Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Review - Exercises - Page 1150: 33



Work Step by Step

Stoke's Theorem states that $\iint_S curl F\cdot dS=\int_C F \cdot dr$ The equation of the plane with the given intercepts is given as: $\dfrac{x}{1}+\dfrac{y}{1}+\dfrac{z}{1}=1$ This gives: $z=1-x-y$ Use definition of curl $F=\nabla \times F$ Consider $F=ai+bj+ck$ That is, curl $F=(\dfrac{\partial c}{\partial x}-\dfrac{\partial b}{\partial z})\hat{i}+(\dfrac{\partial a}{\partial z}-\dfrac{\partial c}{\partial x})\hat{j}+(\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y})\hat{k}$ So, we have : curl $F=-yi-zj-xk$ $\iint_S curl F\cdot dS=\iint_D-y-z-x dA=-\iint_D dA$ Here, $\iint_D dA$ represents the area of the triangle equal to $\dfrac{1}{2}(Height)(Base)=\dfrac{1}{2}(1)(1)=\dfrac{1}{2}$ Hence, $\iint_S curl F\cdot dS=\int_C F \cdot dr=-\iint_D dA=-\dfrac{1}{2}$
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