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 - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 658: 32


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

Work Step by Step

$dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2}=\sqrt{(\sqrt {t})^2+(t^{-1/2})^2}=\sqrt{\dfrac{t^2+1}{t}}$ This implies that $S=\int_{0}^{\sqrt 3} (2 \pi) x ds$ or, $S=\int_{0}^{\sqrt 3} (2 \pi) (\dfrac{2}{3}t^{3/2}) (\sqrt{\dfrac{t^2+1}{t}}) dt=\dfrac{4 \pi}{3}\int_{0}^{\sqrt 3} t\sqrt{t^2+1} dt$ Plug $t^2+1=k \implies (2) dt=dk$ $S=(\dfrac{2 \pi}{3}) \int_{1}^{4} (\sqrt{k}) dk=\dfrac{28 \pi}{9}$
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