Calculus: Early Transcendentals 8th Edition

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

Chapter 16 - Section 16.7 - Surface Integrals - 16.7 Exercise - Page 1133: 24

Answer

$-\dfrac{1712 \pi}{15}$

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. $\iint_S F \cdot n dS=\iint_D \sqrt {x^2+y^2}+z^3 dA$ $=\iint_D \sqrt {x^2+y^2}+(x^2+y^2) \sqrt {x^2+y^2}dA $ $= \int_{1}^{3} \int_0^{2 \pi} (r^2+1) \sqrt {r^2} r d\theta dr$ $= \int_{1}^{3} (r^4+r^2) dr \int_0^{2 \pi} d\theta$ $= 2 \pi \int_{1}^{3}(r^4+r^2) dr$ or, $=2 \pi [\dfrac{r^5}{5}+\dfrac{r^3}{3}]_1^{3}$ Hence, $Flux=-\dfrac{1712 \pi}{15}$
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