Density & Integration.... Help?

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


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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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