The area is given by the surface integral
[tex]\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=9}\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle\int_{u=0}^{u=1}\int_{v=0}^{v=9}\sqrt{u^2+1}\,\mathrm dv\,\mathrm du[/tex]
[tex]=\dfrac9{\sqrt2}+\dfrac92\ln(\sqrt2+1)[/tex]
