Answer
$\dfrac{1}{2}\pi$
Work Step by Step
Here, $(x,y)=(1,1)$
The diagonal points to these points are: $(1,1)=P(\sqrt{1+1},\tan^{-1} \dfrac{1}{1}=P(\sqrt 2,0)$ and $(1,1)=P(\sqrt{1+1},\tan^{-1} \dfrac{1}{1}=P(\sqrt 2, \pi)$
Also, $r=\sqrt {x^2+y^2} \implies r=1$
Therefore, the region described in polar coordinates is:
$0 \leq \theta \leq \pi$ and $0\leq r\leq 1$
This suggests the integral is as follows: $\int_0^\pi \int_0^1 r dr d\theta =\int_0^\pi \dfrac{1}{2} d\theta$
or, $[\dfrac{1}{2}]_0^\pi=\dfrac{1}{2}\pi$
