Double Integral

(1) Review We defined the definite integral of a single variable as a limit involving the Riemann sums



where



are elements in a partition of the interval and is a representative in the subinterval We now apply the same idea to define a definite integral of two variables



over the rectangle

(2) Definition (Riemann Sum) Partition the interval into subintervals and the interval into subintervals. Using these subdivisions, partition the rectangle into cells (subrectangles) and call this partition Choose a representative point from each cell in and form the sum,



where is the area of the -th representative cell. This is called the Riemann sum of with respect to the partition and the cell representation To measure the size of the rectangles in the partition we define the norm of the partition to be the length of the longest diagonal of any of the rectangle in the partition.

(3) Definition (Double Integral) Refine a partition by subdividing the cells in such a way that the norm decreases. When this process is applied to the Riemann sum and the norm decreases to zero, we write



If this limit exists, its value is called the double integral of over the rectangle Indeed, if is defined on a closed, bounded rectangular region in the -plane, then the double integral of over is defined by  



provided this limit exists, in which case, is said to be integrable over

    It can be shown that if is continuous on a rectangle then it must be integrable on , although it is also true that certain discontinuous functions are integrable as well. Moreover, it can also be shown that if the limit that defines the integral exists, then it is unique in the sense that the same limiting value results no matter how the partitions and subintervals are chosen.

(4) Proposition (Properties of Double Integrals) Assume that all the given integrals exist on a rectangular region

(i)  Linearity Rule: For constants and
    

    
(ii) Dominance Rule:  If throughout a rectangular region then
    


(iii) Subdivision Rule:  If the rectangular region of integration is subdivided into two (disjoint) subrectangles and then

    As with single integrals, it is often not practical to evaluate a double integral even over a simple rectangular region by using the definition. Instead, we compute double integrals by a process called iterated integration, which is like partial differentation in reverse. Suppose is continuous over the rectangle Then we write   to denote the integral obtained by integrating with respect to over the interval with held constant. The integral obtained by this partial integration is a function of alone,   which we integrate over the interval to obtain the iterated integral



Similarly, if we integrate first with respect to over holding constant, and then with respect to over we obtain the iterated integral,  



Fubini's Theorem says that these double integrals have the same value, provided is continuous over (in fact, Fubini's theorem say more).

(5) Proposition (Fubini's Theorem) If is continuous over the rectangle then the double integral may be evaluated by either iterated integral; that is,  

(6) Example (Iterated Integrals) Use iterated integration to compute the double integral   where

    Solution. We compute  

(7) Example (Iterated Integrals) Use iterated integration to compute the double integral where

    Solution. We compute  

(8) Example (Iterated Integrals) Use iterated integration to compute the double integral where

    Solution. We compute

    If on the interval the single integral can be interpreted as the area under the curve over The double integral



has a similar interpretation in terms of volume; since it is natural to define the total volume under the surface as the limit of Riemann sums as the norm tends to 0; of course provided, on the rectangular region

(9) Proposition (Volume Interpretation of the Double Integral) If on the rectangular region then the product is the volume of a parallelepiped (a box) with height and base area The Riemann sum  



provides an estimate of the total volume under the surface over and if is continuous, we expect the approximation to improve by using more refined partitions. That is, the volume under over the domain is given by



when on the rectangular region

(10) Example (Volume Interpretation) Find the volume of the solid bounded below by the rectangle in the and above by the graph of

    Solution. We compute  

(11) Example (Volume Interpretation) Find the volume of the solid bounded below by the rectangle in the and above by the graph of

    Solution. We compute  










Note in this example we used the formulas and

    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

(12) 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.

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



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



whenever both integrals exist.

(14) Example (Double Integral over a Region) 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







(15) Example (Double Integral over a Region) 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   

(16) Example (Double Integral over a Region) 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



where This region has an alternate description:  

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

(17) Example (Double Integral over a Region) 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



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.

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Double Integral
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
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