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