Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Section 15.9 - Change of Variables in Multiple Integrals - 15.9 Exercise - Page 1060: 13


$x =u \cos v$ 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 $S=${$(u,v) | 1 \leq u \leq \sqrt 2, 0\leq v \leq \dfrac{\pi}{2}$}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.