Lifting an Elevator 3000lbs with a 12ft Cable: Calculating Work

[turdferguson]

Homework Statement

A freight elevator weighing 3000 pounds is supported by a 12 foot long cable that weighs 14 pounds per linear foot. Approximate the work required to lift the elevator 9 feet by winding the cable onto a winch

Homework Equations

W = int(f)dy

The Attempt at a Solution

The work lifting the elevator is 3000lb(9ft)=27000ft*lb

The winch is lifting a piece of rope dy a distance 12-y from 0 to 9
Each section dy weighs 14dy

(sorry for the lack of latex)
W = int_0-9[14(12-y)]dy = 14int_0-9[(12-y)]dy
W = 14[12y-.5y²]^0-9
W = 945 ft*lb

Total work = 27000+945 = 27,945ft*lb
But the book has an answer of 36,945. The 945 leads me to think I set up the integral correctly, but where did the extra 900 come from?

 

[Dick]

The book's answer is clearly wrong. You knew that, right?