## Calculus, 10th Edition (Anton)

Published by Wiley

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

#### Answer

$\pi$($\frac{\pi}{2}$-1)

#### 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$$(1)^{2}dx- \int \pi(sec^{2}x-1)dx Then we split the ingetral and rearrange the expression: V=2\int \pidx -\pi$$\int$$sec^{2}xdx Now \int$$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)

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