Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.4 - Green's Theorem in the Plane - Exercises 16.4 - Page 979: 19

Answer

$\dfrac{2}{33}$

Work Step by Step

The tangential form for Green's Theorem - work done can be defined as: $W=\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $ $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $ So, $M=2xy^3 ; N=4x^2y^2$ $W=\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial (4x^2y^2)}{\partial x}-\dfrac{\partial (2xy^3)}{\partial y}) dx dy=\iint_{R}$ or, $W= 2xy^2 dx dy=\int_{0}^{1} \int_{0}^{x^3} 2xy^2 dx dy$ So, work done: $W=\int_{0}^{1} [\dfrac{2xy^3}{3}]_{0}^{x^3} dx=\dfrac{2}{33}$
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