Parameterize the surface by
[tex]\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac{5-u\sin v}2\right\rangle[/tex]
with [tex]0\le u\le1[/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=\dfrac{\sqrt5}2u\,\mathrm du\,\mathrm dv[/tex]
and the area of the surface [tex]\mathcal S[/tex] is given by
[tex]\displaystyle\iint_{\mathcal S}\mathrm dS=\frac{\sqrt5}2\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u\,\mathrm du\,\mathrm dv=\frac{\pi\sqrt5}2[/tex]
