Pumping Hot coco out of a inverted cone tank. Work. 
Here is my problem. Ive tried several different ways but still can't seem to get the correct answer. 
Here are my steps... ignore the top portion. Ive tried integrating from 1 to 8, since I'm trying to pump the hot coco out of the tank. I think there is something wrong with how I've set up the integral. Or something about my slice. However, there was a similar problem but with a cylinder. I got that one correct, so I tried to mimc my steps but to no avail. And I've looked at the similar questions here but I still couldn't come up with the answer. any help as to what I'm doing wrong? 

 A: It's a little hard to follow your work (I can't see how you got the constant $14406$), but for what it's worth, this is how I would set this integral up:
Let $y$ denote the height above the cone's tip. We imagine taking a very small cylindrical disk of the chocolate located at height $y$ and asking how much work it takes to move it up to the top. The tiny disk will have tiny volume
\begin{align}
dV &= \pi r^2 \, dy
\end{align}
where $r$ can be computed in terms of $y$ since the conical volume the chocolate fills is geometrically similar to the entire conical tank.
To compute how much work it takes to move this tiny disk up, we need to know how much the disk weighs, and to figure that out, we need to know the mass of the disk, and to figure that out, we need to know the density of the chololate, which we are told is $\rho = 1470$. And of course mass is simply density times volume.
Let $dF$ and $dm$ denote the weight and mass respectively of the tiny disk. Then we have
\begin{align}
dF &= g \cdot dm \\
&= g \cdot \rho \cdot dV \\
&= g \cdot \rho \cdot \pi r^2 \, dy
\end{align}
So that's how much the tiny disk weighs. We need to move it $(8-y)$ meters up to get it over the top of the cone, so the tiny amount of work required is
$$ dW = (8-y) \, dF $$
Now we just need to add up the contribution of all these tiny disks. Can you take it from here?
