Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 305: 22

Answer

$$2\pi$$

Work Step by Step

Given $$x=\sqrt{\sin y},\ \ x=0,\ \ \ y\in [0,\pi ] $$ Then the volume of the solid obtained by rotating the region enclosed by the graph of $ x=\sqrt{y}$ about the $y-$axis is given by \begin{align*} V&=\pi\int_{0}^{\pi}[f(y)]^2dy\\ &= \pi \int_{0}^{\pi}\sin ydy\\ &=-\pi \cos y\bigg|_{0}^{\pi}\\ &=2\pi \end{align*} We use the Mathematica program to plot the solid.
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