Answer
$\dfrac{16}{3}$
Work Step by Step
Re-arrange the given integral as follows:
$V=\int_{0}^{2} [4x-xy^2]_{0}^{1}dy$
This implies that
$V=\int_{0}^{2} 4-y^2 dy=[4y-\dfrac{y^3}{3}]_0^2$
Hence, $V=4(2-0)-(\dfrac{8}{3}-0)=\dfrac{16}{3}$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 14 - Section 14.1 - Double and Iterated Integrals over Rectangles - Exercises - Page 760: 30
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