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: 26



Work Step by Step

Given: $\int_C F \cdot T ds-\int_C F. dr$ Now, $\int_C y dx-x dy=\int_{0}^{\pi/2}(\sin t) (-\sin t dt) -(\cos t) (\cos t dt)=\int_{0}^{\pi/2} (-1) dt$ Thus, $\int_C y dx-x dy=\dfrac{-\pi}{2}$
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