Calculus 8th Edition

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

Chapter 16 - Vector Calculus - Review - Exercises - Page 1190: 28


$32 \pi \sqrt 3$

Work Step by Step

Here we have $dS=\iint_D\sqrt {1+(1)^2 +(1)^2} dy dx=\iint_D\sqrt {3} dy dx$ and $\iint_S x dS=\iint_D (x^2+y^2) \sqrt {4+x+y}\sqrt 3 dx dy$ or, $\int_0^{2\pi}\int_0^{2}r^2(4+r \cos \theta+r \sin \theta)(\sqrt 3)(r dr) d\theta=(\sqrt 3) \int_0^{2\pi}\int_0^2 4r^3+r^4\cos \theta+r^4 \sin \theta dr d \theta$ or, $=(\sqrt 3) \int_0^{2 \pi} [r^4+\dfrac{r^5(\cos \theta+\sin \theta)}{5}]_0^2 d\theta$ or, $=\sqrt 3 [16 \theta+\dfrac{32(\sin \theta-\cos \theta)}{5}]_0^{2 \pi}$ or, $=32 \pi (\sqrt 3)$
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