Answer
$\dfrac{4}{3}$
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, $x=r \cos \theta=2 \implies r= 2\sec \theta$
This information suggests the integral is as follows:
$\int_0^{2\sec \theta} \int_{0}^{\frac{\pi}{4}} r^2 \sin \theta dr d\theta=(\dfrac{8}{3}) \int_{0}^{\frac{\pi}{4}} \sec^3 \theta d\theta$
Thus, $(\dfrac{8}{3}) \int_{0}^{\frac{\pi}{4}} \sec^3 \theta d\theta=(\dfrac{8}{3}) \int_{0}^{\frac{\pi}{4}} \tan \theta \sec^2 \theta d\theta$
or, $(\dfrac{8}{3}) [\dfrac{\tan^2 \theta}{2}]_{0}^{\frac{\pi}{4}}=\dfrac{4}{3}$
