HELP! A cylindrical tank is on the ground (lying on its side), is only half full of water, and... Find work!!?

Question

A cylindrical tank is on the ground (lying on its side), is only half full of water, and has a radius of 1ft and length 5ft. The cylindrical tank has a pipe coming out of the side of the tank (so it is pointing upwards because the tank is on the ground on its side). The spout that points straight upwards is 1.5ft high (so total distance from ground to top of spout is 2r + 1.5ft = 3.5ft). Set up the integral to find out how much work needs to be done to pump the water out of the outlet to the top... water density is 62.5lb/ft^3 ). Do not evaluate integral.

so far I have:

(∫) from 0→1 [(10[1-√(1-x²)])(3.5-x)(62.5)]dx

and the variable x is the the distance from the ground at point 0→x, which points upwards...

Answer 1

I don't believe your integral is correct.

Consider the water at a level x ft above the ground in a cross-section of the tank. We need to find the width of the cylinder at this height. The distance from the centre of the water to the centre of the circle is 1-x ft, and the distance from the centre of the circle to each endpoint of the water is 1 ft. By dividing it up into two congruent right triangles we get the base to be √(1^2 - (1-x)^2) = √(2x - x^2) for each triangle, so the width of the water at this level is 2√(2x - x^2).

So the surface area of the water at this height is this multiplied by the length of the tank, which is 5 ft, so A = 10√(2x - x^2). So the volume element is 10√(2x - x^2) dx and the mass element is 625√(2x - x^2) dx (dm = ρ dV).

The work done is given by W = mgh, where h = 3.5 - x and g = 32 ft/sec^2, so the work element is
dW = 625√(2x - x^2) (32) (3.5-x) dx
= 10000 (7 - 2x) √(2x - x^2) dx

So the integral is
∫(0 to 1) 10000 (7 - 2x) √(2x - x^2) dx.