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: 1

Answer

$15 \pi \ unit^2$

Work Step by Step

The area of the surface generated 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: $$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$ Then $$S=\int_0^4 2\pi (\dfrac{3x}{4}) \sqrt {1 + \dfrac{9}{16}} dx \\=\int_0^4 \pi (\dfrac{3x}{8}) \sqrt {16 +9} dx\\=\dfrac{15\pi}{6}|x^2|_0^4\\=15 \pi \ unit^2$$
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