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 944: 41


$2 \pi -2$

Work Step by Step

Given: $\delta =2-z$ As we know that $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ or, $ds=\sqrt{(0)^2+( -\sin t )^2+(\cos t)^2} dt$ and $ ds=(1)dt= dt$ Now, $I_x=\int_C (y^2+z^2) \delta ds=\int_C (y^2+z^2) (2-z) ds$ or, $I_x=\int_0^{\pi} (cos^2 t+\sin^2 t)( 2-\sin t) dt$ So, $I_x= \int_0^{\pi} (2-\sin t dt)=2 \pi -2$
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