Calculus 8th Edition

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

Chapter 16 - Vector Calculus - Review - Exercises - Page 1189: 17


$-8 \pi$

Work Step by Step

Consider the Green's Theorem: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$ Here, $D$ is the region enclosed inside the counter-clockwise oriented loop $C$. Then, we have $\int_C x^2ydx-xy^2 dy=\iint_D (-y^2-x^2) dA$ Here, the region $D$ is inside the circle $x^2+y^2=4$. Now, in polar co-ordinates we have $\int_C x^2ydx-xy^2 dy=\int_0^{2}\int_0^{2\pi} -r^2(r) d\theta dr$ or, $[\theta]_0^{2\pi} [\dfrac{-r^4}{4}]_0^{2}=(2\pi) \cdot (-4)$ Hence, $\int_C x^2ydx-xy^2 dy=-8 \pi$
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