Reformulation of double integral as an integral over surface element

Question

Is it correct to directly reformulate:

$$I= \iint f(x,y) \,dx\,dy$$

as

$$I= \iint_{Surf} f(\vec{a})\,d\vec{a},$$ $\vec{a}=(x,y)^\top$, transpose to show that we are working with column vectors. I.e., $,d\vec{a}$ would be an infinitesimal surface element.

And the interpretation, as far as I know, is/can be:

• The scalar $f(\vec{a})$ can be interpreted as the density (or some other property, but lets say density) of some object assigned to each surface element $\vec{a}$ ? (And thus the whole double integral being the density of ''something'' on the object's surface?)

• The scalar $f(\vec{a})$ can be interpreted as the height of an object and therefore the whole integral represents the volume of the object.

Please, validate/correct/develop.

(P.S. I know that there is such a thing w.r.t. magnetic flux https://en.wikipedia.org/wiki/Magnetic_flux itself defined as a double integral over a scalar product of the vector field B and an infinitesimal surf element dS, but this is slighlty different)

Answer 1

Ok, since nobody seems to want to answer, I think I can now answer myself. cf. Wikipedia:density

As we can see on the picture $\rho(\vec{r})$ is the density around point $\vec{r}$. The infinitesimal volument element $dV$ can also be written as $dxdydz$ if we wanted to.

So this allows to make the link between both formulations of the OP, except this time in 3D, but it is exactly the same in 2D but for a surface instead of a volume, i guess.