Answer
$16$
Work Step by Step
Consider:
$I= \iint_{R} f(x,y) dA$
or, $= \int_{0}^{2} \int_{0}^{3} (4-y^2) dx dy$
or, $= \int_{0}^{2}[4x-y^2 x]_0^{3} dy$
or, $=\int_{0}^{2} (12-3y^2) dy$
Thus, we have $I=[12y -y^3]_{0}^{2} =24-18=16$
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 15.2 - Double Integrals over General Regions - Exercises 15.2 - Page 883: 61
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