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
Plug $u=\sqrt{x^2+y^2}$
$v=\tan^{-1} \dfrac{y}{x} \implies v=\tan^{-1} \dfrac{y}{x}$
or, $y=x \tan v$
Also, we have $1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$
Now, $u=\sqrt{x^2+y^2}$
or, $u=\sqrt{x^2+x^2 \tan^2 v}=x\sqrt {1+\tan^2 v}=x \sec v=\dfrac{x}{\cos v}$
or, $x =u \cos v$
and $y=(u \cos v )(\dfrac{\sin v}{\cos v})=u \sin v$
Hence, $x =u \cos v$ and $y =u \sin v$
where $S=${$(u,v) | 1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$}
