Problem on review sheet for my son's calculus final:

A right circular canonical tank of height 25 and radius 5 has its vertex at ground level and a vertical axis has oil weighing 50 lbs/cubic foot. Set up an integral that represents the amount of work done in pumping the oil up to the top of the tank. Include an approximation for the work required to move one disk of water to the top, but do not evaluate the integral.

Answer: $\displaystyle \int_{0}^{25}{2 \pi x^2 (25-x)\;dx}$

The way we tried to solve it (imitating an example in the textbook):

Force is weight times volume. Divide the canonical tank into circular slices of height delta y:

$\displaystyle \\\Delta F =(weight)(volume) \\=(50)(\pi x^2 \Delta y)$

A circle with radius 5, centered at (0,0) has the form

$\displaystyle x^2+y^2=25$ which can be solved for

$\displaystyle x^2=25-y^2.$

I don't completely understand how (or why) they picked the center for the example we're working from, so we just chose (0,0).

The distance to the top of the cone is height of the cone minus height of the oil: $\displaystyle 25-y.$

Next, the work incremental is the force incremental times the distance

$\displaystyle \\\Delta W = \Delta F(25-y) \\= (50)(\pi x^2 \Delta y)(25-y) \\=50 \pi (y^3-25y^2-25y+625) \Delta y.$

Integrate from height of 0 to 25, so the work must be

$\displaystyle W=\int_{0}^{25}{50 \pi (y^3-25y^2-25y+625)\;dy}.$

But that doesn't equal the answer given to us.

What's wrong please?

Thanks.