Parameterize the surface (call it [tex]\mathcal S[/tex]) by
[tex]\mathbf s(u,v)=\langle u\cos v,u\sin v,4-u^2\rangle[/tex]
with [tex]0\le u\le2[/tex] and [tex]0\le v\le2\pi[/tex]. Then the surface element is
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]
The area of [tex]\mathcal S[/tex] is then given by the surface integral
[tex]\displaystyle\iint_{\mathcal S}\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du=\dfrac{(17^{3/2}-1)\pi}6\approx36.1769[/tex]
