A double integral, in Calculus III is done using Reimann Sums, dividing the area into rectangles.
Thomas's Calculus says that the norm of the partition is the largest width or height of any rectangle in the partition.
This video, at around minute 6.40 tells met the norm of the partition is the largest area.
Does area work in this context? Since I can then make narrow, long rectangles to get values that are different from the actual volumes.
The usual definition of the norm or mesh of a partition of a region in $\mathbb{R}^n$ is that it is the maximum dimension of any rectangle in the partition. This essentially comes down to the fact that you can control the size of a function "near" a point (via, e.g., continuity), but you can't necessarily control the size of a function far aways—hence there is a need to keep points in any part of a partition "close" to each other.
Note that if you relax this requirement, and only look at the area, bad things can happen. For example, consider a function $f : [0,1]^2 \to \mathbb{R}$ defined by
$$ f(x,y) = \begin{cases} 1 & \text{if $y=0$, and} \\ 0 & \text{otherwise.} \end{cases}$$
Consider a partition by intervals of the form $[x_j,x_j+\varepsilon]\times[0,1]$. The upper Riemann sum on such a partition will always be $1$, while the lower Riemann sum will always be $0$. As this disagree, we might conclude that the function is not integrable, which is a problem.
I will note that the example I gave is not continuous, but a similar continuous example can be obtained by considering a function like
$$ g : [0,1]^2 \to \mathbb{R} : (x,y) \mapsto y. $$