Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.4 The Method of Cylindrical Shells - Exercises - Page 312: 48


$\dfrac{625 \pi}{6}$

Work Step by Step

The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by: $V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$ Now, $V=2\pi \int_{0}^{5} y^2 (5-y) \ dy\\= 2 \pi \int_0^5 (5y^2-y^3) \ dy \\=2 \pi [ \dfrac{5y^3}{3}-\dfrac{y^{4}}{4}]_0^5 \\=2 \pi [ \dfrac{5(5^3)}{3}-\dfrac{(5^{4})}{4}] \\=\dfrac{625 \pi}{6}$
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.