Answer
The total mass is $m \simeq 166.857$ g.
Work Step by Step
We have the parametrization
${\bf{r}}\left( t \right) = \left( {\cos t,\sin t,{t^2}} \right)$, ${\ \ \ }$ for $0 \le t \le 2\pi $
and the mass density $\delta \left( {x,y,z} \right) = \sqrt z $ g/cm.
So,
$||{\bf{r}}'\left( t \right)|| = \sqrt {\left( { - \sin t,\cos t,2t} \right)\cdot\left( { - \sin t,\cos t,2t} \right)} $
$ = \sqrt {{{\sin }^2}t + {{\cos }^2}t + 4{t^2}} $
$ = \sqrt {1 + 4{t^2}} $
Total mass: $m = \mathop \smallint \limits_0^{2\pi } \delta \left( {x,y,z} \right){\rm{d}}s$
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) = \sqrt z $, we get
$m = \mathop \smallint \limits_0^{2\pi } \delta \left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_0^{2\pi } t\sqrt {1 + 4{t^2}} {\rm{d}}t$
Write $u = 1 + 4{t^2}$. So, $du = 8tdt$.
By changing the variable from $t$ to $u$, the integral becomes
$m = \frac{1}{8}\mathop \smallint \limits_1^{1 + 16{\pi ^2}} {u^{1/2}}{\rm{d}}u = \frac{1}{8}\cdot\frac{2}{3}\left( {{u^{3/2}}|_1^{1 + !6{\pi ^2}}} \right)$
$ = \frac{1}{{12}}\left( {{{\left( {1 + 16{\pi ^2}} \right)}^{3/2}} - 1} \right) \simeq 166.857$
So, the total mass is $m \simeq 166.857$ g.
