It depends on how you interpret the symbols.
Your integral could be interpreted as a volume provided that you interpret f(x,y) as a function that defines a surface in 3 dimensional space. So, put z = f(x,y). This defines a surface which projects nicely onto the x-y plane (that is, it never turns vertical). Subdivide the x-y plane into small rectangular segments of sides dx and dy. The segment at position (x,y) has area dx dy , and the height of the surface above the point (x,y) is f(x,y). So the product f(x,y) dx dy is approximately the volume of the rectangular column with base dx dy that reaches up to the surface z = f(x,y). Integration is the process of adding these volumes in the limit as dx and dy tend to zero, and gives the (signed) volume of the vertical solid column defined by the boundary curve on the x-y plane that defines the limits of the integral with vertical walls reaching up to the surface z = f(x,y).
An alternative interpretation is the following. Suppose you have a thin sheet of material (a lamina) with density (more precisely, with mass per unit area) given by f(x,y). Then a small rectangle of material with sides dx and dy will have area dx dy and therefore mass f(x,y) dx dy. Add up the masses of these small rectangles to get the mass of the lamina. In this interpretation, the integral does not describe a 3-dimensional object, but a two dimensional one.
Remember that maths is form without content. The moment you interpret the symbols in a way that associates them with something physical, you are adding content to the maths. A single piece of mathematics is often capable of very different physical or geometrical interpretations. Different physical systems can be described by the same mathematical model.
