Answer
k=$\frac{2}{27}$
Work Step by Step
$\int^{2}_{1} \int^{3}_{1}kx^2y dx dy$
=$\int^{2}_{1}[\frac{k}{3}x^3y]^{x=3}_{x=0}]dx$
=$\int^{2}_{1}9ky dy= \frac{9}{2}ky^]| ^2_1$
=k=$\frac{2}{27}$
=Thus we choose k=$\frac{2}{27}$
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: 31
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