Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - Review - Exercises - Page 1189: 5

Answer

$\int_Cy^{3}dx+x^{2}dy=\frac{4}{15}$

Work Step by Step

We can parameterize the parabola $x=1-y^{2}$ as $C:x=1-y^{2},y=t$ As we move from $(0,-1)$ to $(0,1)$ , $t$ increases from $-1$ to $1$. Therefore, we will integrate from -1 to 1. Note that $dx=-2tdt$ and $dy=dt$ Therefore, $\int_Cy^{3}dx+x^{2}dy=\int_{-1}^{1}(t)^{3}(-2tdt)(1-t^{2})^{2}dt$ $=\int_{-1}^{1}(-2t^{4}+t^{4}-2t^{2}+1)dt$ $=\int_{-1}^{1}(-t^{4}-2t^{2}+1)dt$ $=[(\frac{-t^{5}}{5}-\frac{2t^{3}}{3}+t)]_{-1}^{1}$ $=2(\frac{-3}{15}-\frac{10}{15}+\frac{15}{15})$ Hence, $\int_Cy^{3}dx+x^{2}dy=\frac{4}{15}$
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