Answer
$x =u \cos v$ and $y =u \sin v$
where $S=${$(u,v) | 1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$}
Work Step by Step
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 $S=${$(u,v) | 1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$}
