){\rm{d}}s$
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 $f\left( {x,y} \right) = \delta \left( {x,y} \right) = {x^2}$, we get
$m = \mathop \smallint \limits_0^{2\pi } \delta \left( {x,y} \right){\rm{d}}s = 64\mathop \smallint \limits_0^{2\pi } {\cos ^2}t{\rm{d}}t$
Using the double-angle formula (Section 1.4):
${\cos ^2}x = \frac{1}{2}\left( {1 + \cos 2x} \right)$
we get
$m = \mathop \smallint \limits_0^{2\pi } \delta \left( {x,y} \right){\rm{d}}s = 64\mathop \smallint \limits_0^{2\pi } {\cos ^2}t{\rm{d}}t$
$ = 32\mathop \smallint \limits_0^{2\pi } \left( {1 + \cos 2t} \right){\rm{d}}t$
$ = 32\left( {\mathop \smallint \limits_0^{2\pi } {\rm{d}}t + \mathop \smallint \limits_0^{2\pi } \cos 2t{\rm{d}}t} \right)$
$ = 32\left( {2\pi + \frac{1}{2}\sin 2t|_0^{2\pi }} \right)$
$ = 64\pi $
So, the total mass is $64\pi $ g.
