Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 434: 20


$$A= \frac{22}{15}$$

Work Step by Step

We will integrate over the $y$-axis; one of the limits will be $y=0$, and the other is the intersection of the two curves. $$y^{2}=2-x$$ $$x=2-y^{2}$$ $$y^{4} = 2-y^{2}$$ $$0 = y^{4} + y^{2} -2$$ $$=(y^{2}+2)(y^{2}-1)$$ $$=(y^{2}+2)(y+1)(y-1)$$ $$y=\pm 1$$ $$x=(\pm 1)^{4}=1$$ $y=\sqrt {2-x}$ (which is $x=2-y^{2}$) is on top. $$A = \displaystyle\int_{0}^{1} (2-y^{2} - y^{4})dy$$ $$=\left[2y-\frac{1}{3}y^{3}-\frac{1}{5} \right]^{1}_0$$ $$2-\frac{1}{3} - \frac{1}{5} -0$$ $$=\frac{30-5-3}{15}$$ $$= \frac{22}{15}$$
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.