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