Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.4 - Green''s Theorem - 16.4 Exercise - Page 1160: 10

Answer

$0$

Work Step by Step

Green's Theorem states that: $\oint_CP\,dx+Q\,dy=\iint_{D}(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y})dA$ The parametric representation for the boundary of the ellipse can be written as: $x= \sqrt 2 \cos \theta; y= \sin \theta$ and {$(r, \theta) \in D| 0 \leq r \leq 1; 0 \leq \theta \leq 2 \pi$} We can work out the integrand of the double integral as follows: $\oint_C y^4 dx+2xy^3 dy=\iint_{D}(\dfrac{\partial (2xy^3 )}{\partial x}-\dfrac{\partial (y^4 ) }{\partial y})dA=\iint_{D} -2y^3 dA$ Rewriting the double integral as an iterated integral and solving, we get: $\iint_{D} -2y^3 dA=\int_{0}^{2 \pi} \int_{0}^{1} -2 (r \sin \theta)^3 (r \sqrt 2) dr d \theta$ Consider $I=\int_{0}^{2 \pi} \int_{0}^{1} -2 (r \sin \theta)^3 (r \sqrt 2) dr d \theta \\ =-2 \sqrt 2 (\int_{0}^{2 \pi} \sin^3 \theta d\theta ) ( \int_0^1 r^4 dr)$ Since, $\int_{0}^{2 \pi} \sin^3 \theta d\theta =0 $ Therefore, $\oint_C y^4 dx+2xy^3 dy=-2 \sqrt 2 \times (0) \times \int_0^1 r^4 dr=0$
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