## Calculus: Early Transcendentals 8th Edition

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