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: 18

Answer

Counterclockwise Circulation =$\dfrac{-44}{15}$

Work Step by Step

The tangential form for Green's Theorem - Counterclockwise Circulation can be defined as: $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy$ So, $M=(y+e^x \ln y) ; N=\dfrac{e^x}{y}$ Now, $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial (\dfrac{e^x}{y})}{\partial x}-\dfrac{\partial (y+e^x \ln y)}{\partial y}) dx dy$ or, $\iint_{R} (\dfrac{e^x}{y}-1-\dfrac{e^x}{y}) dx dy =- \iint \space dx \space dy$ or, $=-\int_{-1}^{1} \int_{x^4+1}^{3-x^2} dy dx$ or, $=-\int_{-1}^{1} [y]_{x^4+1}^{3-x^2}$ or, $=-\int_{-1}^{1} (2-x^2-x^4) dx$ or, $=\dfrac{-44}{15}$

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