Flux Calculation for Sphere of Radius R and Detector Acceptance D

[hokutose]

Hi everyone,
I have a simple problem for you. If we have an isotropic particle density n(r,E) where r is the radius and E the energy in a sphere of radius R, What would be the particle flux over a single side of a detector of area A inside the sphere?
What would it be also if the detector's acceptance would be D?
The first approach i have tried is that the flux should be around n*V/4pi where n is the particle density and v the mean velocity of the particles and 4pi the total solid angle.The flux over one side of the detector should be 1/2 of the total flux as it is isotropic.
In the case of the acceptance of the detector, simply multiplying the total flux n*v/4pi by the acceptance would be the flux over the detector´s surface.
Thanx and regards.

 

[hokutose]

I have tried the following course; if the surface detector is squared shape with Area equal to A ,then, integrating over the solid angle from the surface of the detector to a random point of an arbitrary surface concentric and coaxial with the detector(lets say a cube), we have that the acceptance will be pi*A.(The result of integrating the product of the differential area of the surface and the solid angle ).
With this, the flux in the detector would be the total flux,ie n*v/4pi multiplied by pi*A.
I'm not sure if this approach is correct, so any suggestion will be welcome.
Thanks in advanced