Answer
$2$
Work Step by Step
We know that $$ div F=\dfrac{\partial P}{\partial x}i+\dfrac{\partial Q}{\partial y}j $$
$ p=1; Q=1$
So, $$ div F=\dfrac{\partial}{\partial x}[x]+\dfrac{\partial}{\partial y} [y] \\=1+1 \\=2$$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 16: Integrals and Vector Fields - Section 16.8 - The Divergence Theorem and a Unified Theory - Exercises 16.8 - Page 1025: 2
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