Density & Integration.... Help?

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SUMMARY

The discussion focuses on calculating the total mass of sawdust in an inverted conical hole with a depth of 19 meters and a top radius of 16 meters. The density of the sawdust varies with depth, defined by the formula ρ(x) = 2.1 + 1.2e^(-1.2x) kg/m^3. The integral setup for mass calculation involves integrating the product of density and volume from 0 to 19 meters, with the volume expressed in terms of the radius as a function of depth. The correct relationship for the radius is s = (16/19)(19-x), and the integral requires squaring the constant (16/19) to find the total mass accurately.

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  • Knowledge of the geometric properties of cones.
  • Ability to set up and evaluate definite integrals.
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Students in calculus or physics courses, particularly those studying applications of integration in real-world scenarios, and anyone interested in mathematical modeling of variable density materials.

sunnnystrong
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Homework Statement



A hole in the ground in the shape of an inverted cone is 19 meters deep and has radius at the top of 16 meters. The cone is filled to the top with sawdust. The density of the sawdust depends upon the depth, x, following the formula ρ(x) = 2.1 + 1.2e^(-1.2x) kg/m^3. Find the total mass of sawdust in the conical hole.

Homework Equations



mass = density * volume

The Attempt at a Solution


[/B]
So I'm just confused as to how to set up my integral?

I want to find the def. ∫ density*volume from 0 to 9
 

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Using the relationship I posted above ^^^

radius of cone = 16m
height of cone = 19m
s = (16/19)(19-x)
So my integral would be...
 

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Anyone? D:
 
*** Figured it out. all you had to do was square the constant (16/19) as well.
 

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