Answer
$\frac{4}{3}$
Work Step by Step
$\int^2_0 \int ^x_0 ydydx $
=$\int^{\pi/4}_0 \int^{2sec\theta}_0 r^2 sin \theta drd\theta $
=$\frac{8}{3}\int^{\pi/4}_0 tan\theta \sec^2\theta $
=$\frac{4}{3}$
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 15.4 - Double Integrals in Polar Form - Exercises 15.4 - Page 893: 14
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