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 657: 27



Work Step by Step

Here,we have $dS=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ This implies that $S=\int_{0}^{4} \sqrt{t^2+(\sqrt{2t+1})^2} dt=\int_{0}^{4} (t+1) dt$ Therefore, $S=[\dfrac{t^{2}}{2}+t]_{0}^{4} $ or, $S=\dfrac{16}{2}+4=8+4=12$
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