Thomas' Calculus 13th Edition

Published by Pearson

Chapter 16: Integrals and Vector Fields - Section 16.4 - Green's Theorem in the Plane - Exercises 16.4 - Page 979: 21

Answer

$0$

Work Step by Step

The tangential form for Green's Theorem -- Counterclockwise Circulation $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy$ Now, $\oint_C F \cdot n ds= \iint_{R} \dfrac{\partial (x^2)}{\partial x}-\dfrac{\partial (y^2)}{\partial y} dx dy$ or, $= \int_{0}^1 \int_0^{1-x} 2x-2y dx dy$ or, $= \int_{0}^1 \int_0^{1-x} 2x-2y dx dy$ or, $= \int_{0}^1 2x(1-x) -(1-x^2) dx$ or, $= \int_{0}^1 [4x-3x^2-1] dx$ $[2x^2-x^3-x]_0^1=0$

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