As a first step, convert to cylindrical coordinates. The equation [tex]x^2+y^2=9=3^2[/tex] is a clue to set
[tex]x=3r\cos\theta[/tex]
[tex]y=3r\sin\theta[/tex]
then have [tex]0\le r\le1[/tex] and [tex]0\le\theta\le2\pi[/tex]. Meanwhile, the plane equation tells you to use
[tex]4x+4y+z=16\implies z=16-12r\cos\theta-12r\sin\theta[/tex]
So we parameterize the part of the plane within the cylinder with the vector-valued function
[tex]\mathbf s(r,\theta)=(3r\cos\theta,3r\sin\theta,15-12r\cos\theta-12r\sin\theta)[/tex]
The surface integral giving the area of the surface is
[tex]\displaystyle\iint_{\mathcal S}\mathrm dS[/tex]
where [tex]\mathcal S[/tex] denotes the surface in question, and the surface element [tex]\mathrm dS[/tex] is
[tex]\mathrm dS=\left\|\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right\|\,\mathrm dr\,\mathrm d\theta[/tex]
We have
[tex]\dfrac{\partial\mathbf s}{\partial r}=(3\cos\theta,3\sin\theta,-12(\cos\theta+\sin\theta))[/tex]
[tex]\dfrac{\partial\mathbf s}{\partial\theta}=(-3r\sin\theta,3r\cos\theta,12r(\sin\theta-\cos\theta))[/tex]
[tex]\implies\mathrm dS=\|(36r,36r,9r\|\,\mathrm dr\,\mathrm d\theta=9\sqrt{33}r\,\mathrm dr\,\mathrm d\theta[/tex]
So the area is
[tex]\displaystyle\iint_{\mathcal S}\mathrm dS=9\sqrt{33}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta=18\sqrt{33}\,\pi[/tex]
