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: 25



Work Step by Step

Given: $\int_C F \cdot T ds-\int_C F. dr$ Now, $\int_C F \cdot T ds-\int_C F. dr=\int_{4}^1 (x^2) dx-\int_{2}^{-1} y dy$ or, $[\dfrac{x^3}{3}]_{4}^1-[\dfrac{y^2}{2}]_{2}^{-1}=\int_C (x-y) dx +(x+y) dy$ or, $\dfrac{-38}{2}-\dfrac{1}{2}=-\dfrac{39}{2}$
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