Chapter 15: Multiple Integrals - Section 15.4 - Double Integrals in Polar Form - Exercises 15.4 - Page 893: 5

$1\leq r \leq 2\sqrt 3secθ $ and $\frac{\pi}{6}\leq θ \leq \frac{\pi}{2} $

$x^{2}+y^{2}=1^{2}$( Circle equation) $=> r = 1$ $x=2\sqrt 3 (given)$ $=> r = 2\sqrt 3 secθ$ Therefore , $1\leq r \leq 2\sqrt 3secθ $ $y=2(given)$we know that, $y=rsinθ$ $2\sqrt 3 secθ=2cosecθ$ Therefore , $\frac{\pi}{6}\leq θ \leq \frac{\pi}{2} $