Power and intensity of a sound wave

So in the case of a point source of a sound wave in a sphere, I'm wondering why the Intensity decreases with $$r^2$$ because if i setup the formula the Power always increases with $$r^2$$. So these two factors should cancel eachother out no?

So to start $$I = \frac{P}{A}$$ with $$A=4\pi r^2$$

And $$E_{mech}=\frac{1}{2}m\omega^2\Delta s^2$$ with $$\Delta s$$ being the amplitude.

we substitute m with $$m=\rho V_{sphere}$$

$$m=\rho \frac{4}{3}\pi \Delta x^3$$ using $$\Delta x$$ as the radius.

plugging everything in the equation for $$P=\frac{\Delta E}{\Delta t}$$ we get: $$P=\frac{\frac{1}{2}\rho \frac{4}{3}\pi \Delta x^3\omega^2\Delta s^2}{\Delta t}$$ in this we substitute $$v_{x} = \frac{\Delta x}{\Delta t}$$ and get: $$P=\frac{1}{2}\rho \frac{4}{3}\pi \Delta x^2 v_{x}\omega^2\Delta s^2$$}\$

If we substitute this in the equation for I and remember that $$r=\Delta x$$ we get: $$I=\frac{\frac{2}{3}\rho \pi \Delta x^2 v_{x}\omega^2\Delta s^2}{4 \pi \Delta x^2}$$

$$I=\frac{1}{6}\rho v_{x}\omega^2\Delta s^2$$

from which we can clearly see that the intensity does not decline with distance. But ofcourse I know that in reality this is clearly the case. So my question is, what am I doing wrong? or what factor am i forgetting?

In most of these sorts of theories the amplitude decays like $$1/r$$. For example in quantum electrodynamics, the quantum amplitude for a photon to travel through space to a point a distance $$r$$ away is $$\psi(r) = \frac1r e^{2\pi i r/\lambda},$$where $$\lambda$$ is the photon's wavelength. The thing on the left is the magnitude of the amplitude; it diminishing by $$1/r$$ allows the intensity $$|\psi|^2$$ to fall off like $$1/r^2$$; the thing on the right is basically a rotation matrix by angle $$2\pi~r/\lambda$$ and it does all of the wavy stuff that light as an electromagnetic wave does.