Answer
Using Theorem 3 of Section 15.7 and Theorem 4, we prove that the average value $\bar f$ lies between $m$ and $M$.
Then, using the IVT in two variables (Exercise 66), we prove the Mean Value Theorem for Double Integrals (Theorem 5).
Work Step by Step
Let ${\cal D}$ be a closed, bounded, and connected domain.
Recall Theorem 3 of Section 15.7:
If $f\left( {x,y} \right)$ is a continuous function on a closed domain ${\cal D}$, then $f\left( {x,y} \right)$ takes on both a minimum value $m$ and a maximum value $M$. Thus,
$m \le f\left( {x,y} \right) \le M$ ${\ \ \ \ }$ for all $\left( {x,y} \right) \in {\cal D}$
By Eq. (7) of Theorem 4, we have
$m\cdot Area\left( {\cal D} \right) \le \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A \le M\cdot Area\left( {\cal D} \right)$
By definition, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \bar f\cdot Area\left( {\cal D} \right)$
So,
$m\cdot Area\left( {\cal D} \right) \le \bar f\cdot Area\left( {\cal D} \right) \le M\cdot Area\left( {\cal D} \right)$
Therefore, $m \le \bar f \le M$. Hence, the average value $\bar f$ lies between $m$ and $M$.
Now, we prove the Mean Value Theorem for Double Integrals (Theorem 5) using the IVT in two variables (Exercise 66):
Proof.
Since $f\left( {x,y} \right)$ is continuous on ${\cal D}$, by definition:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = \bar f\cdot Area\left( {\cal D} \right)$
We have shown earlier that the average value $\bar f$ lies between $m$ and $M$.
By IVT in two variables (Exercise 66), $f\left( {x,y} \right)$ takes on every value between $m$ and $M$ at some point in ${\cal D}$. Let that point be $P$ such that $f\left( P \right) = \bar f$. Therefore, Eq. (9) of Theorem 5 holds:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}A = f\left( P \right)\cdot Area\left( {\cal D} \right)$
Hence, the Mean Value Theorem for Double Integrals.
