University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 355: 2


The volume of the solid is $16\pi/15$.

Work Step by Step

1) We already know that the cross-section is a circle, whose diameter runs from $y=x^2$ to $y=2-x^2$. As we can see in the given image, the diameter equals the subtraction of $x^2$ from $2-x^2$; in other words, $2-x^2-x^2=2-2x^2$. Therefore, the radius of the cross-sectional circle would be $\frac{2-2x^2}{2}=1-x^2$ Therefore, we can come up with a formula for $A(x)$, which is the area of the cross-sectional circle: $$A(x)=\pi(1-x^2)^2=\pi(1-2x^2+x^4)$$ 3) Limits of integration: From the given image, we see that the circular disks lie on the plane between the points of intersection of 2 parabola. $$x^2=2-x^2$$ $$2x^2=2$$ $$x^2=1$$ $$x=\pm1$$ So the circular disks lie on the plane from $x=-1$ to $x=1$. 4) Integrate $A(x)$ to find the volume of the solid: $$V=\int^1_{-1}\pi(1-2x^2+x^4)dx=\pi\Big(x-\frac{2x^3}{3}+\frac{x^5}{5}\Big)\Big]^1_{-1}$$ $$V=\pi\Bigg(\Big(1-\frac{2}{3}+\frac{1}{5}\Big)-\Big(-1+\frac{2}{3}-\frac{1}{5}\Big)\Bigg)$$ $$V=\pi\Big(\frac{8}{15}-\Big(-\frac{8}{15}\Big)\Big)$$ $$V=\frac{16\pi}{15}$$
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.