what does triple integral represent geometrically? If a single integral represents the area under the curve,
double integral represents volume under the curve,
then what does triple integral represent geometrically?
 A: "Represent geometrically" depends on how you construct the geometric space in your mind.
For a scalar-valued function $f: \mathbb{R}^n\to \mathbb{R}$, if you construct your mental geometric space by putting the output $\mathbb{R}$ on one axis and the input $\mathbb{R}^n$ on other axes, what you get is called "graph".  For example, when $n=2$, you may think of the output $\mathbb{R}$ on the z-axis and the input $\mathbb{R}^2$ on x- and y-axis. Then your graph would be a 3D space.  So if you have triple integral and $n = 3$, your graph is a 4D space and what you are integrating is a hypervolume  (one dimension less of its ambient space).
However, this may not easy to imagine as you may have noticed.  So I would suggest you to construct the geometric space by only putting the input $\mathbb{R}^n$ into the axes.  As for the output $\mathbb{R}$, imagine they are values assigned to each and every single point in this space.  This representation of the function $f$ is called scalar field.  When you are integrating, you are just summing all $f$ weighted by the tiny little volume over the geometric space in your mind.
The following figures show the same function $z=x^2+y^2$ under the representation of graph and scalar field:


A: The premise that a double integral represents a volume is not at all correct, there are only a few cases such as a cylindrically symmetric shape or a shape which can be parameterised as a function of $x$ and $y$ and the plane $z=0$ where just a double integral would give a volume.
A double integral usually represents integration of something over a surface, and a triple integral usually represents integration over a volume. For example if you want to find out the mass of an object, you integrate its density over three dimensions, but if you want to find out the surface area of a body, you do a double integral over a parameterisation of its surface.
