Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 15 - Vector Analysis - 15.2 Exercises - Page 1063: 56

Answer

$\int_C (2x-y)dx+(x+3y)dy = 6$

Work Step by Step

The path is along y axis from $y=0$ to $y=2$ Hence, $x=0$ and $dx = 0$ on every point $\therefore \int_C (2x-y)dx+(x+3y)dy \\ = \int_{y=0}^23ydy = [\frac{3y^2}{2}]_0^2 = 6$
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