Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.1 Integration in Two Variables - Exercises - Page 847: 43

Answer

$m = \frac{3}{4}$

Work Step by Step

We have $f\left( {x,y} \right) = mx{y^2}$, where $m$ is a constant, and the domain ${\cal R} = \left[ {0,1} \right] \times \left[ {0,2} \right]$. Evaluate the double integral $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A$. Write $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{y = 0}^2 \mathop \smallint \limits_{x = 0}^1 mx{y^2}{\rm{d}}x{\rm{d}}y$ $ = m\mathop \smallint \limits_{y = 0}^2 \left( {\mathop \smallint \limits_{x = 0}^1 x{\rm{d}}x} \right){y^2}{\rm{d}}y$ $ = m\mathop \smallint \limits_{y = 0}^2 \left( {\frac{1}{2}{x^2}|_0^1} \right){y^2}{\rm{d}}y$ $ = \frac{m}{2}\mathop \smallint \limits_{y = 0}^2 {y^2}{\rm{d}}y$ $ = \frac{m}{2}\left( {\frac{1}{3}{y^3}|_0^2} \right)$ $ = \frac{{4m}}{3}$ For $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = 1$, we solve the equation $\frac{{4m}}{3} = 1$. So, $m = \frac{3}{4}$. Thus, the value of $m$ such that $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = 1$ is $m = \frac{3}{4}$.
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