Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises 16.2 - Page 955: 14



Work Step by Step

Given: $y=t^2 \implies dy=2t dt$ Now, $\int_C \dfrac{x}{y} dx=\int_1^2 (\dfrac{t}{t^2})(2t) dt$ $\int_1^2 [2] dt=[2t]_1^2$ or, $=4-2$ Thus, $\int_C \dfrac{x}{y} dx=2$
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