University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 32


$\dfrac{28 \pi}{9}$

Work Step by Step

Since, $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $dS=\sqrt{(t^{1/2})^2+(t^{-1/2})^2}=\sqrt{\dfrac{t^2+1}{t}}$ or, $S=\int_{0}^{\sqrt 3} (2 \pi) x ds=\int_{0}^{\sqrt 3} (2 \pi) (2/3t^{3/2}) (\sqrt{\dfrac{t^2+1}{t}}) dt$ or, $S=\dfrac{4 \pi}{3}\int_{0}^{\sqrt 3} t\sqrt{t^2+1} dt$ Suppose $p=t^2+1 \implies dp=2 dt$ $S=\dfrac{2 \pi}{3}\int_{1}^{4} \sqrt{p} dp=\dfrac{28 \pi}{9}$
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