Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.8 - Stokes'' Theorem - 16.8 Exercise - Page 1200: 8

Answer

$\dfrac{1}{24}$

Work Step by Step

Stokes' Theorem can be defined as: $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr $ and $curl \ F=(x-y) i-yj +k$ Since, $D$ is the triangle formed by the vertices $(0,0), (1/3,0)$ and $(0,1/2)$, then we have: $\iint_{S} curl \space F \cdot dS=\iint_{D} 3x-5y+1 dA$ The equation of joining points $(1/3,0)$ and $(0,1/2)$ is: $\dfrac{x}{(1/3)}+\dfrac{y}{(1/2)}=1$ and $y=\dfrac{1-3x}{2}$ So, $\iint_{D} 3x-5y+1 dA=\int_{0}^{1/3} [3xy -\dfrac{5y^2}{2}+y]_0^{(1-3x)/2} dx$ Consider $y=\dfrac{1-3x}{2}$ $$\iint_{D} 3x-5y+1 dA=\int_{0}^{1/3} [\dfrac{3x-9x^2}{2} -\dfrac{45x^2}{8}+\dfrac{3x}{2}] \ dx \\=\dfrac{1}{8} \int_0^{1/3} [24x -81x^2] \ dx \\=\dfrac{1}{8}[12x^2-27 x^3]_0^{1/3 }\\=\dfrac{1}{8} (\dfrac{4}{3}-1) \\=\dfrac{1}{24}$$
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