University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 15 - Section 15.1 - Line Integrals - Exercises - Page 827: 25

Answer

$\dfrac{ 5 \sqrt 5+7\sqrt 2-1}{6}$

Work Step by Step

Here, we have $\int_C f(x,y,z) ds=\int_{C_1} f(x,y,z) ds+\int_{C_2} f(x,y,z) ds$ $\int_C f(x,y,z) ds=\int_{0}^{1} 2t \sqrt {1+4t^2} dt+\int_{1}^{0}(t+\sqrt t) \sqrt 2 dt$ This implies that $\dfrac{1}{4} [\dfrac{2(1+4t^2)^{3/2}}{3}]_0^1+\sqrt 2 [\dfrac{t^2}{2}+\dfrac{2t^{3/2}}{3}]_1^0=\dfrac{ 5 \sqrt 5+7\sqrt 2-1}{6}$
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