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MI.5 Applications of Double Integrals

We have already seen on application of double integrals: computing volumes. This section will allow us to explore the physical applications such as computing mass, electric charge, center of mass, and moment of inertia. We will see that these physical ideas are also important when applied to probability density functions of two random variables.

Density and Mass

We know how to compute the center of mass of a thin plate or lamina with constant density. But now, equipped with the double integral, we can consider a lamina with variable density. Suppose the lamina occupies a region D of the xy-plane and its density at a point in D is given by $\rho (x,y)$ where $\rho$ is a continuous function on D. This means that:

where $\Delta m$ and $\Delta A$ are the mass and area of a small rectangle that contains and the limit is taken as the dimensions of the rectangle approach 0.

Therefore we arrive at the definition of total mass in the lamina. All one has to do is find the double integral of the density function.

Physicists also consider other types of density that can be treated in the same manner. For example, if an electric charge is distributed over a region D and the charge density is given by $\sigma (x,y)$ at a point in D, the total charge of Q is given by:

Moments and Centers of Mass

Now lets find the center of mass of a lamina with density function $\rho (x,y)$ that occupies a region D. Recall that to find the center of mass we first have to find the moment of a particle about an axis, which is defined as the product of its mass and its directed distance from the axis. For double integrals this is done in the following way:

The moment of the entire lamina about the x-axis:

Similarly, the moment about the y-axis:

As before, we define the center of mass so that and The physical significance is that the lamina behaves as if its entire mass is concentrated at its center of mass. Thus, the lamina balances horizontally when supported at its center of mass.

The coordinates of the center of mass of a lamina occupying the region D and having density function $\rho (x,y)$ are:

Where the mass, , is given by:

Moment of Inertia

The moment of inertia of a particle of mass m about an axis is defined to be $mr^{2}$ , where r is the distance from the particle to the axis. We extend this concept to a lamina with density function $\rho (x,y)$ and occupying a region D by proceeding as we did for ordinary moments: we use the double integral:

The moment of inertia of the lamina about the x-axis:

Similarly the moment about the y-axis is:

It is also of interest to consider the moment of inertia about the origin, also called the polar moment of inertia:

Also notice the following:

Probability

In a previous section we considered the probability density function of a continuous random variable X. This means that $f(x)\geq 0$ for all x, and the probability that X lies between and is found by integrating from to

Now we consider a pair of continuous random variables X and Y, such as the lifetimes of two components of a machine or the height and weight of an adult female chosen at random. The joint density function of X and Y is a function of two variables such that the probability that (X,Y) lies in region D is:

In particular, if the region is a rectangle, the probability that X lies between a and b and Y lies between c and d is

Picture such as on page 1035

The probability that X lies between a and b and Y lies between c and d is the volume that lies above the rectangle and below the graph of the joint density function. Because probabilities aren't negative and are measured on a scale from 0 to 1, the join density function has the following properties:

The double integral over $\U{211d} ^{2}$ is an improper integral defined as the limit of double integrals over expanding circles or squares and we can write:

Expected Values

If X and Y are random variables with joint density function , we define the X-mean and Y-mean, also called the expected values of X and Y, to be:

Notice how closely the expressions for $\mu _{1}$ and $\mu _{2}$ resemble the moments M $_{x}$ and M $_{y}$ of a lamina (such as discussed above). We can think of probability as being like continuously distributed mass. We calculate probability the way we calculate mass - by integrating a density function. And because the total "probability mass" is 1, the expression for and above show that we can think of the expected values of X and Y, $\mu _{1}$ and $\mu _{2}$ , as the coordinates of the "center of mass" of the probability distribution. In the next example. we deal with normal distribution. In this, a single random variable is normally distributed if its probability density function is of the form:

where $\mu$ is the mean and $\sigma$ is the standard deviation.

Example

A factory produces (cylindrically shaped) roller bearings that are sold as having a diameter of 4.0 cm and a length of 6.0 cm. In fact the diameters X are normally distributed with mean 4.0 cm and standard deviation .01 cm while the lengths Y are normally distributed with mean 6.0 cm and standard deviation .01 cm. Assuming that X and Y are independent. write the joint density function and graph it. Find the probability that a bearing randomly chosen from the production line has either length or diameter that differs from the mean by more than .02 cm.

Solution

We are given that X and Y are normally distributed with $\mu _{1}=4.0~cm$ and $\mu _{2}=6.0~cm,$ and So the individual density functions for X and Y are:

Since X and Y are independent, the joint density function is the product:

A graph of this function is show below:

$\bigskip$

graph:

Lets first calculate the probability that both X and Y differ from their means by less than .02 cm. using a calculator or computer to estimate the integral, we have:

Then the probability that either X or Y differs from its mean by more than .02 cm is approximately:

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