Answer
$\frac{16}{3}$
Work Step by Step
V=$\int \int f(x,y)dA$
=$\int^{1}_{0} \int^{2}_{0}(4-y^2)dy dx$
=$\int^{1}_{0}[4y-\frac{1}{3}y^3]^2_0dx$
=$\int^{1}_{0}(\frac{16}{3})dx$
=$[\frac{16}{3}dx]$
=$\frac{16}{3}$
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: 30
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