Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 656: 63

Answer

$S={\frac{6\pi a^{2}}{5}}$

Work Step by Step

$\begin{gathered}\frac{d x}{d \theta}=-3 a \cos ^2(\theta) \sin (\theta) \\ \frac{d y}{d \theta}=3 a \sin ^2(\theta) \cos (\theta) \\ \left(\frac{d x}{d \theta}\right)^2+\left(\frac{d y}{d \theta}\right)^2=9 a^2\left[\cos ^4(\theta) \sin ^2(\theta)+\sin ^4(\theta) \cos ^2(\theta)\right] \\ =9 a^2 \sin ^2(\theta) \cos ^2(\theta)\left[\cos ^2(\theta)+\sin ^2(\theta)\right] \\ =9 a^2 \sin ^2(\theta) \cos ^2(\theta) \\ \sqrt{\left(\frac{d x}{d \theta}\right)^2+\left(\frac{d y}{d \theta}\right)^2=} \sqrt{9 a^2 \sin ^2(\theta) \cos ^2(\theta)}=3 a \sin (\theta) \cos (\theta) \\ S=2 \pi \int_0^{\pi / 2} 3 a^2 \sin (\theta)^4 \cos (\theta) d \theta=\left.\frac{6 \pi a^2}{5} \sin (\theta)^5\right|_{t=0} ^{t=\pi / 2}=\frac{6 \pi a^2}{5}\end{gathered}$
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.