Double Integrals Over More General Regions

The relationship between double integrals and iterated integrals (Fubini's Theorem) over more general regions is detailed. The distinction between a double integral and an iterated integral in two variables is explained. Indeed, the double integral is often evaluated by converting it to an equivalent iterated integral, which is usually easier to compute; but nonetheless double integrals and iterated integration are distinct concepts. Essentially, an iterated integral is like the inverse of mixed partial differentiation and a double integral is a direct extension of the Riemann integral (Riemann sums) of one variable to functions of two independent variables.

A type I, or vertically simple region is a region of the plane that can be described by the inequalities

where and are continuous functions of on Similarly, a type II, or horizontally simple region , in the plane is a region that can be described by the inequalities

where and are continuous functions of on

Definition (Double Integral over a Region) Let be a function that is continuous on the region that can be contained in a rectangle Define the function on as if is in and 0 otherwise. If is integrable over , we say that is integrable over , and the double integral of over is defined as

The function may have discontinuities on the boundary of but if is continuous on and the boundary of is fairly "well behaved", then it can be shown that exists and hence that exists. This procedure is valid for the type I and type II simple regions.

Proposition (Double Integral over a Region) If is a type I region, then

double integral over a more general region _gr_37.gif]

whenever both integrals exist. Similarly, for a type II region

whenever both integrals exist.

Example (Double Integral over a Region) Compute the double integral over the nonrectangular region.

(a) Compute the double integral

    Solution. The region is horizontally simple or a type II region as   and   We compute

This region could also be considered as a vertically simple or a type I region as   and We compute

double integral over a more general region _gr_47.gif]

(b) Compute the double integral

    Solution. The region is vertically simple or a type I region as   and   We compute

The region is also horizontally simple or a type II region as and We compute

(c) Evaluate the integral

    Solution. If we try to evaluate the integral as it stands, we are faced with the task of first evaluating But this is impossible to do so in finite terms since is not an elementary function. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. We have

double integral over a more general region _gr_58.gif]

where This region has an alternate description:

Thus we can express the double integral as an iterated integral in the reverse order:

(d) Evaluate   where is the region bounded by the line and the parabola

    Solution. The region is both a type I and a type II region, but the description of as a type I is more complicated because the lower boundary consists of two parts. Therefore we express as a type II region:

Then the double integral becomes

double integral over a more general region _gr_71.gif]

If we had expressed as a type I region, then we would have obtained

but this would have involved more work than the first part.