Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 18 - Fundamental Theorems of Vector Analysis - 18.1 Green's Theorem - Exercises - Page 983: 3

Answer

$$\int_C y^2 \ dx+x^2 \ dy=0$$

Work Step by Step

Green's Theorem states that: $\int_C fdx+g dy=\iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA$ Here, we have $f=y^2 ; g =x^2$ Now, $\dfrac{\partial g}{\partial x} =2x$ and $\dfrac{\partial f}{\partial y}=2y$ Therefore, $ \iint_D (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y})dA=\iint_D (2x-2y) \ dx \ dy = 2 \iint_D x \ dx \ dy -2 \iint_D y \ dx \ dy $ But by symmetry $ \iint_D x \ dx \ dy = \iint_D y \ dx \ dy=0 $ Thus, $\int_C y^2 \ dx+x^2 \ dy=0$
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