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 329: 6

Answer

$36 \pi$

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^{3} (2 \pi) \cdot (x)[\dfrac{9x}{(x^3+9)^{1/2}}) dx$ Suppose $a=x^3+9 \implies da=3x^2dx$ Now, $V=2 \pi \times \int_{9}^{36} [3 a^{-1/2} da$ or, $=6 \pi \times [2 a^{1/2}]_{9}^{36}$ or, $=36 \pi$
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