Matching the accessible phase volume of an ideal gas to Boltzmann's equation

I have been attempting to calculate the entropy increase of a monatomic ideal gas upon addition of a small amount of heat, and have been unable to resolve an apparent contradiction between the requirement that $$dS = dQ/T$$ and Boltzmann's law, $$dS \propto d\log \Omega$$. I describe my dillemma below.

A monatomic ideal gas of constant total energy, $$H$$, and with $$n/3$$ particles each of unit mass has Hamiltonian given by:

$$H = \sum_{i=1}^n \frac{p_i^2}{2}.$$

This may be re-written as

$$(\sqrt{2H})^2 \equiv R^2 = \sum_{i=1}^n p_i^2$$

Written this way, we see that $$R=\sqrt{2H}$$ is the radius of an $$n$$-sphere which the $$p_i$$ coordinates of the phase space of the system are restricted to. We refer to the accessible volume of the system within the phase space as $$\Omega$$ (this accessible volume also includes the position coordinates, however we may ignore them, as they will remain unchanged throughout this analysis, and these will only contribute a constant factor $$\propto V^n$$ in the integral over the accessible phase space).

According to Boltzmann, the entropy of a system is related to the accessible phase volume, $$\Omega$$, by

$$S = k_B \log \Omega$$

And, in particular, for a small change in the entropy, $$dS \propto d\log \Omega$$. Now suppose our monatomic ideal is at unit temperature, and suppose we add a small amount of heat, $$dQ$$, to the gas while keeping its volume constant. We also suppose the energy supplied is so small the temperature may be treated as constant. This will increase the total energy, $$H$$, by $$dH = dQ$$, and for a gas at unit temperature this will increase the entropy by $$dS = dQ = dH$$ as well. Hence we expect the accessible volume of phase space to increase in proportion with $$d\log \Omega$$. The accessible region of the momentum phase space will now be an $$n$$-sphere of slightly greater radius:

$$(\sqrt{2(H + dH)})^2 = \sum_{i=1}^n p_i^2$$

Because the radius of the accessible $$n$$-sphere has increased, the accessible volume of the phase space, $$\Omega$$, will also have increased. In particular, the surface area of a unit sphere is proportional to $$R^{n-1}$$, and so the increase in the volume of the accessible phase space upon the addition of heat $$dQ = dH$$ will be

$$d\Omega \propto (2H+2dH)^{(n-1)/2} - (2H)^{(n-1)/2} \approx ndH = ndS$$

However, Boltzmann's law leads us to expect that the accessible volume will increase as

$$dS = d\log \Omega.$$

Effectively, this naive calculation shows a proportionality between the phase volume and the heat supplied to the system (albeit with a very large constant):

$$d\Omega \propto dS$$

whereas Boltzmann's law suggests an exponential dependence:

$$d\Omega \propto d(\exp S )$$

I'm rather confused about this. I had expected that Boltzmann's law would work in this simplest-possible-case. Why isn't the phase volume growing exponentially as energy is added, as Boltzmann's law predicts?