You have the parametrization $r(v,\theta)=(3vcos(\theta),3vsin(\theta),2\theta)$. Now by simple calculation:
$r_v=(3cos(\theta),3sin(\theta),0)$
$r_\theta=(-3vsin(\theta),3vcos(\theta),2)$
Now you need to calculate the cross product of the vectors $r_v\times r_\theta$. Also note that $\sqrt{x^2+y^2}=3v$. So then your integral will be:
$\int_0^1\int_0^\frac{\pi}{2}3v||r_v\times r_\theta|| d\theta dv$.