## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 469: 38

#### Answer

$$28.12$$

#### Work Step by Step

Since \begin{align*} S&=2\pi \int_{a}^{b}f(x) \sqrt{1+[f'(x)]^2}dx\\ &= 2\pi \int_{0}^{4}x^2\sqrt{1+4x^2}dx\\ \end{align*} Let $2x=\tan \theta \to 2dx= \sec^2\theta d\theta$ and by using $$\int \sec ^{n}(x) \mathrm{d} x=\frac{n-2}{n-1} \int \sec ^{n-2}(x) \mathrm{d} x+\frac{\sec ^{n-2}(x) \tan (x)}{n-1}$$ we get \begin{align*} 2 \pi \int_{0}^{4} x^{2} \sqrt{1+4 x^{2}} d x&=2 \pi \int_{0}^{\tan ^{-1} x^{-1}} \frac{\tan ^{2} \theta}{4} \sqrt{1+\tan ^{2} \theta} \frac{\sec ^{2} \theta}{2} d \theta\\ &=2 \pi \cdot \frac{1}{8} \int_{0}^{\tan ^{-1} 8} \tan ^{2} \theta \sec \theta \sec ^{2} \theta d \theta \\ &=\frac{\pi}{4} \int_{0}^{\tan ^{-1} \frac{1}{8}} \tan ^{2} \theta \sec ^{3} \theta d \theta \\ &=\frac{\pi}{4} \int_{0}^{\tan ^{-1} 8}\left(\sec ^{2} \theta-1\right) \sec ^{3} \theta d \theta \\ &=\frac{\pi}{4} \int_{0}^{\tan ^{-1} 8}\left(\sec ^{5} \theta-\sec ^{3} \theta\right) d \theta\\ &= \frac{\pi}{4} \cdot\left[\frac{\sec ^{3}\left(\tan ^{-1} 8\right) \tan \left(\tan ^{-1} 8\right)}{4}-\frac{1}{8} \sec \left(\tan ^{-1} 8\right) \tan \left(\tan ^{-1} 8\right)-\left(-\frac{1}{8} \ln \left(\sec \left(\tan ^{-1} 8\right)+\tan \left(\tan ^{-1} 8\right)\right)\right)\right] \\&=\frac{\pi}{4}\left[\left(2 \sec ^{3}(82.87)-\sec (82.87)-\frac{1}{8} \ln (\sec (88.87))\right)\right]\\ &\approx 28.12 \end{align*}

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