University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 6 - Section 6.1 - Volumes Using Cross-Sections - Exercises - Page 356: 36


$$=\dfrac{ \pi}{2}$$

Work Step by Step

We need to integrate the integral to compute the volume. We have: $$Area=\pi r^2=\pi (\dfrac{\sqrt {2y}}{y^2+1})^2=\dfrac{2 \pi y}{(y^2+1)^2}$$ Now, $$Volume= \int_{0}^{1} \dfrac{2 \pi y}{(y^2+1)^2} \space dy $$ Let us consider $u =t^2+1$ and $dt=2y \space dy$ Now, $$Volume= \pi \int_{1}^2 \dfrac{dt}{t^2} \\ = \pi (-\dfrac{1}{t})_{1}^{2} \\=\dfrac{ \pi}{2}$$
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