Answer
We choose the point $P = \left( {\frac{1}{3},2\sqrt 2 } \right) \in {\cal D}$ that satisfies $x{y^2} = \frac{8}{3}$, such that $f\left( P \right) = \bar f$.
Work Step by Step
We have $f\left( {x,y} \right) = x{y^2}$ on ${\cal D} = \left[ {0,1} \right] \times \left[ {0,4} \right]$. We can consider ${\cal D}$ as a vertically simple region whose description is given by
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,0 \le y \le 4} \right\}$
Next, we evaluate the double integral of $f\left( {x,y} \right)$ on ${\cal D}$ as an iterated integral:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^1 \left( {\mathop \smallint \limits_{y = 0}^4 x{y^2}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{1}{3}\mathop \smallint \limits_{x = 0}^1 x\left( {{y^3}|_0^4} \right){\rm{d}}x$
$ = \frac{{64}}{3}\mathop \smallint \limits_{x = 0}^1 x{\rm{d}}x$
$ = \frac{{32}}{3}\left( {{x^2}|_0^1} \right)$
$ = \frac{{32}}{3}$
Since the domain ${\cal D}$ is a rectangle: ${\cal D} = \left[ {0,1} \right] \times \left[ {0,4} \right]$, the area of ${\cal D}$ is $Area\left( {\cal D} \right) = 4$.
By Theorem 5, the Mean Value Theorem for Double Integrals, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = f\left( P \right)\cdot Area\left( {\cal D} \right)$,
where $f\left( P \right) = \bar f$.
Therefore, $\frac{{32}}{3} = 4f\left( P \right)$. So, $f\left( P \right) = \bar f = \frac{8}{3}$.
Thus, we obtain a curve $x{y^2} = \frac{8}{3}$ whose point $P \in {\cal D}$ is the point such that the average $f\left( P \right) = \bar f$ of $f\left( {x,y} \right) = x{y^2}$ on ${\cal D} = \left[ {0,1} \right] \times \left[ {0,4} \right]$. For example, we can choose the point $P = \left( {\frac{1}{3},2\sqrt 2 } \right) \in {\cal D}$ that satisfies $x{y^2} = \frac{8}{3}$, such that $f\left( P \right) = \bar f$. This is illustrated in the figure attached.
