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 956: 27



Work Step by Step

Given: $\int_C F \cdot T ds-\int_C F. dr=\int_C xy dx +(y-x) dy$ Now, $\int_C xy dx +(y-x) dy=\int_C x(2x-1) dx + \int_C (2x-1-x) (2 )dx$ or, $\int_{1}^{2} 2x^2 +x-2 dx=[\dfrac{2x^3}{3}+\dfrac{x^2}{2}-2x]_1^2$ Thus, $(\dfrac{16}{3})+2-4-(\dfrac{2}{3}) -(\dfrac{1}{2})-2=\dfrac{25}{6}$
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