University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 339: 83


The area of the region is $56/15$.

Work Step by Step

$x+4y^2=4$ and $x+y^4=1$ for $x\ge0$ - For the curve $x+4y^2=4$, we can rewrite into $$x=-4y^2+4$$ - For the curve $x+y^4=1$, we can rewrite into $$x=-y^4+1$$ 1) Draft the graph: The graph is drafted below. We would use integration with respect to $y$ in this exercise. 2) Find limits of integration: To find the limits of integration, we will find the intersection points with respect to $y$. $$-4y^2+4=-y^4+1$$ $$y^4-4y^2+3=0$$ $$(y^2-1)(y^2+3)=0$$ $$y=-1\hspace{1cm}\text{or}\hspace{1cm}y=1$$ Therefore, the upper limit is $1$ and the lower one is $-1$. 3) Find the area: The region from $y=-1$ to $y=1$ is bounded on the right by $x=-4y^2+4$ and on the left by $x=-y^4+1$. So according to definition, the area of the region is $$A=\int^1_{-1}(-4y^2+4+y^4-1)dy=\int^1_{-1}(y^4-4y^2+3)dy$$ $$A=\frac{y^5}{5}-\frac{4y^3}{3}+3y\Big]^1_{-1}$$ $$A=\Big(\frac{1}{5}-\frac{4}{3}+3\Big)-\Big(-\frac{1}{5}+\frac{4}{3}-3\Big)$$ $$A=2\Big(\frac{1}{5}-\frac{4}{3}+3\Big)$$ $$A=2\times\frac{28}{15}=\frac{56}{16}$$
Small 1548973198
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.