Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Section 6.3 - Volumes by Cylindrical Shells - 6.3 Exercises - Page 454: 12

Answer

$\displaystyle{V=16\pi}$

Work Step by Step

$\displaystyle{y = \frac{12 \pm \sqrt{\left(-12\right)^2-4(3)(9)}}{2(3)}}\\ y=1\qquad y=3$ $\displaystyle{V=\int_{1}^{3}(2\pi y)\left(-3y^2+12y-9\right)\ dy}\\ \displaystyle{V=2\pi\int_{1}^{3}12y^2-9y-3y^3\ dy}\\ \displaystyle{V=2\pi\left[4y^3-\frac{9}{2}y^2-\frac{3}{4}y^4\right]_{1}^{3}}\\ \displaystyle{V=2\pi\left(\left(4(3)^3-\frac{9}{2}(3)^2-\frac{3}{4}(3)^4\right)-\left(4(1)^3-\frac{9}{2}(1)^2-\frac{3}{4}(1)^4\right)\right)}\\ \displaystyle{V=16\pi}$
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.