In the previous notes on iterated integrals we saw that they can be used to find the area of a plane region by letting the integrand be a constant- valued function, for example, f(x,y) = 1. The value of the area of the plane region would also be the value of the volume of a solid that has that plane region as its base and a constant height of 1 unit. For example, , has a value of 1 which is the measure of the area of the rectangular region bounded by the graphs of the limits of integration, namely, x = 0 x = 1 y = 0 and y = 1. The value of this iterated integral is also the measure of the volume of a cube whose base is the afore mentioned rectangle and whose height is given by the integrand function, f(x,y) = 1 . This raises the question of whether or not the iterated integral, , has a value that measures the volume of a solid with the afore mentioned rectangle and whose height is given by the non-constant function

The answer, as expected, is yes.

The next example should further illustrate the relationship of interated integral to measure of volume of certain solids. The example is chosen so that a value can be found by Calculus I integration on a formula for cross-sectional area; then we can solve with an iterated integral. An assumption is made in the following example, namely: if a formula is present for the area of a cross-section of a solid at any point along an axis through the solid, then an integration on that formula over the axis will produce the measure of the volume of that solid. I tell you that because not all Calculus I students have done that problem.

( The example mentions double integral but is about iterated integral. We'll get to the relationship of double integral after this example. )

See 14.2 Evaluation of Double Integral

In the previous notes on Iterated Integrals I referred you to some review of what a Definite Integral or Riemann Integral is. With that in mind, we now look at the definition for a Double Integral.

See 14.2 Double Integrals and Volume

From the discussion of Iterated Integrals above, we now suspect that if the region on which the Double Integral is defined is bounded by "curves"(straight lines are included as curves here )with equations that read as x = a and x = b and y = f(x) and y = g(x) with f(x)<g(x) OR y= c and y = d and x = h(y) and x = q(y) where h(y)<q(y) then the Double Integral can be evaluated without the use of limit work, but instead by an Iterated Integral.( The significance of the boundary equations for the region is that they be as in the Calculus I problems on finding area measure.) This is stated in Fubini's Theorem which is in the previous "link" above See 13.2 Evaluation of Double Integral

For an example of the use of the this Theorem: ( Notice the function for this example was previouly used and that the difference in the following example is the region in the xy plane for the double integral is an ellipse rather than a unit square. The equations for the boundaries of the ellipse are x = -2 and x = 2 and and 

See 14.2 Example 3

Look at the following solved problems in the Exercises at the end of 14.2 : 5,7,13,17,21,23 Then
do the following unsolved problems as homework: 8,14,16,18,8,22,24,26