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Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 64

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Find the volume obtained by rotating the region bounded by $y = \sec x$ and $y = \cos x$ from $\displaystyle 0 \leq x \leq \frac{\pi}{3}$ about $x$-axis.

By using vertical strips, notice that if you slice the figure, you'll get a cross section of a washer with outer radius $r_o = 1 + \sec x$ and inner radius $r_i = 1 + \cos x$. So, the cross sectional area is computed by subracting the outer circle to the inner circle. A washer = $A_{\text{outer}} - A_{\text{inner}} = \pi (1 + \sec x)^2 - \pi (1 + \cos x)^2$. Thus, the volume is...

$ \begin{equation} \begin{aligned} V = \int^b_a A (x) dx &= \int^{\pi/3}_0 \left[ \pi (1 + \sec x)^2 - \pi (1 + \cos x)^2 \right] dx\\ \\ &= \int^{\pi/3}_0 \left( 1 + 2 \sec x + \sec ^2 x - 1 - 2 \cos x - \cos^2 x \right) dx\\ \\ &= \int^{\pi/3}_0 \left( \sec^2 x + 2 \sec x - \cos^2 x - 2 \cos x\right) dx && \text{ recall that } \cos^2x = \frac{1+ \cos(2x)}{2}\\ \\ &= \int^{\pi/3}_0 \left(\sec^2 x + 2\sec x + \left[ \frac{1}{2} + \frac{\cos(2x)}{2} \right] - 2 \cos x \right) dx\\ \\ &= \pi \left[ \tan x + 2 \ln |\sec x + \tan x| - \frac{1}{2} x - \frac{1}{2} \left( \frac{1}{2} \right) \sin (2x) - 2 \sin x\right]^{\pi/3}_0\\ \\ &= 5.95 \text{ cubic units} \end{aligned} \end{equation} $