Double Integrals over Rectangular Regions
Volumes of solids
For a function of one variable with positive values i.e. the graph of the function is above the x-axis, its definite integral over an interval is the area under the graph on such interval. We can generalize the definition of definite integral to functions of two variables as follows:
Let be a function of two variables such that on the domain i.e. a rectangular region in . Then we divide into rectangular subregions using lines parallel to x and y axes.
For , let and be the length and width of the rectangular subregion respectively. Then the area of is . Take any point in and consider the rectangular box with base and height . Then its volume is
And we can approximate the volume under the graph of over by the sum of volumes of all these rectangular boxes/prisms:
As , and if limit of the above "Riemann sum" exists for all partitions of and for all choices of within those partitions, then the limit is the volume under the graph over and is called the double integral of over , denoted as follows:
In the following applet, we cut evenly into identical rectangular regions i.e. and is chosen to be the lower left corner of the rectangular region. As get larger and larger, the sum of the rectangular boxes/prisms will get closer and closer to the volume under the graph over .
As we know, the definite integral is still well-defined even when the graph of is not entirely above the x-axis. The area below the x-axis can be regarded as the "negative area". Similarly, the double integral is well-defined for any function . The portion of volume below the xy-plane can be regarded as the "negative volume".
Iterated integrals
How can we compute a double integral? It turns out that we have the following extremely theorem to help us computing any double integral over a rectangular region:
Fubini's theorem: Suppose on . Then
The integrals on the right hand side are called iterated integrals. The integral means we integrate the function with respect to i.e. is regarded as a fixed value. What you obtain after integrating is a function of and we then integrate it with respect to .
Example: Compute , where .
Answer:
By Fubini's theorem, we have
Alternatively, we compute the following iterated integral:
Exercise: Compute , where .
Exercise: When you use Fubini's theorem to compute , where , which iterated integral would you choose?
More about double integrals
The following are some nice properties of double integrals:
- , for any real number
- , where is the disjointed union of and .
- If on , then .
- .
- If , then , where .