Answer
$\dfrac{1}{8}\pi$
Work Step by Step
Conversion of polar coordinates and Cartesian coordinates are as follows:
a)$r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$
b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$
c) $x=r \cos \theta$
d) $y=r \sin \theta$
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 $(0,1)=P(\sqrt{1+1},\tan^{-1} \dfrac{1}{1})=P(\sqrt 2, \dfrac{1}{2}\pi)$
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:
$\int_0^{\frac{1}{2}\pi} \int_0^1 (x^2+y^2) r dr d\theta =\int_0^{\frac{1}{2}\pi} \dfrac{1}{4} d\theta$
or, $[\dfrac{1}{4}\theta]_0^{\frac{1}{2}\pi}=\dfrac{1}{8}\pi$
