I am trying to understand my teacher's example of a Work problem.

Cut to the chase here's a picture of the problem:

goo.gl/photos/AqZ6ENmHLg6heixD7

While I understand how to integrate fairly well, I'm still confused on how exactly my teacher set up the work problem.

Since work = force * distance

The distance would be (y + 3) [which makes since because the engine is 3 feet above the fuel tank.]

The length would be 3 [makes sense], and somehow, the width is 2 - (y/3)??? Where did that come from???

I understand that we're trying to integrate by dy, but I'm still clueless on how the teacher got that width equation. Nevertheless, the integral would lead to:

dF = 53.1(3)(2 - (y/3)) dy

W = ∫ 53.1(3)(2 - (y/3)) dy

W = 3106.35 ft-lb

So again, what exactly is "y" representing when it comes to the width? Where did that equation come from?
I think it might've been a way to flip the trapezoidal graph, but I'm not completely sure.

Any help is greatly appreciated!

Correction, I typed the integral wrong! Forgot to include the distance:

dF = 53.1(3)(2 - (y/3)) dy

dw = dF * Distance

W = ∫ 53.1(3)(2 - (y/3)) (y + 3) dy

W = 3106.35 ft-lb

the width equation is due to the taper of the tank

as y goes from 0 to 3, the width goes from 2 to 1

the volume of the fuel increment is
... l * w * dy
... the force increment is the volume times the fuel density

To understand how your teacher set up the work problem, let's analyze the equation for the width: 2 - (y/3).

In the context of the problem, the width represents the breadth of the fuel tank at a specific height (y) above the bottom.

To visualize this, imagine the fuel tank as a trapezoidal shape, with the top side being 2 feet long and the bottom side being 3 feet long. The height (y) measures the vertical distance from the bottom of the tank to a particular level.

Now, as you correctly stated, the engine is located 3 feet above the fuel tank. This means that at y = 0 (the bottom of the tank), the width is 2 feet (the top side of the trapezoidal shape).

As you move upwards, the width of the tank gradually decreases because the top side of the trapezoid shortens. At y = 3 (the level of the engine), the width becomes 0 because the top side has fully disappeared.

To represent this variation in width mathematically, your teacher used the equation 2 - (y/3).

Let's break it down:
- The term 2 represents the initial width of the tank (at y = 0).
- The term (y/3) represents the reduction in width as y increases. As y increases, the value of (y/3) also increases, causing the width to decrease.

So, overall, the equation 2 - (y/3) represents the width of the fuel tank at a height (y) above the bottom, taking into account the gradual reduction in width.

Regarding the integration by dy, it is used to sum up infinitesimally small strips of work as we move up from the bottom of the fuel tank to the level of the engine. Each strip of work, represented by dF, is the force exerted by a slice of fuel multiplied by the distance it is lifted.

By integrating over the entire height range, from 0 to 3, you obtain the total work done, which in this case was calculated to be 3106.35 ft-lb.

So, in summary, "y" represents the vertical distance (height) above the bottom of the fuel tank, and the equation for the width, 2 - (y/3), describes how the width of the tank decreases as you move up. This equation is fundamental in setting up the work problem and determining the force and distance for each infinitesimally small slice of fuel.