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