Answer
1
Work Step by Step
V=$\int\int f(x,y)dA$
=$\int^{1}_{0} \int^{1}_{0}(2-x-y)dy dx$
=$\int^{1}_{0}[2y-xy-\frac{1}{2}y^2]^1_0dx$
=$\int^{1}_{0}(\frac{3}{2}-x)dx$
=$[\frac{3}{2}x-\frac{1}{2}x^2]^1_0$
=1
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 15: Multiple Integrals - Section 15.1 - Double and Iterated Integrals over Rectangles - Exercises 15.1 - Page 875: 27
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