Setting up a work problem. pumping gasoline

bakin
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I just have a quick question. I'm supposed to find the work required to pump gas out of a cone, but am having trouble finding the radius of the cone.

The cone measures 10 feet in height. I broke it up and set it up using similar triangles, and this is what I got:
4l5h449.jpg

It doesn't seem right though. And, if it is right, where do I go from there? Any help is appreciated :)
 
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if the height is 10, then y must be the difference from 10, or is that straight from the book?
 
Ok here is the full problem.

Gasoline weighing 42lb/ft(cubed) is contained in a buried conical tank shown below. (above). Set up an appropriate integral, and find the work required to pump all the gas to ground level.

The tank is 5 feet below ground, i just didn't draw it on the diagram.

edit: so do you think it's 10-y ?
 
Because 10 is the height of the cone. I'm trying to find the radius of the cone. I know how to do the pumping problems, but I'm just stuck on how to find the radius that you'll end up plugging into the integral.
 
I don't think I have all the information for the problem. I did a similar problem that gave me the radius of the top of the tank and I was able to do it. I'll check tonight to see if I'm missing something.
 
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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