Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.2 - Line Integrals - 16.2 Exercise: 1

Answer

$\frac{4}{3}(10^{\frac{3}{2}} − 1)$

Work Step by Step

$x = t^{2},y=2t, 0 ≤ t≤ 3$ $\int_{C}y ds$ = $\int_{0}^{3}2t\sqrt {{(\frac{dx}{dt})^{2}}+{(\frac{dy}{dt})^{2}}}ds$= $\int_{0}^{3}2t\sqrt {{(2t)^{2}}+{(2)^{2}}}ds$= $\int_{0}^{3}2t\sqrt {{4(t)^{2}}+{4}}ds$= $\int_{0}^{3}4t\sqrt {{t^{2}}+{1}}ds$= $\frac{4}{3}(t^{2}+1)^{\frac{3}{2}}|_{0}^{3}$= $\frac{4}{3}(10^{\frac{3}{2}} − 1)$
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