Example 1:
We are to evaluate the double integral

over the rectangle .

Here is our standard double integral expression. First and foremost, notice that unlike single variable calculus, there are two ‘components’ to any double integral problem, the function f(x,y) we are integrating, in this case , and the region R. It doesn’t make much sense if we are to find the double integral of a function, yet without a region R specified making it unclear as to what region to integrate over.

Using our previous theorem, the value of the double integral may be obtained by either iterated integrals

 or

bearing in mind that for a rectangular region (which is in this case), the order of the repeated integrals does not matter.

Using the former iterated integrals, we have

A point to know is that after the first integration, we are substituting the limits into y as shown by the y = 0. Since there are two variables in the square bracket, this is to make it unambiguous as to which variable we are substituting the limits into.

With a quick check with the later iterated integral, one would reach the same answer.

Example 2:
We shall now use a double integral to find the volume of the solid that is bounded above by the plane z = 4 – x – y  and below by the rectangle .

The above graph gives us a feel of the geometrical meaning of the double integral. The function z = 4 –x – y represents the plane which bounds the solid from above and region R is the rectangle which bounds the solid from below. One may say that region R is the projection of the plane onto the x-y plane.

Again, we would get the exact same result if we integrated with respect to y first and then with respect to x.