## Calculus 8th Edition

$x =u \cos v$ and $y =u \sin v$ where $S=${$(u,v) | 1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$}
Here, we have $u=\sqrt{x^2+y^2}$ and $v=\tan^{-1} \dfrac{y}{x}$ and $y=x \tan v$ Therefore, we get $1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$ Further, $u=\sqrt{x^2+x^2 \tan^2 v}=x\sqrt {1+\tan^2 v}$ This implies that $u=x \sec v=\dfrac{x}{\cos v}$ or, $x =u \cos v$ and $\tan v=\dfrac{y}{u \cos v} \implies y=u \sin v$ Hence, $x =u \cos v$ and $y =u \sin v$ where $S=${$(u,v) | 1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$}