Parameterize the surface by
[tex]\mathbf s(u,v)=\langle u\cos v,u\sin v,u^2\rangle[/tex]
with [tex]0\le u\le5[/tex] and [tex]0\le v\le2\pi[/tex]. Then
[tex]\mathrm dS=\|\mathbf s_u\time\mathbf s_v\|=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]
so 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=5}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv=\dfrac{(101^{3/2}-1)\pi}6[/tex]
