Answer
$\dfrac{\pi \sqrt 5}{2}$
Work Step by Step
Here, we have $y+2z=2$ and $x^2+y^2=1$
Thus, $x=r \cos \theta, y=r \sin \theta $ $y+2z=2 \implies z=(\dfrac{2-r \sin \theta}{2})$
Then, we have $r(r,\theta)=(r \cos \theta) i+(r \sin \theta) j+(\dfrac{2-r \sin \theta}{2}) k$
Then $r_r=( \cos \theta) i+(\sin \theta) j-(\dfrac{ \sin \theta}{2}) k$
and
$r_{\theta}=(-r \sin \theta) i+(r \cos \theta) j-(\dfrac{r \cos \theta}{2}) k$
Also, $|r_r \times rr_{\theta}|=\dfrac{\sqrt 5r}{2}$
where $0 \le r \le 1$; $0 \le \theta \le 2\pi$
Thus, the area is given by $A=\int_0^{2 \pi} \int_0^{1} \dfrac{\sqrt 5r}{2} dr d \theta$
or, $A=\int_0^{2 \pi} [ \dfrac{\sqrt 5r^2}{4}]_0^{1} d \theta=\dfrac{\pi \sqrt 5}{2}$
