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 16.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises 16.2 - Page 955: 16

Answer

$2\sqrt 3-4$

Work Step by Step

Re-write as: $\int_C \sqrt {x+y} \ dx=\int_{C_{2}} \sqrt {x+y} \ dx+\int_{C_{2}} \sqrt {x+y} \ dx + \int_{C_{3}} \sqrt {x+y} \ dx....(1)$ For $C_1$: $\int_{C_{1}} \sqrt {x+y} \ dx=\int_{0}^1 \sqrt {x+3x} \ dx = 2\int_0^1 \sqrt x d x= \dfrac{4}{3}$ For $C_2$: $\int_{C_{2}} \sqrt {x+y} \ dx=\int_{1}^0 \sqrt {x+3} \ dx = \int_1^0 (x+3)^{1/2} dx=[\dfrac{2(x+3)^{3/2}}{3}]_0^1 =2 \sqrt 3 - \dfrac{16}{3}$ For $C_3$: $\int_{C_{3}} \sqrt {x+y} \ dx=\int_{C_3} \sqrt {x+y} (0) =0$ Equation (1) becomes: $\int_C \sqrt {x+y} \ dx= \dfrac{4}{3}+2\sqrt 3 - \dfrac{16}{3}+0=2\sqrt 3-4$
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