Pumping a cylindrical tank

A storage tank is a right circular cylinder 20ft long and 8 ft in diameter with its axis horizontal. If the tank is half full of olive oil weighing 57 lb/ft^3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank.

My work

I set the bottom of the tank at the point (0,0) & then drew a circle with a radius of 4.

dV=2x(20)dy

then solved the the circle's equation for x, x=$\displaystyle \sqrt{(16-y^2)}$

dV=40$\displaystyle \sqrt{(16-y^2)}$

F(y)=57(40)$\displaystyle \sqrt{(16-y^2)}$

10-y should be the distance the work must do

W=2280 $\displaystyle \int$(10-y)$\displaystyle \sqrt{(16-y^2)}$

Then I distributed the (10-y)

W=22800$\displaystyle \int$$\displaystyle \sqrt{(16-y^2)}$- 2280$\displaystyle \int$y$\displaystyle \sqrt{(16-y^2)}$

For part two, I set u=16-y^2 & got

W=22800$\displaystyle \int$$\displaystyle \sqrt{(16-y^2)}$+ 1140$\displaystyle \int$y$\displaystyle \sqrt{(16-y^2)}$

This is the part where I get lost, I did it a bit differently from the solutions manual, but at this point the manual shows inserting 4$\displaystyle \pi$ as follows

22800(4$\displaystyle \pi$)+ 1140$\displaystyle \int$y$\displaystyle \sqrt{(16-y^2)}$ (evaluated from 0 to -4), I tried from (0 to 4) in my solution.

I don't understand why the solutions manual is doing this particular step, it says its the area of a semicircle, but can anyone explain why I use it in this problem & where the integral disappears to?