Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.6 - Surface Integrals - Exercises 16.6 - Page 1001: 37

Answer

$-32$

Work Step by Step

We have: $F \cdot n =2uv+3v^2 -12$ and $\iint_{S} F \cdot n \ d S=\iint_{S} (2uv+3v^2 -12) \ dS $ or, $=\int_{-2}^{2} \int_{0}^{1} (2uv+3v^2 -12) \ du \ dv$ or, $= \int_{-2}^{2} [u^2 v+3uv^2-12u]_{0}^{1} \ dv$ or, $=\int_{-2}^{2} (v+3v^2-12) \ dv$ or, $=[\dfrac{v^2}{2}+v^3 -12 v]_{-2}^2 $ or, $=(\dfrac{(2)^2}{2}+(2)^3 -12 \times 2) -(\dfrac{(-2)^2}{2}+(-)^3 -12 \times (-2)) $ or, $=-32$
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