Double Integrals: Applications & Examples

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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Double integrals extend the possibilities of one-dimensional integration. In this lesson, we will focus on the application of the double integral for finding enclosed area, volume under a surface, mass specified with a surface density, first and second moments, and the center of mass.

Background on Integrals

Sometimes we can take a concept in one dimension and apply it to a higher dimension. The line in one dimension becomes the surface in two dimensions. Extending this idea to the realm of calculus integration, the single integral (which uses one variable) becomes the double integral (which uses two variables). Many of the same rules for evaluating single integrals apply here, so if you're unfamiliar with those rules, you may want to review some of our lessons on evaluating single intervals first. In this lesson, we look at methods for evaluating the double integral and some of the interesting applications.

Finding the Area of a Bounded Region

Look at a rectangle, of length 4 and width 2, in the x-y plane.

We can bound this rectangle using the lines x = 2, x = 6, y = 1 and y = 3.


Finding this area using a double integral:


The inner integral:


The double integral now becomes this:


Let's do another area example.


Identify the curves bounding this figure.


As before, the area is given by this:


The inner integral (which has limits defined by curves which bound the region) is an integration on x. We fix a y and look at which curves bound the x values.


For a fixed y, the values of x range from x = y to x = 2 - y. Note that the formulas for the curves have been rewritten so that x is the subject.

Our inner integral is now:


The double integral becomes:


The limits on the outer integral (which uses the numerical limits of the region rather than curves) are the numerical boundaries for the variable y. These limits for y are 0 and 1.

Our double integral now becomes:


Note that we can also use double integrals for finding areas of bounded regions that form more complex shapes, which may not be as familiar as rectangles or triangles. As long as we're careful about defining the limits for the inner and outer integrals, we can follow the same general steps to find the area.

Exploring the Ordering of Integration

When partitioning a region, we consider rectangular elements of dimension Δx by Δy. The product of these rectangular dimensions gives us a small area. As this small area becomes infinitely smaller, the Δ's become differentials. That is, Δx becomes dx and Δy becomes dy.

What if we change the order of the differentials? Since multiplication is commutative, dx dy = dy dx. The area element could be written as dy dx instead of dx dy. Intuitively, the result should remain the same. However, the way we perform the integration calculation will change.

Reworking the last example with the inner integral now on y means that fixing an x produces two regions.


For a fixed x in region 1, y is bounded by y = 0 and y = x.

In region 2, for a fixed y, the bounds are y = 0 and y = 2 - x.

Now, our area integral has two parts:


Substituting the limits of integration:


As expected, the answer is the same.

Our conclusion is that although the answer stays the same, one differential ordering may lead to a more difficult calculation.

Finding a Volume Under the Surface

Finding the areas of bounded regions is one of the more basic applications of double integrals, but moving into a higher dimension also allows us to explore volume. Think of it this way: if the single integral is the area under a curve, then the double integral can be interpreted as the volume under a surface as we add a dimension.

Let f(x,y) determine our surface. For example,


f(x,y) is the height above the x-y plane. For instance, at the origin, x = 0 and y = 0. Thus, at the origin, the surface height is 2.

Let's consider a square in the x-y plane. In particular,


A plot of this box with a slanted roof is shown as:


The volume under this surface is:


First, we calculate the inner integral:


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