Moving oil out of a conical shaped tank I need some help finishing out a Calculus problem. I'm not sure how $d$ works (single value, or integral) at the end.

A conical shaped tank, with its apex pointing upward is one fourth full of motor oil. That is, the initial depth of the oil in the center of the tank is 2.5 feet. The radius of the circular base of the tank if 4 feet. The height of the tank is 10 feet. Given that motor oils weighs $32\frac{lbs}{cubic foot}$, find the work required to pump the oil of of the top of the tank.

So, I know that
$$ v = \int_{0}^{2.5}{\pi\left(\frac{-2y + 20}{5}\right)^2}{dy} $$
And thus mass
$$ m = \int_{0}^{2.5}{\pi\left(\frac{-2y + 20}{5}\right)^2}{dy} \times \frac{32lbs}{ft^3} $$
And work is
$$ W = mgd $$
$$ W = \int_{0}^{2.5}{\pi\left(\frac{-2y + 20}{5}\right)^2}{dy} \times \frac{32lbs}{ft^3} \times 32.174 \frac{ft}{s^2} \times d $$
I'm stuck at what $d$ is. Is it 10? 7.5? A new integral? Or does it get put into the first integral?
Thanks!
 A: For this kind of problem, I would suggest drawing the conical shaped tank in a cartesian coordinate system with the origin at the point where the oil is being pumped at the top of the tank. The trick is to point the $x$-axis downwards, parallel to the height of the tank, and point the $y$-axis sideways, parallel to the radius of the tank.
Then, find an equation for the radius of the conical shaped tank with respect to $x$. If you look at the cartesian coordinate system, the shape looks like a slope intercepting the $y$-axis at the origin and passing by the point $(10,4)$. So the equation is
$$
(y - 4) = \frac{4}{10}(x - 10) \\
y = \frac{2x}{5} \\
$$
An element of volume is then a circular disk having thickness $\Delta_i x$ and radius $f(x_i)$, where $f(x) = y$. The volume of this element is
$$
\Delta_i V = \pi [f(x_i)]^2 \Delta_i x
$$
Given the motor oil weighs $32 \, lbs/ft^3$, then $32 \Delta_i V$ is the force required to pump an element and $32 \Delta_i V x_i$ is the work needed to pump it to the top of the tank. Thus
$$
\begin{eqnarray}
W &=& \lim_{||\Delta|| \to 0} \sum_{i=1}^n 32 \pi [f(x_i)]^2 x_i \Delta_i x \\
&=& 32 \pi \int_{7.5}^{10} \! [\frac{2x}{5}]^2 x \, \textrm{d}x \\
&=& \frac{128}{25} \pi \int_{7.5}^{10} \! x^3 \, \textrm{d}x \\
\end{eqnarray}
$$
