Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 21


$\dfrac{76 \pi}{3}$

Work Step by Step

Here, $\dfrac{dx}{dt}= t$ ; $y=2t$ or, $\dfrac{dy}{dt}= 2$ Surface area: $S=\int_0^{\sqrt 5} 2\pi (2t) \sqrt {t^2+4} dt$ Consider $t^2+4=p$ or, $2t=dp$ Thus, $S=\int_4^{9} 2\pi (p)^{1/2} dp$ Hence, $S=2\pi [(\dfrac{2}{3}) p^{(3/2)}]+(4)^{9}=\dfrac{76 \pi}{3}$
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