phase evolution of MRI signal during RF excitation

Question

I am trying to compute the phase evolution of a proton spin in the presence of a slice selection gradient. I have the following properties for the slice selection gradient and the RF excitation:

[Slice selection gradient]
Peak amplitude: 0.015 T/m
Duration: 0.0042 s.

[RF excitation]: Sinc pulse
Duration: 0.004 s.
flip angle: 30 degrees
center frequency: -2000 Hz

The reason the two durations are slightly different is that the RF pulse begins when the gradient has reached its peak amplitude (after a small ramp time).

Now, I want to track the phase evolution of the signal during the whole process. To do so, I have discretized the signal in time steps of $10^{-6}$ seconds and what makes sense to me is to numerically integrate the frequency over time.

i.e.

$$ \phi(z) = \int \gamma \ G_z z \ dt $$

and I will do this for the whole time when the gradient and the RF pulse is applied (including the ramp up and ramp down time). Does this sound correct?

The reason I ask is that in the book "Handbook for MRI pulse sequence" there is an alternate formula in terms of the RF bandwidth and the isodelay parameter and the phase is given as:

$$ \Delta \phi = 2 \pi \Delta f \Delta t_1 $$

where $\Delta f$ is the bandwidth and $\Delta t_1$ is the isodelay parameter. I am unable to develop any intuition for this formula. So using the method I described before and this one, are they equivalant in terms of the final phase evolution? If yes, I think the numerical integration might be more accurate as it takes the phase dispersion during the gradient ramp time into account?

[EDIT]: I implemented this and what I find is that the phase dispersion that I compute using numerical integration from the beginning of the RF pulse is twice as long as the second formula (which sort of makes sense as the isodelay is approximately half the length of the RF pulse). Why do we not compute the phase dispersion from the beginning of the RF pulse?

Another thing I wanted to clarify was the purpose of doing this. Is this because the MRI slice that will be excited has a certain thickness and the frequency variation along this thin slice would cause the spins to get out of phase with each other and eventually result in signal loss? However, if we assume a true 2D infinitesimally thin slice, this would not be necessary as all the spins in the slice would experience the same frequency and not go out of sync.