Calculating Work for Buried Gas Tank

[Sheneron]

Homework Statement

A cylindrical gasoline tank with radius 4ft and length 15 ft is buried under a service station. The top of the tank is 10ft underground. Find the amount of work needed to pump all the gasoline in the tank to a nozzle 3 ft above the ground. Gasoline weighs 42lb/ft^3.

The Attempt at a Solution

So the way I have started is by setting a cylinder with the specified radius centered at (0,0). With that I know the top is at (0,4), bottom is at (0,-4) and the place it needs to reach is y = 17 (13ft above the top of the cylinder).

I divide the cylinder into strips of thickness delta y. So I know I need to find the weight of each of those strips and multiply it by the distance it must travel (17-y).

I have the density so I really now just need the volume of those strips. Volume = length * width * thickness. I have the thickness (delta y), the length (15), but I need the width.

I can't figure out how to find the width of the strips. I know that at y = 0 the width is 8 and at y=4 and y = -4 the width is 0.

Something like
width = 2\sqrt{16-y^2}
would tell me the width with respect to y, but that would only be true from y=-4 to y=4, and it needs to stop after that.

Any advice would be appreciated thanks.

 

[Sheneron]

Wait tell me if this would work:

width = 2\sqrt{16-y^2}
thickness = \Delta(y)
length = 15

weight = 2\sqrt{16-y^2} * \Delta(y) * 15 * 42

and the height = 17 - y
so the integral would be:

\int_{-4}^{4} 1260\sqrt{16-y^2}(y-17) dy

 

[HallsofIvy]

One important thing you did not specify: is the axis of the cylinder vertical or horizontal? You seem to be assuming it is horizontal. If so then, yes, width= \sqrt{16- y^2}. You will only be integrating from y= -4 to 4 so the fact that it "has to stop at 4" is irrelevant. Looks to me like your final integral is correct.