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
We need to convert polar ordinates to $x$ and $y$ values.
Let us consider $u=\sqrt{x^2+y^2}$ and $v=\tan^{-1} \dfrac{y}{x}$
$v=\tan^{-1} \dfrac{y}{x} \implies y=x \tan v$
Therefore, $1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$
Now, $u=\sqrt{x^2+y^2}=\sqrt{x^2+x^2 \tan^2 v}=x\sqrt {1+\tan^2 v}=x \sec v$
and $u=x \sec v=\dfrac{x}{\cos v}$
This gives: $x =u \cos v$
From $y=x \tan v$, we have $\tan v=\dfrac{y}{u \cos v}$
This gives: $y=u \cos v \dfrac{\sin v}{\cos v}=u \sin v$
Hence, $x =u \cos v$ and $y =u \sin v$
where $S=${$(u,v) | 1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$}
