Subsections

# The Definite Integral, Area and Center of Mass

## Introduction

There are two main ways to think of the definite integral. The easiest one to understand is as a means for computing areas (and volumes). The second way the definite integral is used is as a sum. That is, we use the definite integral to add things up''. Here are some examples.
• Computing net or total distance traveled by a moving object.
• Computing average values, e.g. centroids and centers of mass, moments of inertia, and averages of probability distributions.
This lab is intended to introuduce you to Maple commands for computing integrals, including applications of integrals.

## Definite and indefinite integrals with Maple

The basic Maple command for computing definite and indefinite integrals is the int command.

To compute the indefinite integral

with Maple:
> int(x^2,x);

Note that Maple does not include a constant of integration. Suppose you wanted to compute the following definite integral with Maple.

The command to use is:
> int(x^2,x=0..4);


## Finding area between two curves

If you want to find the area bounded by the graph of two functions, you should first plot both functions on the same graph. You can then find the intersection points using either the solve or fsolve command. Once this is done, you can calculate the definite integral in Maple. An example below illustrates how this can be done in Maple by finding the area bounded by the graphs of and :
> f := x-> -x^2+4*x+6;
> g := x-> x/3+2;
> plot({f(x),g(x)},x=-2..6);
> a := fsolve(f(x)=g(x),x=-2..0);
> b := fsolve(f(x)=g(x),x=4..6);
> int(f(x)-g(x),x=a..b);


## Center of Mass

In designing mechanisms or structures, one often has to deal with distributed forces, that is, forces that do not act at a discrete, finite set of points. The most common example of a distributed force is the force of gravity, which acts on all parts of any body of matter. Other examples are pressure in fluids and electrostatic forces, though there are many others.

One of the basic useful principles of analyzing distributed forces is the idea of replacing them with a single, aggregate force that acts at a single point and is somehow equivalent to the original distributed force. This may not always be possible, but this technique has found great use in engineering and science. As a simple example, suppose we have gravity acting on a solid plate of uniform thickness and density, but irregular shape. Finding the equivalent force is really the problem of finding the point where we could exactly balance the plate. This balance point is often called the center of mass of the body.

For symmetric objects, the balance point is usually easy to find. For example, the balance point of an empty see-saw is the exact center. Similarly, the balance points for rectangles or circles are just the geometrical centers. For non-symmetric objects, the answer is not so clear, but it turns out that there is a fairly simple algorithm involving integrals for determining balance points.

We begin by restricting our attention to thin plates of uniform density. In Engineering and Science, this type of object is called a lamina. For mathematical purposes, we assume that the lamina is bounded by , , , and , with . Then the book gives the following formulas for the coordinates of the center of mass.

To see where these formulas come from, we need the following result. Suppose we have a composite body that is made up of two masses and and that we know the centers of mass are and . Then according to the principles of mechanics, the center of mass of the composite body can be determined from the following equations.

This result can be easily generalized to the case where we have a composite body made up of masses to give the following result.

To go from these formulas to the integral formulas presented earlier, we first partition the interval into subintervals

Then we approximate the lamina on each subinterval with a rectangle by letting be the midpoint of the subinterval and using as the top of the rectangle and as the bottom. This gives a rectangle of width and height . Assuming that the density is 1, we have the following for the mass, , and center of mass, of the rectangle.

Using these relations, we get the following equations.

The right hand sides of these equations should be easy to recognize as Riemann sums, so that when we take limits as goes to infinity, we get the following in our equations for the center of mass of our lamina.

These are easy to rearrange into the equations given in the text.

We end this section with an example, including Maple commands, for computing the center of mass. Suppose you have a lamina bounded by the curves , whose area was computed above and you want to compute the center of mass. If you are not sure about the relative positions of the two curves, it is a good idea to plot them both. Assuming constant density , then area and mass would be equivalent.

> plot([f(x),g(x)],x=-5..5);
> mass:=int(f(x)-g(x),x=a..b);


Now we are ready to compute the center of mass. Using labels can help you organize your calculations and avoid mistakes. Computing the mass separately as shown above lets you check it. If you get a negative value for the mass, something is wrong and you have to check what you have done. A common mistake is reversing the order of the functions.

> xbar := int(x*(f(x)-g(x)),x=a..b)/mass;
> ybar := 1/2*int(f(x)^2-g(x)^2,x=a..b)/mass;
> plot([f(x),g(x),[[xbar,ybar]]],x=a..b,style=[line,line,point],color=black);

The plot of both functions on the same graph along with the point helps to confirm that your point is in the physical center of the lamina.

## Exercises

1. Use Maple to compute the each of the following definite integrals:
A)
(Note the syntax for the function is .)
B)
(Note: to square a trig function put the ^2 after the angle in parentheses)

2. Find the area of the region bounded by the curves and .

3. Find the coordinates of the center of mass of the lamina bounded by the curves and . Plot both functions on the same graph along with the point to see if your answer makes sense.

4. Find the coordinates of the center of mass of the lamina bounded by the curves and . Plot both functions on the same graph along with the point to see if your answer makes sense.

5. Find the coordinates of the center of mass of the lamina bounded by the curves and . Plot both functions on the same graph along with the point to see if your answer makes sense.

6. Find the coordinates of the center of mass of the lamina bounded by the curves and . Plot both functions on the same graph along with the point to see if your answer makes sense.