I have no idea what the "official" answer is, if any, and this might just be a repeat of Glen_b's answer with more calculus-y language (I don't know), but the whole idea is to be analogous to the continuous version of the arithmetic mean, right? Well the arithmetic mean of a finite number of elements can be defined as:
$$\sum_{i=1}^n \frac{1}{n} x_i$$
For the arithmetic mean of the values of f(x) of a continuous region of real-number x values starting from a and ending in b, we instead write:
$$\frac{\int_a^b f(x)dx}{b-a}$$
The logic I'm missing here is explaining exactly what x values we're summing together, which is something to do with Riemann sums, but the point is that we can do it.
Similarly the geometric mean could be written as:
$$\prod_{i=1}^n x_i^{\frac{1}{n}}
= e^{\ln\left(\prod_{i=1}^n x_i^{\frac{1}{n}}\right)}
= e^{\sum_{i=1}^n \frac{1}{n} \ln(x_i)}
= \exp\left( \sum_{i=1}^n \frac{1}{n} ln(x_i) \right)$$
This is just the definition of an arithmetic mean stuck in an exponent. Thus, it would be intuitive at least to generalize and say that the "geometric mean" of the values of f(x) over a continuous region of real-number x values starting with a and ending in b is:
$$\exp\left(\frac{\int_a^b \ln(f(x))dx}{b-a}\right)$$
As Glen_b rightly pointed out, this only works if f(x) is always non-negative, because ln(x) is undefined for negative x values, but how often to people use geometric means on negative numbers anyway? It's only even possible if there are an odd number of values you're finding the mean of or if their product is positive, despite some being negative, and even then, I'm not sure what significance the number you would get would have.
To generalize to higher dimensions, i.e. functions of more than one variable, we can either observe that a-b is the length of the line we integrated across, or observe that:
$$\frac{\int_a^b f(x)dx}{b-a} = \frac{\int_a^b f(x)dx}{\int_a^b dx}$$
Thus for a double, triple, quadruple, etc. integrals of the values of a function of multiple variables over a continuous region, we divide by the area of that integral by area, volume, or whatever sort of hypervolume of the region we're integrating across:
$$\frac{\iint_A f(x,y) dx dy}{\iint_A dx dy}$$
$$\frac{\iiint_V f(x,y,z) dx dy dz}{\iiint_V dx dy dz}$$
$$\frac{\iiiint_{HV} f(x_1,x_2,x_3,x_4) dx dy dz}{\iiiint_{HV} dx_1 dx_2 dx_3 dx_4}$$
$$etc.$$
Thus, we can just do the same with continuous "geometric means":
$$\exp\left(\frac{\iint_A \ln(f(x,y)) dx dy}{\iint_A dx dy}\right)$$
$$\exp\left(\frac{\iiint_V \ln{f(x,y,z)} dx dy dz}{\iiint_V dx dy dz}\right)$$
$$\exp\left(\frac{\iiiint_{HV} \ln(f(x_1,x_2,x_3,x_4)) dx dy dz}{\iiiint_{HV} dx_1 dx_2 dx_3 dx_4}\right)$$
$$etc.$$
