Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 286: 12


$$4\pi $$

Work Step by Step

The area is given by \begin{align*} A&=\int_{-2}^{2}\left(y_{1}(x)-y_{2}(x)\right) d x\\ &=2 \int_{-2}^{2} \sqrt{4-x^{2}} d x\\ &=4 \int_{0}^{2} \sqrt{4-x^{2}} d x \end{align*} Let $x=2\sin u, \ \ dx = 2\cos udu$. Then \begin{align*} A &=4 \int_{0}^{2} \sqrt{4-x^{2}} d x\\ &=4\int_{0}^{\pi/4} 4\cos^2 udu\\ &=8\int_{0}^{\pi/4} [1+\cos2 u]du\\ &=8[u+\frac{1}{2}\sin 2u]\bigg|_{0}^{\pi/2}\\ &=4\pi \end{align*}
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.