Use a parametrization to express the area of the surface s as a double integral. then, evaluate the integral to find the area of the surface. s is the portion of the plane y + 3 z = 2 inside the cylinder x^2 + y^2 = 1.

The cylinder gives you a hint as to where to start with a parameterization [tex]\mathbf s(u,v)=\langle x(u,v),y(u,v),z(u,v)\rangle[/tex]. A sensible choice would be to set

[tex]x=u\cos v[/tex]
[tex]y=u\sin v[/tex]

so that

[tex]y+3z=2\implies z=\dfrac{2-u\sin v}3[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le2\pi[/tex]. Then

[tex]\mathbf s_u\times\mathbf s_v=\left\langle0,\dfrac u3,u\right\rangle[/tex]

so the surface element is

[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\dfrac{\sqrt{10}u}3\,\mathrm du\,\mathrm dv[/tex]

So the area of the surface is

[tex]\displaystyle\iint_{\mathcal S}\mathrm dS=\frac{\sqrt{10}}3\int_{v=0}^{v=2\pi}\int_{u=0}^{u=1}u\,\mathrm du\,\mathrm dv=\frac{\sqrt{10}\pi}3[/tex]