## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 16 - Section 16.7 - Surface Integrals - 16.7 Exercise - Page 1133: 26

#### Answer

$-\dfrac{32\pi}{3}$

#### 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, $D$ is projection of $S$ onto xz-plane. $\iint_S F \cdot n dS=-\iint_D 2\sqrt {4-x^2-y^2} dA$ $= \int_{0}^{2 \pi} \int_0^{2} -2r \sqrt {4-r^2} dr d\theta$ Plug $a=4-r^2 \implies da=-2r dr$ $= \int_{0}^{2 \pi} \int_4^{0} \sqrt a da d\theta$ $= \int_{0}^{2 \pi} [-d \theta] \times \int_4^{0} \sqrt a da$ $=[-2 \pi -0] \times [(2/3) (4^{3/2}-0]$ Hence, $Flux=-\dfrac{32\pi}{3}$

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