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 362: 10

Answer

$\text{The volume is}$ \begin{align} V = \sqrt 3 -1 \end{align}

Work Step by Step

$\text{It is given that}$ \begin{align} y = \sec x; x = \frac{\pi}{4} \ and \ x = \frac{\pi}{3} \end{align} $\text{To find the volume, we have to find the area of the cross section.}$ $\text{It is given that the cross section is squre, therefore:}$ \begin{align} A(x) = y^2 = \sec^2 x \end{align} $\text{Thus, the volume is}$ \begin{align} V = \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \sec^2 x \ dx = \left[\tan x \right]_{\frac{\pi}{4}}^{\frac{\pi}{3}} = \sqrt 3 -1 \end{align}
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