Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 6: Applications of Definite Integrals - Practice Exercises - Page 363: 5


$\dfrac{ 72 \pi }{35}$

Work Step by Step

The area of each cross-section triangle is given as: $A=\pi r^2 =\pi (\sqrt x- \dfrac{x^2}{8}) ^2 $ We integrate the integral to calculate the volume as follows: $V= \int_{0}^{4} \pi (\sqrt x- \dfrac{x^2}{8}) ^2 dx$ or, $=\pi [\dfrac{x^2}{2}-\dfrac{x^{7/2}}{14} +\dfrac{x^5}{320}]_0^4$ or, $=\pi (8-\dfrac{64}{7} +\dfrac{16}{5}) $ or, $=\dfrac{ 72 \pi }{35}$
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.