Answer
A function $F\left( {x,y} \right)$ satisfying $\frac{{{\partial ^2}F}}{{\partial x\partial y}} = 6{x^2}y$ is
$F\left( {x,y} \right) = {x^3}{y^2} + C$,
where $C$ is an integration constant.
Using the result of Exercise 50, we evaluate
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} 6{x^2}y{\rm{d}}A = 16$
Work Step by Step
We have $\frac{{{\partial ^2}F}}{{\partial x\partial y}} = 6{x^2}y$ and ${\cal R} = \left[ {0,1} \right] \times \left[ {0,4} \right]$.
To find $F\left( {x,y} \right)$ we take the double integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} \frac{{{\partial ^2}F}}{{\partial x\partial y}}{\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} 6{x^2}y{\rm{d}}y{\rm{d}}x$
$ = \mathop \smallint \limits_x^{} \left( {\mathop \smallint \limits_y^{} 6{x^2}y{\rm{d}}y} \right){\rm{d}}x$
$ = \mathop \smallint \limits_x^{} \left( {3{x^2}{y^2}} \right){\rm{d}}x$
$ = {x^3}{y^2} + C$
So, $F\left( {x,y} \right) = {x^3}{y^2} + C$, where $C$ is an integration constant.
Next, we evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} 6{x^2}y{\rm{d}}A$
Recall from the result of Exercise 50:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = F\left( {b,d} \right) - F\left( {a,d} \right) - F\left( {b,c} \right) + F\left( {a,c} \right)$
where $f\left( {x,y} \right) = \frac{{{\partial ^2}F}}{{\partial x\partial y}}$ and ${\cal R} = \left[ {a,b} \right] \times \left[ {c,d} \right]$.
We have $f\left( {x,y} \right) = \frac{{{\partial ^2}F}}{{\partial x\partial y}} = 6{x^2}y$.
Since ${\cal R} = \left[ {0,1} \right] \times \left[ {0,4} \right]$, using the result of Exercise 50 we get
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} 6{x^2}y{\rm{d}}A$
$ = F\left( {1,4} \right) - F\left( {0,4} \right) - F\left( {1,0} \right) + F\left( {0,0} \right)$
$ = 16$
Notice that the constant of integration $C$ cancels out in the calculation above.
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} 6{x^2}y{\rm{d}}A = 16$.
