Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.7 Surface Integrals - 16.7 Exercises - Page 1173: 26



Work Step by Step

Since, $\iint_S F \cdot dS=\iint_D [P \dfrac{\partial h}{\partial x}-Q+R\dfrac{\partial h}{\partial z}]dA $ where, $D$ shows the projection of $S$ onto xz-plane. $\iint_S F \cdot n dS=-\iint_D 2\sqrt {4-x^2-y^2} dA= \int_{0}^{2 \pi} \int_0^{2} -2r \sqrt {4-r^2} dr d\theta$ Suppose $4-r^2=p; dp=-2r dr$ $\iint_S F \cdot n dS= \int_{0}^{2 \pi} \int_4^{0} \sqrt p d d\theta= \int_{0}^{2 \pi} [-d \theta] \times \int_4^{0} \sqrt p dp$ Hence, we have $\iint_S F \cdot n dS=[-2 \pi -0] \times [(\dfrac{2}{3}) (4^{3/2}-0)]=(-2 \pi) (\dfrac{2}{3}) (4^{3/2})=-\dfrac{32\pi}{3}$
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