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: 13


$3 \sqrt{14}$

Work Step by Step

As we know that $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ Here, $ds=\sqrt{(-1)^2+( -3 )^2+(-2)^2} dt \implies ds= \sqrt {14} dt$ Line integral: $l=\int_C (x+y+z) ds$ or, $(\sqrt{14}) \int_{0}^{1} (1-t+2-3t+3-2t) dt=\sqrt{14}\int_{0}^{1} (2-2t) dt$ or, $=3 \sqrt {14} (2-1)$ Thus, $l=\int_C (x+y+z) ds=3 \sqrt{14}$
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