More volume with more mass? Regarding to the Schwarzschild solution, is there more volume with more mass?
Lets take for example a space shuttle (spheric for simplicity) in an stationary orbit at the position $r_1$.
The black hole around which it is orbitting has a mass $M$.
Now, the mass is increased (not interesting, how). $M_2=2\cdot M_1$for example.
The position $r_1$ is kept the same. Doesn't the volume of the spaceshuttle increase due to the mass increase and curvature increase?
According to a comment, the question might be similar to the question whether the volume of the space shuttle increases while moving nearer to the center of the black hole. The curvature there is stronger then outside.
As the metric is $ds^2 = -Bdt^2 + Adr^2 + \text{angular terms}$,
and A is increasing while decreasing r, the volume should increase either.
However, this seems to be dependent on the sign convention: If (+---) is used instead of (-+++), then the volume is decreasing, isn't it? I read somewhere, that Einstein used (+---) and now we are using (-+++) mostly and it is pure convention - but isn't the volume change important to classify?
Addendum: To make the question clearer I want to change it to the following thought experiment:
Mankind finally managed to develop a material which completely isolates from gravity (please, don't ask me how). Real hoverboards (see "Back to the future" 2) are used all-around the planet and, at last, the following experiment is done regularly in the physics undergrads course:
They got an empty cube of this gravity-isolating material. In the middle of the cube, there is a (non-isolating) sphere which can be filled with a material of high gravitational mass. The volume of the cube is measured with and without the mass in the middle (probably by simply pressing water into it). Of course, the undergrads need to be very precise with the volume measurement, carefully keeping temperature the same and everything. Do they measure more volume inside the cube if the middle-sphere is filled with mass?
For clarification: volume is positive here, mass is also positive here. The material of the cube is magical. It just stays where it is and isolates gravity.
(Hm, ok, I see: "It stays where it is" regarding inside the cube is different from "it stays where it is" regarding the room with the students. So, to define even this: It stays where it is regarding the room with the students)
 A: I'm not sure this fully answers your question, but I hope this can help to shed some light on the issue.
Instead of a spherical spaceship I consider a rod in free fall towards a Schwarzschild black hole.

In the Schwarzschild coordinates, the rod occupies the space between $r=r_1$ and $r=r_2$. The length $L$ of the rod can be found by considering the amount of time it would take a light signal to move from one end of the rod to the other. We have
$$L = c \int_{t_1}^{t_2} \text{ d}t,$$
where we assume that the light moves from $(t_1,r_1)$ to $(t_2,r_2)$.
If the light moves in a purely radial direction, we can describe its path by the coordinate functions $t(\lambda)$ and $r(\lambda)$. The equation of motion $ds^2 =0$ then takes the form
$$ g_{tt} \left(\frac{dt}{d\lambda}\right)^2 + g_{rr} \left(\frac{dr}{d\lambda}\right)^2 = 0,$$
which we can rewrite as
$$\left(\frac{dt}{dr}\right)^2 = -\frac{g_{rr}}{g_{tt}}.$$
The length of the rod is then
$$L =  c \int_{r_1}^{r_2} \frac{dt}{dr} \text{ d}r
= c \int_{r_1}^{r_2} \sqrt{-\frac{g_{rr}}{g_{tt}}} \text{ d}r,$$
where I have taken the positive square root because $r_2 > r_1$.
Notice that the length is independent of the signature of the metric, so whether you work with the (-+++) or (+---) metric is purely conventional and will not change the physics.
For the Schwarzschild metric, we obtain explicitly
$$L = r_2 - r_1 + r_s \ln\left(\frac{r_2 - r_s}{r_1 - r_s}\right) > r_2 - r_1.$$
Now what happens if you magically, instantaneously increase the mass of the black hole? I think the length $L$ of the rod stays the same (I'm here assuming that the rod is infinitely stiff), but that it would now "appear shorter" to the distant observer - i.e. it would no longer occupy the entire space between $r_1$ and $r_2$.

Applying this reasoning to the case of a spaceship, the astronauts inside the ship should not feel its volume change when mass is added to the black hole. However, a distant observer might see that the spaceship now looks smaller than it did before.
In this analysis I have assumed that the rod/ship can be considered to be stationary for the duration of our experiment. I am not sure if all the same conclusions would hold if the rod/ship were in motion during the experiment.
A: @BarrierRemoval, you must be aware of what kind of curvature do you speak. Schwarzschild vacuum spacetime is indeed curved. Its Riemann curvature $R^{\rho}_{\sigma \mu \nu}$  is non-zero. On the other hand, its Ricci $R_{\mu \nu}$, and correspondingly $R$ curvatures are zero. Therefore, for a co-moving observer a spaceship does not change its volume. For reference see "Physical and Geometric Interpretations of the Riemann Tensor, Ricci Tensor, and Scalar Curvature", by Lee C. Loveridge https://arxiv.org/abs/gr-qc/0401099v1
A: For a Schwarzschild solution, the Ricci tensor vanishes identically. As the space shuttle moves towards the black hole (or if there is some change in the black hole's mass due to matter falling in) the volume of the shuttle won't change, since volume changes are governed by the Ricci tensor. The Weyl tensor (i.e. the components of the Riemann tensor which are not on the Ricci tensor) does not vanish, but it leads only to shape deformations. Hence, while the shuttle will undergo spaghettification, its volume remains constant while the shape changes. A similar statement will hold for any object in a region where the stress-energy tensor vanishes. However, there is no way to determine what would happen to the cube: since it is speculative, so is any affirmation we can make regarding it. Gravity as we understand can't be isolated. If you are assuming it can be isolated, then we are no longer working with GR and there is no meaning in discussing what would happen from the point of view of GR.
As for the sign conventions, they are just conventions. If you change your preferred metric signature, you must also change the signs on remaining formulae so that the predictions are still the same. Nevertheless, this is meaningless for a Schwarzschild solution, since the Ricci tensor vanishes and there is never any change in volume of infalling objects.
A: The volume of cube will change most probably.
If there is matter or negative matter added into the center of the, special gravity isolated cube. The cube itself is made of exotic matter as it defies gravity in most probability, as normal matter can't defy gravity. In your question , it must be determined whether the spherical mass is
M , -M.
This type of a situation could also result in negative volume in the universe.
The radius gets larger. That’s “how” it changes.
Oh, did you want to know “why” the radius of the Event Horizon changes when you add mass to a Black Hole? Well, then, why didn’t you say so?
The more mass a singularity (or any object) has, the greater its gravity. The greater its gravity, the farther away from the singularity is the point where light cannot escape the gravitational pull of the singularity. And, since the Event Horizon is defined as the maximum distance from a singularity where light cannot escape the gravitational pull of the singularity; that means the radius of the Event Horizon gets larger.
The equation is,
*R_schwarzschild=2GM/c^2 *
and if you understand equations you can see how R is directly proportional to M.
