Call the surface [tex]S[/tex]; then the area of [tex]S[/tex] is given by the surface integral
[tex]\displaystyle\iint_S\mathrm d\mathbf S[/tex]
Parameterize the surface by
[tex]\mathbf r(u,v)=\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=u^2\cos^2v-u^2\sin^2v=u^2\cos2v\end{cases}[/tex]
with [tex]2\le u\le4[/tex] and [tex]0\le v\le2\pi[/tex]. The surface element is given by
[tex]\mathrm d\mathbf S=\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\mathrm d\mathbf S=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]
So the area is
[tex]\displaystyle\iint_S\mathrm d\mathbf S=\int_{v=0}^{v=2\pi}\int_{u=2}^{u=4}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle2\pi\int_{w=17}^{w=65}\sqrt w\,\frac{\mathrm dw}8[/tex]
where [tex]w=1+4u^2\implies\mathrm dw=8u\,\mathrm du[/tex]
[tex]=\displaystyle\frac\pi4\frac23w^{3/2}\bigg|_{w=17}^{w=65}[/tex]
[tex]=\dfrac\pi6\left(65^{3/2}-17^{3/2}\right)\approx237.69[/tex]
