Calculus, 10th Edition (Anton)

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

Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 60



Work Step by Step

When applying the method of disk to find V(I+II) and V(II), we find: V= V(I+II) - V(II)= $\int$ $\pi$$y_{1}^{2}$dx - $\int$ $\pi$$y_{2}^{2}$dx --We know, $y_{1}$=1 and $y_{2}$=tanx & x∈[0, $\frac{\pi}{4}$]. So V=$\int$ $\pi$$(1)^{2}$dx- $\int$ $\pi$$tan^{2}x$dx We apply the identity $tan^{2}x$=$sec^{2}x$-1: V=$\int$ $\pi$$(1)^{2}$dx- $\int$ $\pi$($sec^{2}x-1)$dx Then we split the ingetral and rearrange the expression: V=2$\int$ $\pi$dx -$\pi$$\int$$sec^{2}x$dx Now $\int$$sec^{2}x$dx=tanx, so V=(2$\pi$x-$\pi$tanx)|($x_{i}$=0, $x_{f}$=$\frac{\pi}{4}$)=$\pi$($\frac{\pi}{2}$-1)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.