Answer
$\sqrt 2$
Work Step by Step
Re-arrange the given integral as follows:
$V=\int_{0}^{\pi/2} [2 \sin x \sin y]_{0}^{\pi/4}dx$
This implies that
$V=\int_{0}^{\pi/2} \sqrt 2\sin x dx$
Hence, $V=\sqrt 2 [-\cos x]_{0}^{\pi/2}=\sqrt 2$
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: 29
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