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: 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, $ds=\sqrt{(1)^2+( 3 )^2+(1)^2} dt \implies ds= \sqrt {3} dt$ Line integral: $l=\int_C \dfrac{\sqrt 3}{x^2+y^2+z^2} ds$ or, $ (\sqrt{3}) \int_{1}^{\infty} \dfrac{\sqrt 3}{t^2+t^2+t^2} dt=\int_{1}^{\infty} \dfrac{3}{3t^2} dt$ Thus, $l=[\dfrac{-1}{t}]_{1}^{\infty}=1$
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