Understanding how to compute the space volume for an ideal gas

There is a system of N non-interacting particles (Ideal Gas). The Hamiltonian of a system of free particles is given by:

$$H = \sum_{i=1}^{N}\frac{p_{i}^2}{2m} + \sum_{i=1}^{N} \psi(q_i)$$

where to the kinetic term we have added a confining potential:

$$\psi(q_i) = \begin{cases} 0 & q_i \in V\\ \infty & otherwise\end{cases}$$

which keeps the particles inside the volume $$V$$.

First of all, some definitions:

$$\Omega (E, V, N) = \int \frac{d\Gamma}{N!\hbar^3N} \Theta(E - H(\Gamma))$$

Where $$d\Gamma = dp_1...dp_Ndq_1...dq_N$$ (p is momentum and q position) and $$\Theta$$ is the step function.

I want to compute the microcanonical phase space volume for an ideal gas. To do so I have to solve the following integral:

$$\Omega (E, V, N) = \int \frac{d\Gamma}{N!\hbar^3N} \Theta(E - H(\Gamma)) = \int q_1\int q_2\int q_N\int \frac{dp_1...dp_N}{N!\hbar^3N}\Theta(E - \sum(\frac{p_{i}^2}{2m}))$$

I know the solutions follows as:

$$\int q_1\int q_2\int q_N\int \frac{dp_1...dp_N}{N!\hbar^3N}\Theta(E - \sum(\frac{p_{i}^2}{2m})) = \frac{V^N}{N!\hbar^3N}\int_{\sum \frac{p_{i}^2}{2m} \leq E} dp_1...dp_N = \frac{V^N}{N!\hbar^3N} V'_{3N}\sqrt{2mE}$$

But I do not really understand how can we go from having the step function to just the integral over $$dp_1...dp_N$$.

Consider the definition of the step function:$$\Theta(x)=\cases{1,x>0\\0,x<0}$$The step function is equal to $$1$$ if all particles are inside the volume $$V$$ and if $$\sum \frac{p_{i}^2}{2m} \leq E$$. If one particle is outside, $$\psi_i(q_i)=\infty$$, so if everything else is finite you get $$\Theta(-\infty)=0$$. That's why the integral over each of $$dq_i$$ yields a factor of $$V$$. If now the sum over the kinetic energies is grater than $$E$$, the $$\Theta$$ function is also $$0$$, so if you write $$\int dp_1...dp_n=\int_{\sum \frac{p_{i}^2}{2m} \leq E} dp_1...dp_n\Theta(E-\sum \frac{p_{i}^2}{2m})+\int_{\sum \frac{p_{i}^2}{2m} \gt E} dp_1...dp_n\Theta(E-\sum \frac{p_{i}^2}{2m})$$ then the second integral is $$0$$, while the first integral is just $$\int_{\sum \frac{p_{i}^2}{2m} \leq E} dp_1...dp_n$$