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 - Section 6.2 - Volumes Using Cylindrical Shels - Exercises 6.2 - Page 330: 17

Answer

$\dfrac{8 \pi}{3}$

Work Step by Step

We need to use the shell model as follows: $V=\int_p^{q} (2 \pi) \cdot (\space radius \space of \space shell) ( height \space of \space Shell) \space dx$ $ \implies V= \int_{0}^{2} (2 \pi) \cdot y (2y-y^2) dy$ Now, $V=2 \pi (\dfrac{2y^3}{3}-\dfrac{y^4}{4})_{0}^{2}$ or, $=2 \pi \times (\dfrac{16}{3}-4)$ or, $=\dfrac{8 \pi}{3}$
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