## Calculus 8th Edition

Published by Cengage

# Chapter 16 - Vector Calculus - 16.7 Surface Integrals - 16.7 Exercises - Page 1173: 22

#### Answer

$\pi$

#### Work Step by Step

Since, $\iint_S F \cdot dS=\iint_S F \cdot n dS$ wWhen the surface is orientable then the flux through a surface is defined as: $\iint_S F \cdot dS=\iint_S F \cdot n dS$ where, $n$ is the unit vector and $dS=n dA=\pm \dfrac{(r_u \times r_v)}{|r_u \times r_v|}|r_u \times r_v| dA=\pm (r_u \times r_v) dA$ $\iint_S F \cdot n dS=\iint_D (vi+u \sin v j+u \cos v k) \cdot (\sin v i-\cos v j+uk) dA= \int_{0}^{1} \int_0^{\pi} [v \sin v -u \times \sin v \times \cos v +u^2 \cos v ] dv du$ $\iint_S F \cdot n dS= \int_{0}^{1} [(- v \cos v + \sin v) +\dfrac{u \cos 2 v}{4} +u^2 \sin v]_0^{\pi} du= \int_{0}^{1} \pi du=[\pi^2/2]_0^1\pi=\pi$

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