Answer
$$4 \space \pi \sqrt 2$$
Work Step by Step
We have $ r_r= \cos \theta \space i+\sin \theta \space j- \cos \theta k ; \\ r_{\theta}=-r \sin \theta \space i+r \cos \theta \space j+\space r \sin \theta \space k $
Also, $|r_r \times r_{\theta}|=r \sqrt 2$
and $0 \le r \le 2$ and $0 \le \theta \le 2\pi $
Now, $$ Area=\int_0^{2 \pi} \int_0^{2} r \sqrt 2 \space dr \space d \theta \\=\int_0^{2 \pi} [ 2 \space \sqrt 2] \space d \theta\\=4 \space \pi \sqrt 2$$
