Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.6 Surface Area - 6.6 Exercises - Page 457: 11

Answer

$\dfrac{\pi}{8} (e^8+e^{-8}+16)$$

Work Step by Step

Let us consider that $f(x)$ is a non-negative function which defines a continuous first derivative on the interval $[a,b]$. The area of the surface obtained when the graph of a function $f(x)$ on the interval $\left[ {a,b} \right]$ is revolved about the $x$- axis can be expressed as: $$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$ We have: $f(x)=y=\dfrac{1}{4}(e^{2x}+e^{-2x}) \implies f'(x)=\dfrac{1}{2}e^{2x}-\dfrac{1}{2} e^{-2x}$ Then $$S=\int_{-2}^{2} 2\pi (\dfrac{1}{4}(e^{2x}+e^{-2x})) \sqrt {1 + (\dfrac{1}{2}e^{2x}-\dfrac{1}{2} e^{-2x})^2} dx \\=2 \pi \times \dfrac{1}{8} \int_{-2}{2}(e^{2x}+e^{-2x}) \times (e^{2x} +e^{-2x}) \ dx \\=\dfrac{\pi}{4} \times [\dfrac{e^{4x}}{4}-\dfrac{e^{-4x}}{4}+2x]_{-2}^2\\=\dfrac{\pi}{4} \times [(\dfrac{e^{4(2)}}{4}-\dfrac{e^{-4(2)}}{4}+4)-(\dfrac{e^{4(-2)}}{4}-\dfrac{e^{-4(-2)}}{4}-4)]\\=\dfrac{\pi}{8} (e^8+e^{-8}+16)$$
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