In Double Integrals over Rectangular Regions, we studied the concept of double integrals and examined the tools needed to compute them. We learned techniques and properties to integrate functions of two variables over rectangular regions. We also discussed several applications, such as finding the volume bounded above by a function over a rectangular region, finding area by integration, and calculating the average value of a function of two variables.
In this section we consider double integrals of functions defined over a general bounded region D D on the plane. Most of the previous results hold in this situation as well, but some techniques need to be extended to cover this more general case.
An example of a general bounded region D D on a plane is shown in Figure 5.12. Since D D is bounded on the plane, there must exist a rectangular region R R on the same plane that encloses the region D , D , that is, a rectangular region R R exists such that D D is a subset of R ( D ⊆ R ) . R ( D ⊆ R ) .
Figure 5.12 For a region D D that is a subset of R , R , we can define a function g ( x , y ) g ( x , y ) to equal f ( x , y ) f ( x , y ) at every point in D D and 0 0 at every point of R R not in D . D .
Suppose z = f ( x , y ) z = f ( x , y ) is defined on a general planar bounded region D D as in Figure 5.12. In order to develop double integrals of f f over D , D , we extend the definition of the function to include all points on the rectangular region R R and then use the concepts and tools from the preceding section. But how do we extend the definition of f f to include all the points on R ? R ? We do this by defining a new function g ( x , y ) g ( x , y ) on R R as follows:
g ( x , y ) = < f ( x , y ) if ( x , y ) is in D 0 if ( x , y ) is in R but not in D g ( x , y ) = < f ( x , y ) if ( x , y ) is in D 0 if ( x , y ) is in R but not in D
Note that we might have some technical difficulties if the boundary of D D is complicated. So we assume the boundary to be a piecewise smooth and continuous simple closed curve. Also, since all the results developed in Double Integrals over Rectangular Regions used an integrable function f ( x , y ) , f ( x , y ) , we must be careful about g ( x , y ) g ( x , y ) and verify that g ( x , y ) g ( x , y ) is an integrable function over the rectangular region R . R . This happens as long as the region D D is bounded by simple closed curves. For now we will concentrate on the descriptions of the regions rather than the function and extend our theory appropriately for integration.
We consider two types of planar bounded regions.
A region D D in the ( x , y ) ( x , y ) -plane is of Type I if it lies between two vertical lines and the graphs of two continuous functions g 1 ( x ) g 1 ( x ) and g 2 ( x ) . g 2 ( x ) . That is (Figure 5.13),
A region D D in the x y x y plane is of Type II if it lies between two horizontal lines and the graphs of two continuous functions h 1 ( y ) and h 2 ( y ) . h 1 ( y ) and h 2 ( y ) . That is (Figure 5.14),
Figure 5.13 A Type I region lies between two vertical lines and the graphs of two functions of x . x .
Figure 5.14 A Type II region lies between two horizontal lines and the graphs of two functions of y . y .
Consider the region in the first quadrant between the functions y = x y = x and y = x 3 y = x 3 (Figure 5.15). Describe the region first as Type I and then as Type II.
The region D is drawn between two functions, namely, y = the square root of x and y = x3." width="503" height="277" />
When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region D D is bounded above by y = x y = x and below by y = x 3 y = x 3 in the interval for x in [ 0 , 1 ] . x in [ 0 , 1 ] . Hence, as Type I, D D is described as the set < ( x , y ) | 0 ≤ x ≤ 1 , x 3 ≤ y ≤ x >. < ( x , y ) | 0 ≤ x ≤ 1 , x 3 ≤ y ≤ x >.
However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region D D is bounded on the left by x = y 2 x = y 2 and on the right by x = y 3 x = y 3 in the interval for y in [ 0 , 1 ] . [ 0 , 1 ] . Hence, as Type II, D D is described as the set < ( x , y ) | 0 ≤ y ≤ 1 , y 2 ≤ x ≤ y 3 >. < ( x , y ) | 0 ≤ y ≤ 1 , y 2 ≤ x ≤ y 3 >.
Consider the region in the first quadrant between the functions y = 2 x y = 2 x and y = x 2 . y = x 2 . Describe the region first as Type I and then as Type II.
To develop the concept and tools for evaluation of a double integral over a general, nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to decide the limits of integrations in double integrals. As a first step, let us look at the following theorem.
Suppose g ( x , y ) g ( x , y ) is the extension to the rectangle R R of the integrable function f ( x , y ) f ( x , y ) defined on the region D D , where D D is inside R R . Sample regions are as shown in Figure 5.12. Then g ( x , y ) g ( x , y ) is integrable and we define the double integral of f ( x , y ) f ( x , y ) over D D by
∬ D f ( x , y ) d A = ∬ R g ( x , y ) d A . ∬ D f ( x , y ) d A = ∬ R g ( x , y ) d A .The right-hand side of this equation is what we have seen before, so this theorem is reasonable because R R is a rectangle and ∬ R g ( x , y ) d A ∬ R g ( x , y ) d A has been discussed in the preceding section. Also, the equality works because the values of g ( x , y ) g ( x , y ) are 0 0 for any point ( x , y ) ( x , y ) that lies outside D , D , and hence these points do not add anything to the integral. However, it is important that the rectangle R R contains the region D . D .
As a matter of fact, if the region D D is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle R R containing the region.
For a function f ( x , y ) f ( x , y ) that is continuous on a region D D of Type I, we have
∬ D f ( x , y ) d A = ∬ D f ( x , y ) d y d x = ∫ a b [ ∫ g 1 ( x ) g 2 ( x ) f ( x , y ) d y ] d x . ∬ D f ( x , y ) d A = ∬ D f ( x , y ) d y d x = ∫ a b [ ∫ g 1 ( x ) g 2 ( x ) f ( x , y ) d y ] d x .
Similarly, for a function f ( x , y ) f ( x , y ) that is continuous on a region D D of Type II, we have
∬ D f ( x , y ) d A = ∬ D f ( x , y ) d x d y = ∫ c d [ ∫ h 1 ( y ) h 2 ( y ) f ( x , y ) d x ] d y . ∬ D f ( x , y ) d A = ∬ D f ( x , y ) d x d y = ∫ c d [ ∫ h 1 ( y ) h 2 ( y ) f ( x , y ) d x ] d y .
The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate f ( x , y ) f ( x , y ) with x x being held constant and the limits of integration being g 1 ( x ) and g 2 ( x ) . g 1 ( x ) and g 2 ( x ) . In the inner integral in the second expression, we integrate f ( x , y ) f ( x , y ) with y y being held constant and the limits of integration are h 1 ( x ) and h 2 ( x ) . h 1 ( x ) and h 2 ( x ) .
Evaluate the integral ∬ D x 2 e x y d A ∬ D x 2 e x y d A where D D is shown in Figure 5.16.
First construct the region D D as a Type I region (Figure 5.16). Here D = < ( x , y ) | 0 ≤ x ≤ 2 , 1 2 x ≤ y ≤ 1 >. D = < ( x , y ) | 0 ≤ x ≤ 2 , 1 2 x ≤ y ≤ 1 >. Then we have
∬ D x 2 e x y d A = ∫ x = 0 x = 2 ∫ y = 1 / 2 x y = 1 x 2 e x y d y d x . ∬ D x 2 e x y d A = ∫ x = 0 x = 2 ∫ y = 1 / 2 x y = 1 x 2 e x y d y d x .
Figure 5.16 We can express region D D as a Type I region and integrate from y = 1 2 x y = 1 2 x to y = 1 , y = 1 , between the lines x = 0 and x = 2 . x = 0 and x = 2 .
Therefore, we have
∫ x = 0 x = 2 ∫ y = 1 2 x y = 1 x 2 e x y d y d x = ∫ x = 0 x = 2 [ ∫ y = 1 / 2 x y = 1 x 2 e x y d y ] d x Iterated integral for a Type I region. = ∫ x = 0 x = 2 [ x 2 e x y x ] | y = 1 / 2 x y = 1 d x Integrate with respect to y using u -substitution with u = x y where x is held constant. = ∫ x = 0 x = 2 [ x e x − x e x 2 / 2 ] d x Integrate with respect to x using u -substitution with u = 1 2 x 2 . = [ x e x − e x − e 1 2 x 2 ] | x = 0 x = 2 = 2 ∫ x = 0 x = 2 ∫ y = 1 2 x y = 1 x 2 e x y d y d x = ∫ x = 0 x = 2 [ ∫ y = 1 / 2 x y = 1 x 2 e x y d y ] d x Iterated integral for a Type I region. = ∫ x = 0 x = 2 [ x 2 e x y x ] | y = 1 / 2 x y = 1 d x Integrate with respect to y using u -substitution with u = x y where x is held constant. = ∫ x = 0 x = 2 [ x e x − x e x 2 / 2 ] d x Integrate with respect to x using u -substitution with u = 1 2 x 2 . = [ x e x − e x − e 1 2 x 2 ] | x = 0 x = 2 = 2
In Example 5.12, we could have looked at the region in another way, such as D = < ( x , y ) | 0 ≤ y ≤ 1 , 0 ≤ x ≤ 2 y >D = < ( x , y ) | 0 ≤ y ≤ 1 , 0 ≤ x ≤ 2 y >(Figure 5.17).
This is a Type II region and the integral would then look like
∬ D x 2 e x y d A = ∫ y = 0 y = 1 ∫ x = 0 x = 2 y x 2 e x y d x d y . ∬ D x 2 e x y d A = ∫ y = 0 y = 1 ∫ x = 0 x = 2 y x 2 e x y d x d y .
However, if we integrate first with respect to x , x , this integral is lengthy to compute because we have to use integration by parts twice.
Evaluate the integral ∬ D ( 3 x 2 + y 2 ) d A ∬ D ( 3 x 2 + y 2 ) d A where = < ( x , y ) | − 2 ≤ y ≤ 3 , y 2 − 3 ≤ x ≤ y + 3 >. = < ( x , y ) | − 2 ≤ y ≤ 3 , y 2 − 3 ≤ x ≤ y + 3 >.
Notice that D D can be seen as either a Type I or a Type II region, as shown in Figure 5.18. However, in this case describing D D as Type I I is more complicated than describing it as Type II. Therefore, we use D D as a Type II region for the integration.
which has points of intersection (6, 3), (1, negative 2), and (0, negative 3). There are vertical lines in the shape, and it is noted that this is a type I region: integrate first with respect to y. In figure b, a region is bounded by x = y2 minus 3 and x = y + 3, which has points of intersection (6, 3), (1, negative 2), and (0, negative 3). There are horizontal lines in the shape, and it is noted that this is a type II region: integrate first with respect to x." width="891" height="372" />
Choosing this order of integration, we have
∬ D ( 3 x 2 + y 2 ) d A = ∫ y = −2 y = 3 ∫ x = y 2 − 3 x = y + 3 ( 3 x 2 + y 2 ) d x d y Iterated integral, Type II region. = ∫ y = −2 y = 3 ( x 3 + x y 2 ) | y 2 − 3 y + 3 d y Integrate with respect to x . = ∫ y = −2 y = 3 ( ( y + 3 ) 3 + ( y + 3 ) y 2 − ( y 2 − 3 ) 3 − ( y 2 − 3 ) y 2 ) d y = ∫ −2 3 ( 54 + 27 y − 12 y 2 + 2 y 3 + 8 y 4 − y 6 ) d y Integrate with respect to y . = [ 54 y + 27 y 2 2 − 4 y 3 + y 4 2 + 8 y 5 5 − y 7 7 ] | −2 3 = 2375 7 . ∬ D ( 3 x 2 + y 2 ) d A = ∫ y = −2 y = 3 ∫ x = y 2 − 3 x = y + 3 ( 3 x 2 + y 2 ) d x d y Iterated integral, Type II region. = ∫ y = −2 y = 3 ( x 3 + x y 2 ) | y 2 − 3 y + 3 d y Integrate with respect to x . = ∫ y = −2 y = 3 ( ( y + 3 ) 3 + ( y + 3 ) y 2 − ( y 2 − 3 ) 3 − ( y 2 − 3 ) y 2 ) d y = ∫ −2 3 ( 54 + 27 y − 12 y 2 + 2 y 3 + 8 y 4 − y 6 ) d y Integrate with respect to y . = [ 54 y + 27 y 2 2 − 4 y 3 + y 4 2 + 8 y 5 5 − y 7 7 ] | −2 3 = 2375 7 .
Sketch the region D D and evaluate the iterated integral ∬ D x y d y d x ∬ D x y d y d x where D D is the region bounded by the curves y = cos x y = cos x and y = sin x y = sin x in the interval [ −3 π / 4 , π / 4 ] . [ −3 π / 4 , π / 4 ] .
Recall from Double Integrals over Rectangular Regions the properties of double integrals. As we have seen from the examples here, all these properties are also valid for a function defined on a nonrectangular bounded region on a plane. In particular, property 3 3 states:
If R = S ∪ T R = S ∪ T and S ∩ T = ∅ S ∩ T = ∅ except at their boundaries, then
∬ R f ( x , y ) d A = ∬ S f ( x , y ) d A + ∬ T f ( x , y ) d A . ∬ R f ( x , y ) d A = ∬ S f ( x , y ) d A + ∬ T f ( x , y ) d A .
Similarly, we have the following property of double integrals over a nonrectangular bounded region on a plane.
Suppose the region D D can be expressed as D = D 1 ∪ D 2 D = D 1 ∪ D 2 where D 1 D 1 and D 2 D 2 do not overlap except at their boundaries. Then
∬ D f ( x , y ) d A = ∬ D 1 f ( x , y ) d A + ∬ D 2 f ( x , y ) d A . ∬ D f ( x , y ) d A = ∬ D 1 f ( x , y ) d A + ∬ D 2 f ( x , y ) d A .
This theorem is particularly useful for nonrectangular regions because it allows us to split a region into a union of regions of Type I and Type II. Then we can compute the double integral on each piece in a convenient way, as in the next example.
Express the region D D shown in Figure 5.19 as a union of regions of Type I or Type II, and evaluate the integral
∬ D ( 2 x + 5 y ) d A . ∬ D ( 2 x + 5 y ) d A .The region D D is not easy to decompose into any one type; it is actually a combination of different types. So we can write it as a union of three regions D 1 , D 2 , and D 3 D 1 , D 2 , and D 3 where, D 1 = < ( x , y ) | − 2 ≤ x ≤ 0 , 0 ≤ y ≤ ( x + 2 ) 2 >, D 1 = < ( x , y ) | − 2 ≤ x ≤ 0 , 0 ≤ y ≤ ( x + 2 ) 2 >, D 2 = < ( x , y ) | 0 ≤ y ≤ 4 , 0 ≤ x ≤ ( y − 1 16 y 3 ) >, D 2 = < ( x , y ) | 0 ≤ y ≤ 4 , 0 ≤ x ≤ ( y − 1 16 y 3 ) >, D 3 = x , y - 4 ≤ y ≤ 0 , - 2 ≤ x ≤ y - y 3 16 . D 3 = x , y - 4 ≤ y ≤ 0 , - 2 ≤ x ≤ y - y 3 16 . These regions are illustrated more clearly in Figure 5.20.
Here D 1 D 1 is Type I I and D 2 D 2 and D 3 D 3 are both of Type II. Hence,
∬ D ( 2 x + 5 y ) d A = ∬ D 1 ( 2 x + 5 y ) d A + ∬ D 2 ( 2 x + 5 y ) d A + ∬ D 3 ( 2 x + 5 y ) d A = ∫ x = −2 x = 0 ∫ y = 0 y = ( x + 2 ) 2 ( 2 x + 5 y ) d y d x + ∫ y = 0 y = 4 ∫ x = 0 x = y − ( 1 / 16 ) y 3 ( 2 + 5 y ) d x d y + ∫ y = −4 y = 0 ∫ x = −2 x = y − ( 1 / 16 ) y 3 ( 2 x + 5 y ) d x d y = ∫ x = −2 x = 0 [ 1 2 ( 2 + x ) 2 ( 20 + 24 x + 5 x 2 ) ] d x + ∫ y = 0 y = 4 [ 1 256 y 6 − 7 16 y 4 + 6 y 2 ] d y + ∫ y = −4 y = 0 [ 1 256 y 6 − 7 16 y 4 + 6 y 2 + 10 y − 4 ] d y = 40 3 + 1664 35 − 1696 35 = 1304 105 . ∬ D ( 2 x + 5 y ) d A = ∬ D 1 ( 2 x + 5 y ) d A + ∬ D 2 ( 2 x + 5 y ) d A + ∬ D 3 ( 2 x + 5 y ) d A = ∫ x = −2 x = 0 ∫ y = 0 y = ( x + 2 ) 2 ( 2 x + 5 y ) d y d x + ∫ y = 0 y = 4 ∫ x = 0 x = y − ( 1 / 16 ) y 3 ( 2 + 5 y ) d x d y + ∫ y = −4 y = 0 ∫ x = −2 x = y − ( 1 / 16 ) y 3 ( 2 x + 5 y ) d x d y = ∫ x = −2 x = 0 [ 1 2 ( 2 + x ) 2 ( 20 + 24 x + 5 x 2 ) ] d x + ∫ y = 0 y = 4 [ 1 256 y 6 − 7 16 y 4 + 6 y 2 ] d y + ∫ y = −4 y = 0 [ 1 256 y 6 − 7 16 y 4 + 6 y 2 + 10 y − 4 ] d y = 40 3 + 1664 35 − 1696 35 = 1304 105 .
Now we could redo this example using a union of two Type II regions (see the Checkpoint).
Consider the region bounded by the curves y = ln x y = ln x and y = e x y = e x in the interval [ 1 , 2 ] . [ 1 , 2 ] . Decompose the region into smaller regions of Type II.
