Question

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.

Answer 1

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

$x=u\cos v$
$y=u\sin v$

so that

$y+3z=2\implies z=\dfrac{2-u\sin v}3$

with $0\le u\le1$ and $0\le v\le2\pi$. Then

$\mathbf s_u\times\mathbf s_v=\left\langle0,\dfrac u3,u\right\rangle$

so the surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\dfrac{\sqrt{10}u}3\,\mathrm du\,\mathrm dv$

So the area of the surface is

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