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 16.1 - Line Integrals - Exercises 16.1 - Page 943: 10


$-\sqrt 2$

Work Step by Step

As we know that $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ Here, $ds=\sqrt{1^2+(-1)^2+0^2} dt \implies ds= \sqrt 2 dt$ Line integral: $l=\int_C (x-y+z-2) ds$ or, $\int_0^1 (t-(1-t)+1-2) \sqrt 2 dt=\sqrt 2\int_0^1 (2t-2) dt$ Thus, $l=\int_C (x-y+z-2) ds=\sqrt 2(1-2)=-\sqrt 2$
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