How can the weight of the cone influence the work done?

[Sarangalex]

Homework Statement

A great conical mound of height h is built by the slaves of an oriental moarch, to commemorate a victory over the barbarians. If the slaves simply heap up uniform material found at ground level, and if the total weight of the finished mound is M, show that the work they do is (1/4)hM.

Homework Equations

dW = dF(distance)
W = ∫ρ*dV*(distance)*dx

The Attempt at a Solution

I said dF is equal to ρ*dV and the distance is x.
dV should be equal to ∏r²h*dx.

I just really don't know what to do from this point. What does the given M have to do with anything?

 

[HallsofIvy]

"dx" is not "distance moved", it is the height to which a particular "piece" (dV) of Earth has to be raised. You are told that M is the mass of the finished cone so M is the integral of \rho dV. dV cannot be "\pi r^2 h dx", that has the units of distance⁴, not distance³. The full volume of a cone of height h and radius R is (1/3)\pi R^2h. Since that has weight M, your density is \rho= 3M/(\pi R^2h.

If the final cone has height h and base radius R, then the radius of the cone at height z is r= (R/h)z so each cross section would be a disk of are \pi((R/h)z)^2 and you can "build" the cone out of disks of volume \pi(R^2/h^2)z^2dz where dz is the "thickness of each cone". Multiply that by \rho to get the weigth of that "layer" of Earth and by z for the heigth to which it was lifted. That gives the work done in lifting that particular "layer" of earth. Integrate to find the total work.