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: 18

Answer

$$\dfrac{8\pi}{3} (5\sqrt 5-2 \sqrt 2)$$

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: $$S = \int_a^b 2\pi f(x) \sqrt {1 + [f'(x)]^2} dx $$ Rewrite the function as: $y=\dfrac{x^2}{4} \implies f(y)=2y^{1/2}$ We have: $f(y)=2y^{1/2} \implies f'(y)=2 \times \dfrac{1}{2 \sqrt y}$ Then $$S=\int_{1}^{4} 4\pi \sqrt {y+1} \ dy \\=4 \pi [\dfrac{2}{3}(y+1)^{3/2}]_1^4\\=4 \pi \times \dfrac{2}{3} \times [(1+4)^{3/2}-(1+1)^{3/2}] \\=\dfrac{8\pi}{3} (5\sqrt 5-2 \sqrt 2)$$
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