Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1124: 1


$\dfrac{4}{3}(10^{3/2} − 1)$

Work Step by Step

Consider $I=\int_{C}y ds=\int_{0}^{3}(2t) \times \sqrt {{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}ds=\int_{0}^{3}2t\sqrt {{(2t)^{2}}+{(2)^{2}}}ds= \int_{0}^{3}2t\sqrt {{4t^2}+{4}}ds=\dfrac{4}{3}|(t^{2}+1)^{3/2}|_{0}^{3}=\dfrac{4}{3}(10^{3/2} − 1)$
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