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 362: 4

Answer

$\dfrac{ 72}{5}$

Work Step by Step

The area of each cross-section triangle is given as: $y^2=36-24 \sqrt {6x} +36 x-4 \sqrt 6 x^{3/2} +x^2$ We integrate the integral to calculate the volume as follows: $V= \int_{0}^{6} [36-24 \sqrt {6x} +36 x-4 \sqrt 6 x^{3/2} +x^2] dx$ or, $= [36x-16 \sqrt {6}x^{3/2} +18x^2 -\dfrac{8 \sqrt 6 x^{5/2}}{5} + \dfrac{x^3}{3}]_0^6$ or, $=\dfrac{ 72}{5}$
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.