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: 7

Answer

$-1$

Work Step by Step

Stokes' Theorem can be defined as: $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr $ Now, $$\iint_{S} curl \space F \cdot dS=\iint_{D} -(-2z)(-1) -(-2x) (-1) -2y dA$$ Consider $1-x-y=z$ $$\iint_{S} curl \ F \cdot dS=\iint_{D} -2+2x+2y-2x-2y dA$$ Since, $D$ is the triangle formed by the vertices $(0,0), (1,0)$ and, $(0,1)$, then we have: $$\iint_{S} curl F \cdot dS \\=\iint_{C} F \cdot dr \\=-2 \iint_{D} dA \\=-2 \times \dfrac{1}{2} \\=-1$$
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