Parameterize the part of the paraboloid within the cylinder - I'll call it [tex]S[/tex] - by
[tex]\mathbf r(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle=\left\langle u\cos v,u\sin v,3u^2\right\rangle[/tex]
with [tex]0\le u\le2[/tex] and [tex]0\le v\le2\pi[/tex]. The region's area is given by the surface integral
[tex]\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=2}\int_{v=0}^{v=2\pi}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+36u^2}\,\mathrm du[/tex]
Take [tex]w=1+36u^2[/tex] so that [tex]\mathrm dw=72u\,\mathrm du[/tex], and the integral becomes
[tex]=\displaystyle\frac{2\pi}{72}\int_{w=1}^{w=145}\sqrt w\,\mathrm dw[/tex]
[tex]=\displaystyle\frac\pi{36}\frac23w^{3/2}\bigg|_{w=1}^{w=145}[/tex]
[tex]=\dfrac\pi{54}(145^{3/2}-1)\approx101.522[/tex]
