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: 31


$\dfrac{\pi a^3}{6}$

Work Step by Step

We have: $|\nabla g|=\sqrt {4x^2+4y^2 +4z^2 }= 2a$ and $\iint_{S} F \cdot n=\dfrac{z^2}{a}$ We set up the integral and solve the flux of $F$ as follows: $\iint_{S} F \cdot n \ d \theta =\iint_{R} (\dfrac{z^2}{a}) (\dfrac{a}{z}) \ dA$ or, $=\iint_{R} \ z \ dA$ or, $=\int_{0}^{\pi/2} \int_{0}^{a} \sqrt {a^2-(x^2+y^2)} \ dx \ dy$ or, $= \int_{0}^{\pi/2} \int_0^a \sqrt {a^2-r^2} \ dr \ d \theta $ or, $= \dfrac{\pi a^3}{6}$
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