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: 15


$\dfrac{5 \sqrt 5+9}{6}$

Work Step by Step

Consider the integral such as: $\int_C f(x,y,z) ds=\int_{C_1} f(x,y,z) ds+\int_{C_2} f(x,y,z) ds$ and $\int_C f(x,y,z) ds=\int_{0}^{1} (2t)\sqrt{1^2+(2t)^2+(0)^2} dt+\int_{0}^{1} (2-t^2)\sqrt{0^2+0^2+1^2} dt$ $\int_C f(x,y,z) ds=\int_{0}^{1} (2t) \sqrt {4t^2+1} dt+\int_{0}^{1} (2-t^2)dt=(\dfrac{1}{6})[(4t^2+1)]_{1}^{5}+[2t-\dfrac{t^3}{3}]_{0}^{1}$ Thus, $\int_C f(x,y,z) ds=\dfrac{(5 \sqrt 5-1)}{6}+\dfrac{5}{3}=\dfrac{5 \sqrt 5+9}{6}$
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