We apply the procedure of “Slice, Approximate, Integrate” to model physical situations.

*geometric*applications of the procedure “Slice, Approximate, Integrate.” Indeed, it can be used to find lengths, areas, volumes. This procedure is not limited to modeling only geometric situations. Many problems from other STEM fields requires the same technique. This section examines many of these situations. First, we make a more generalized observation of the philosophy behind the “Slice, Approximate, Integrate” procedure.

### A broader perspective

Let’s take a step back and try to think about each of the situations we’ve used the “Slice, Approximate, Integrate” procedure to model. For each quantity of interest - length, area, or volume- we used an object whose length, area, or volume we could calculate.

- To find the area between curves, we know how to find the area of rectangles, so we approximate each slice by a rectangle.
- For solids of revolution, we can calculate the volume of washers or shell, so we approximate each slice by a washer or a shell.
- For length of curves, we can find the length of a line segment, so we approximate each slice by a line segment.

In physics, we take measurable quantities from the real world and attempt to find meaningful relationships between them. These relationships are often expressed using formulas, but these come with assumptions or restrictions. If these requirements are not met, we can still apply the “Slice, Approximate, Integrate” procedure to compute physical quantities of interest.

### Mass of a wire with variable density

Given a physical object, its density is a measure of how the mass of the object is
distributed. In the case where the density of the object is *constant*, we have the
following.

Three dimensions:

Two dimensions:

One dimension:

We can (and often do) approximate physical objects, like wires, as one dimensional or thin sheets as two dimensional.

Sometimes the linear density of an object can vary from one part of the object to
another. In this case, density will be a non-constant function, which we will represent
by the Greek letter, *rho*, (pronounced “rho”). In this case, the above formula no
longer allows us to compute the mass of the wire. Let’s tackle this scenario with a
motivating example.

What should we do? Let’s try the “Slice, Approximate, Integrate” procedure.

Step 1: Slice We cut the wire into numerous pieces. One such piece of width is shown.

Step 2: Approximate We know how to calculate the mass of a one dimensional object is when the density is constant. This is a rather poor approximation for the entire rod, but if each slice is very small, it is a fairly good approximation. For example, consider the slice that extends from to . Using the formula , we find to decimal places: and . In this case, the density doesn’t vary too much over a small slice, so treating the density as a constant along the slice is a good approximation.

Since we know how to calculate the mass of an object with constant density, we thus approximate that

Since the density along is slice is constant, we find that the mass of a single slice is given by

where is an -value along the slice where the density is approximated.

Let denote the -value used to determine the density of the -th slice, and be the width of the -slice. The total approximate mass of the rod is thus

Step 3: Integrate The usual procedure converts this approximate mass into the exact mass; indeed the definite integral will perform the simultaneous limiting process of shrinking the width of each slice while adding up the contributions from all of them. Since the mass of a single slice was found to be , the total exact mass is found the same way as before.

Now we can compute the mass of the wire.

There was nothing particularly special about this example. In fact, we can summarize the results in a formula.

Without performing any calculations, which half of the rod has more mass?

We now find the total mass of the rod. It is still true that

but since is now a piecewise function, we will need to compute the integral in pieces.

Let’s now find the exact value of so that the mass of the wire from to is exactly half of the total mass of the wire.

Since we expect that , we can find by setting:

and use the bottom piece for in the second integral. Recall that we found that the total mass of the wire is units, we have the following.

Solving for , we find .

### Work

One of the most important concepts in physics is that of **work**, which measures the
change in energy that occurs when a force moves an object over a certain
displacement. For those familiar with physics, the Work-Energy Theorem is a
powerful tool for studying situations in Newtonian mechanics (such as a box sliding
down an incline plane).

For a *constant* force acting on a particle over a displacement *in the direction of
displacement*, work is given by the formula:

Denoting work by , force by , and displacement by , we can write:

*constant*over the motion! There are many examples of important forces in physics that are not constant:

- The force required to stretch a spring units from its equilibrium position is given by , where is a constant of proportionality.
- The
*attractive gravitational*force between two objects of masses and that are separated by a distance of units is , where is a constant. - The
*electrostatic*force between two particles with charges and that are separated by a distance of units is , where is a constant.

In what follows, we will look at two scenarios in which the formula does not immediately apply. They result when either the force required to move an object is not constant, or when different parts of the object we move must be moved different distances. They can be summarized by:

#### Work done by a non-constant force on a particle

What will happen if the work done over the displacement is not constant? We study another motivating example that leads to a more general result.

Step 1: Slice We divide the displacement into numerous pieces. One such piece of width is shown.

Step 2: Approximate The only instance in which we know how to calculate the work is when force is constant. If the force is continuous and each slice is very small, it is a good approximation. Indeed, since we know how to calculate when is constant by the formula , we approximate that the force along each piece is constant,

Since the force along is slice is constant, we find that the work done by the force along to displace the object along the single slice is given by

where is the -value along the slice where the force is approximated.

Since there are slices used, let denote the -value used to determine the force along the -th slice, and be the width of the -slice. The total approximate work done by the force is thus

Step 3: Integrate The usual procedure converts this approximate work into the exact work; indeed the definite integral will perform the simultaneous limiting process of shrinking the width of each slice while adding up the contributions from all of them. Indeed, since the work done by the force along a single slice was found to be , the total exact force is found by

Let’s try some examples to see the formula in action.

**spring constant**.

#### Work done by a constant force on a collection of particles

In the last problems, a non-constant force acted on a single particle. Another scenario occurs when a constant force acts on a collection of particles. The following example explores this situation in detail.

Note that we need to overcome the force exerted by gravity to lim the liquid, which in this case is a constant .

Note that if we were lifting the entire tank meters, there would be no need for calculus since each particle of water would be moved the same distance. Here, different parts of the water must be moved different distances, Indeed, the water at the top of the tank has a much shorter distance to travel than the water at the bottom of the tank.

Step 1: Slice A key observation is that each particle of water at the same height must be moved the same distance, so we will slice the tank into vertical pieces. Setting at the base of the tank, and examine a piece at height .

Step 2: Approximate We know how to compute the work done by a constant force if we displace an object a certain distance. Here, the force is constant, but each particle in the slice must be moved a different distance. If the slice is thin enough, however, we can approximate that each particle within it must be moved the same distance. That is, we can treat the slice as a single object and find the work required to move it using

Note here that the slice is at a height and needs to be limed to a height of , so .

The amount of force is small because the slice width is small. We must express in terms of since we sliced with respect to . First, note that gravity exerts a force of on the slice. Since the density of the liquid is constant, we have . Note here that the cross-sections of the tank are circular, so . We note that the cross-sectional area is constant no matter at what height the slice is drawn, so .

Putting it all together, we find

Rearranging this gives the approximate work required to move a single slice of liquid.

Step 3: Integrate We can now find the exact work in the usual way. We need to add up the contributions of all of the slices of liquid, which begin at and extend to . Thus, the exact work is given by

Evaluating this integral is easier if we factor out the first.

As usual, the process here can be generalized into another formula.

where is the acceleration due to gravity and is the distance that all particles at a given height must be moved.

### Final thoughts

The technique of “Slice, Approximate, Integrate” can be used to solve physical problems as well as geometric ones. These are not the only examples of physical problems that can be modeled and solved by using this technique. The exercises will explore other problems, and you will run into many more in other STEM courses.

“Mathematics is the abstract key which turns the lock of the physical universe” - John Polkinghome