Our study of Double Integral and Fubini's Theorem is analogous to your previous study of Integtral Calculus. For a Riemann Integral(Definite Integral) an interval [a,b] in a real number line is partitioned  for a Double Integral a region in a plane will be partitioned . For each of these Integrals a limit is found, if it exists, it is said to be the value for the Integral. For each of these, a shortcut sometimes exists for finding the limit. For Definite Integrals the limit is sometimes used to define a measure of area for a bounded region in a plane. For the Double Integral the value is sometimes used to define a measure of volume for a bounded solid in 3-space.

Let's, very briefly, recall some work on Definite Integral. Then we'll move ahead to the Double Integral.

See 4.3 Definite Integral, pg. 273

The Fundamental Theorem of Integral Calculus has us evaluating a Definite Integral by subtracting two values of a function referred to as an anti-derivative. The following is a more formal statement :

See 4.4 Fundamental Theorem of Calculus , pg. 282

As we begin the study of Multiple Integration, the Iterated Integral problem will be analogous to the Fundamental Theorem problem. While we will not attempt to prove Fubini's Theorem concerning the relationship of Iterated Integrals and Double Integrals, we will be able, by use of appropriately chosen problems, to " see " that such a theorem might exist.

We begin with the Iterated Integral ( some might say, an integral of an integral ---- we will only allow one variable at a time, like in the case of partial derivative ) .

See 14.1 Example 1

See 14.1 Example 2

If we let the integrand be constant valued, that is, f(x,y) = 1 , then an iterated integral will do the " job " of defining a measure of area, a job previously assigned to the Definite Integral in Cal I ) .

In the following problem the Definite Integral from Cal I would have been:

Now look at the problem using an Iterated Integral.

See 14.1 Example 4

Now that we have an Iterated Integral to find the area of a bounded region, we can " see " that the limits of integration ( the order of integration, if you like ) can be reversed. We do this by not changing the integrand; we only change the limits; we should get the same answer for the area ( after all, the region has not changed ) .

See 14.1 Example 5

Look at the following solved problems in the Exercises at the end of 14.1 : 1,3,11,17,21,25,29,33,41

Then do the following unsolved problems as homework: 2,4,12,18,22,26,30,34,42