University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Practice Exercises - Page 593: 22

Answer

$2\pi(2-\dfrac{3\sqrt 2}{4})$

Work Step by Step

We are given that $x=t^2+(1/2t) \implies \dfrac{dx}{dt}= 2t-\dfrac{1}{2t^2}$ and $y=4\sqrt t \implies \dfrac{dy}{dt}= \dfrac{2}{\sqrt t}$ The surface area is given as follows: $S=\int_{1/\sqrt 2}^{1} 2\pi (t^2+(1/2t)) \sqrt { (2t-\dfrac{1}{2t^2})^2 +(\dfrac{2}{\sqrt t})^2} dt$ Then, $S=(2 \pi)[\dfrac{1}{2}t^4+\dfrac{3}{2} t-\dfrac{1}{8}t^{-2}]_{1/\sqrt 2}^{1}$ Thus, $S=2\pi(2-\dfrac{3\sqrt 2}{4})$
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