Anyone have any suggestions with these Calc 2 work problems?

Question

I'm not sure if I went about them correctly, but here are the two problems. Any help anyone could provide would be greatly appreciated. Thanks!

--Problem 1--

Find the work done in raising 1,500 pounds of ore from a mine that is 1,500 feet deep. The rope weighs 2 pounds per foot.

1.) The work of the ore is simply W = F * d = 1,500 * 1,500 = 2250000 ft-lbs.
2.) The work of the rope is the integral from 0 to 1,500 of 2x dx = 2250000.
3.) Then add the two together for a total work of 4.5 * 10^6 ft-lbs.

Even if I set Step 2 up like integral[0,1500] of (2)(1,500-x) dx...I still get the same answer.

--Problem 2--

Find the work done in filling an upright cylindrical tank of radius 3 feet and height 10 feet with a liquid that weighs 50 pounds per cubic foot. Assume the liquid is pumped into a hole in the bottom of the tank.

1.) V = pi*r^2*h, so V = 9*pi*delta x
2.) W = integral[0,10] of 50 * 9 * pi * x dx = 450 * pi * x dx = 70,685.83
ft-lbs.

Answer 1

Problem 1
========
Seems to be the correct approach. Why do you doubt your answer?

Alternatively you could combine the two parts into one integrand; the force needed at any stage of the process is

F = 1500 + 2x

and you would integrate INT (1500 + 2x) dx.

Problem 2
========
The work is equal to the total gravitational energy of the liquid after filling. A layer of height dy has

area: A = pi r^2 = 9 pi
volume: dV = A dy = 9 pi dy,
mass: dm = 50 * dV = 450 pi dy
grav. energy: dE = m g y = 450 pi y dy (g = 1)

so

W = INT dE = INT 450 pi y dy = 225 pi y^2 from 0 to 10,

so you find

W = 225 pi * 10^2 = 22500 pi



Looks like you are doing great!