## 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: 34

#### Answer

$\int_{a}^b f(x) \ dx =- \oint_{C} y \ dx$ (Verified)

#### Work Step by Step

Since, $\int_{a}^b f(x) \ dx =\int_{a}^b \int_{0}^f(x) \ dy \ dx =\iint_{R} \ dx \ dy$ The tangential form for Green's Theorem is given as: The counterclockwise circulation is: $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy$ and $Area = \iint_{R} \ dx \ dy = \iint_{R} 0+1 \ dx \ dy= \iint_{R} (\dfrac{\partial (0)}{\partial x} +\dfrac{\partial (y)}{\partial y}) dx dy \\= \oint_{C} 0 \ dy - y \ dx$ Thus, we have: $\int_{a}^b f(x) \ dx =- \oint_{C} y \ dx$ The result has been proved.

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