Answer
$2 -\sqrt 3$
Work Step by Step
Here, $1 \leq y \leq x, \leq \dfrac{\pi}{4}$
The lower limit of $\phi$ is $\arctan (\sqrt 3)=\dfrac{\pi}{6}$
Here, $x=r \cos \theta=\sqrt 3 \implies r= \sqrt 3 \sec \theta$
This information suggests the integral is as follows:
$\int_{\csc \phi}^{\sqrt 3 \sec \theta} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} r dr d \phi=\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} (\dfrac{3}{2} \sec^2 \phi-\dfrac{1}{2} \csc^2 \phi) d\phi$
Thus, $[(\dfrac{3}{2}) \tan \phi +\dfrac{1}{2} \cot \phi]_{\frac{\pi}{6}}^{\frac{\pi}{4}}=(\dfrac{3}{2})(1-(1/\sqrt 3))+ (\dfrac{1}{2})(1-\sqrt 3)=2 -\sqrt 3$
