Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 363: 42

Answer

$\text{The volume is}$ \begin{align} V = \pi \end{align}

Work Step by Step

$\text{It is given that}$ \begin{align} y = \sqrt x; y = 0 \ and \ x = 4 \end{align} $\text{To find the volume, we have to find the area of the cross section.}$ $\text{It is given that the cross section is semicircle, therefore:}$ \begin{align} A(x) = \frac{1}{2} \pi r^2 = \frac{1}{2} \pi \left(\frac{\sqrt x}{2} \right)^2 = \frac{\pi x}{8} \end{align} $\text{Thus, the volume is}$ \begin{align} V = \int_0^4 \frac{\pi x}{8} \ dx = \frac{\pi}{8} \left[\frac{x^2}{2} \right]_0^4 = \pi \end{align}
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