It depends on how you define work. Work is sometimes defined as pressure times a change in volume ($p\Delta V$, or $\int p dV$ if the pressure is not constant), which is equivalent to force times distance, as John Rennie defines it in his answer. In this case it's necessary for the volume to change, purely from the definition.

However, another way to define work is something along the lines of "a change in energy that is not due to the transfer of heat or thermal radiation." In this case it is not necessary for the volume to change. A particular example where the word "work" is used in this way is in electronics, when one calculates the work involved in charging a capacitor as $\int V dQ$, where $V$ here is the voltage rather than the volume, and Q is the charge. In this case neither the volume of the capacitor nor the battery charging it changes, but work is said to be done because the internal energy of the capacitor has changed reversibly, without a transfer of heat.

To put it more mathematically, consider the fundamental equation of thermodynamics:
$$
d U = T dS - pdV + \sum_i \mu_i dN_i + \dots
$$
(where the "$\dots$" can include many other optional terms, including a $V dQ$ one for a capacitor). The right-hand side of this equation represents all the ways in which a system's energy can change. The first term ($T dS$) represents a transfer of heat. Some people define "work" as just the second term ($-p dV$), whereas others define it as the sum of all the other terms apart from the first one. Thus, with the second definition you can have work that's associated with a change in volume, but you can also have work that's associated with a change in the chemical composition, charge, or any other conserved quantity.

The question is a little hard to define because, as pointed out in the comments, it's not clear what the "outside volume" is, or how to define it in curved space to compare it to regular space. Still, I think there is a sense in which the answer is "yes".

A regular sphere has area $A = 4 \pi R^2$ and volume $V = 4 \pi R^3/3$, giving a relation $V = (1/6)\sqrt{A^3/\pi}$. We may ask whether there is some situation where a sphere would have a larger volume for a given area. If we imagine that someone standing on the surface of this sphere can only measure the area, they would infer the volume using the above formula, and they would be surprised to know that the interior volume is in fact larger.

There is a situation in which this can happen. For many years, cosmologists thought that the most likely shape for our universe was that of a 3-sphere, which at a fixed cosmological time has a metric given by

$$ds^2 = \frac{dr^2}{1-r^2} + r^2 (d\theta^2 + \sin^2 \theta d\varphi^2)$$

in suitable coordinates. The area of a sphere at a given radius is still $A = 4\pi R^2$, but following the standard methods of differential geometry the volume is

$$V = 4\pi \int_0^R dr\ \frac{r^2}{\sqrt{1-r^2}} = 2\pi \left(\arcsin R - R \sqrt{1-R^2}\right).$$

You can see by plotting that for any $R$, this volume is larger than the one given by the usual formula, even though the area is the same.

Edit in response to your edit: you ask whether it's possible to fit an elephant in a mouse sized box. Unless our conception of space radically changes (again), then the answer is clearly **no**. A mouse sized box is a box in which a mouse fits, i.e., its interior volume is that of a mouse. You're asking whether an object can have an interior volume larger than its interior volume; I hope it's clear that this is not possible. You can, however, fit an elephant in a box with the **surface area** of a mouse.

## Best Answer

The key for producing a nice tone, is a mix of two facts: first, offering the air stream a somehow geometrical regular hole, where a stationary wave can be born with a certain frequency and other possible frequencies are filtered out, and second, a low enough stream velocity, so that no turbulences can happen and the flow is ordered (laminar).

The whistle is produced as an intermediate solution between having a hole that is small enough for the precise stationary wavelengths to happen, and big enough for the air not to flow at much too high a speed. That is why people with their fingers in their mouth can whistle louder, because, by putting the fingers, they create a richer and bigger aperture than a simple, small hole and so air can flow slowly enough but, at the same time, the fingers combined with the lips maintain the involved geometrical distances conveniently small, so that the stationary waves can happen.

When you blow too strongly or through an irregular-shaped hole, air flows in a disordered way, thus vibrating with thousands different modes, and that is why you hear the typical "hiss" sound, because a hiss is a mix of lots of frequencies (technically called White Noise).

Physics is not physics without, at least, a little maths. So now, follow me. You surely know Newton's second law:

$ma = F$

First, we assume that we take the force per unit volume, so that we use the density instead of the mass. And, at the same time, we write the acceleration as the derivative of the velocity:

$\rho \frac{dv}{dt} = f$

Because, as you know, the acceleration is the amount of change of the velocity per unit time. When we deal with fluid mechanics, that amount of change is due to two terms: one deals with the change in time of the velocity in a fixed point of space, and the other is due to the change of the velocity from one point to another. That is written in this way:

$\rho (\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}) = f$

(For those of you who see it for the first time, don't worry about the triangle, it is kind of a sophisticated derivative. Just follow what the terms mean)

The $f$ as you know, stands for forces (due to weight, springs, etc). In fluid dynamics, we like to give a special role to two kind of forces, so that we make them appear separately in the equation:

$\rho (\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}) = -\nabla p + \frac{1}{\mathrm{Re}} \nabla^2 \mathbf{v} +f$

The term with the $p$ is due to the change in pressure from one point to another. The other one with the $\mathrm{Re}$ is due to the viscosity, i.e. a measure of strong the fluid tries to avoid changes in shape (honey has a higher viscosity than water). This is the Navier-Stokes equation for an incompressible flow (the air is compressible, but in the range of speeds involved in a whistle, it can be very good approximated by this equation - for the purists: the Navier Stokes equation is experimentally found to hold, up to a good degree of approximation, in turbulent flows too).

The $Re$ stands for "Reynolds Number". It is a somewhat heuristic quantity that goes proportional to the velocity but inversely proportional to viscosity. It depends too on the geometrical dimensions of the problem, so it is not straightforward to derive its value. The important fact when you whistle is that, if you blow strongly, and thus increase the velocity of air, therefore having a big Reynolds Number, then the forces due to viscosity become less important in the equation (because the viscose term has $\mathrm{Re}$ in the denominator). The viscose forces are the ones that most help maintaining the flow geometrically ordered (laminar). When their contribution is not dominant, the flow becomes disordered (turbulent). A turbulent flow has no mechanical properties that are stable enough in time for a stationary wave to establish, so that your nice whistle vanishes and is replaced by a randomly fluctuating mix of thousands frequencies, that is, the white noise of the hiss...