Volume and surface of resolution

[alext87]

I am doing some analysis of results from an impact crater caused by steel balls. I am measuring the diameter of the impact crater and need to link it to the volume of the crater using the volume of resolution. Can anybody help I have a formula that I derived it is correct?

 

Volume of moved by impact = (2/3) π r3 – (1/12) π(4r2-D2)3/2

where r is the radius of the steel ball and D is the diameter of the crater.

This i think is linking the diameter of impact crater to the volume moved...

 

I also have two other calculus problems:

I need to find out the surface area of the crater in terms of the diameter of the impact crater can anybody help???

 

Then I need the rate of change of surface area with respect to diameter and then to depth???

 

These problems are beyond my mathematical skills...

[insane_alien]

don't you also need the depth of the crater to work out its volume?

[alext87]

As (diameter of crater/2)^2 = (radius of the steel ball)^2 + Radius of steelball - depth of crater)^2

 

which is just the equation of a circle.

 

There is a definite relationship so that if you know diameter of the crater you also know the depth.