1. TBF 122 - Generall Mathematics - II Lecture – 7 : Double Integrals
Prof. Dr. Halil İbrahim Karakaş
Başkent University

2. Antiderivatives.
Let F, G and f be functions of two variables and let D be a set which is contained in the domain of each of these functions.
If Fx(x,y)=f(x,y) for all (x,y)D, then the function F is called an antiderivative of f with respect to x over D.
Similarly, If Gy(x,y)=f(x,y) for all (x,y)D, then the function G is called an antiderivative of f with respect to y over D.
When one talks about antiderivatives, if no reference is made to a set D, then the largest set over which the property of being antiderivative is taken as D.
Example 1. For the functions ,
, the function F is an antiderivative of f with respect to x, because
and
the function G is an antiderivative of f with respect to y, because

3. As for antiderivatives of functions of one variable, we hwve:
If F1 and F2 are two antiderivatives with respect to x of the same function f , then there exists a function B (of one variable) such that F2(x,y)= F1(x,y) + B(y).
If G1 and G2 are two antiderivatives with respect to y of the same function f , then there exists a function C of one variable) such that G2(x,y)= G1(x,y) + C(x).
Obviously, B and C may be constant functions in the above statements.
The collection of all antiderivatives of f with respect to x is called the indefinite integral of f with respect to x and it is denoted by
Thus the if F is an antiderivative of f with respect to x , then there is a function B (of one variable) such that

4. The collection of all antiderivatives of f with respect to x is called the indefinite integral of f with respect to y and it is denoted by
Thus the if G is an antiderivative of f with respect to y , then there is a function C (of one variable) such that
Summarizing the above, we have
To find the indefinite integral of a function with respect to a variable, other variables are regarded as constants and integration is performed with respect to that variable as in the case of functions of one variable.

5. Example.
Indefinite integrals of with respect to x and y :
Indefinite integrals of with respect to x and y :
Example.

6. It is possible to consider definite integrals with respect to each variable by means of indefinite integrals. If F is an antiderivative of f with respect to x, then
If G is an antiderivative of f with respect to y, then
Example.

7. When we evaluate a definite integral, we should make sure that the integrand and its antiderivatives are defined in the interval of the limits of integration. Otherwise, the integral is undefined.
Example.
Example.
u=y , dv=exydy
du=dy , v=(1/x)exy
As we see in the above examples, the definite integral of a function with respect to one of the variables gives an expression in the other variable. In other words, definite integral of a function of two variables with respect to one of the variables gives a function of two variables where the independent variable is the other variable. Hence after evaluating definite integral of a function of two variables with respect to a variable, we may think of the definite integral of the obtained function with respect to the remaining variable. Namely we can think of iterated integrals.

8. is evaluted by first finding the integ-ral in square brackets with respect to x and then integrating the obtained expression with respect to y. As we have seen before,
Example. The iterated integral
and thus
Let us evaluate the same integral by reversing the order of integration (note that when we reverse the order of integration the limits of integration also change):

9. We saw in the last example that the value of the iterated integral has not changed when the order of integration is reversed. Let us have another example
Example.
We again obtained the same result when we reversed the order of integration. Actually , one always obtains the same result when one reverses the order of integration in an iterated integral. We will take thia for granted and use it to define double integrals over rectangular regions.
From now on we will express the iterated integral without writing the square bracket because the order of integration will be clear even if we omit square brackets. Thus
and

10. The set D = {(x,y) : a x  b , c  y  d}
Double Integrals over Rectangular Regions. The definition we will give here is not the most general definition of the double integral over a rectangular region; however, it is equivalent to the general definition for all the functions that we will consider.
All the functions we will consider have the following property: For any a, b, c, d  ℝ with a < b, c < d , we have x y
(b,d)
(0,0)
(a,d)
(b,c)
(a,c) a b c d D
The set D = {(x,y) : a x  b , c  y  d}
where a, b, c, d  ℝ and a < b, c < d ( see the figure on the right) is called the rectangular region defined by a, b, c and d.
The double integral of a function f of two variables over the rectangular region
D = {(x,y) : a x  b , c  y  d}
is defined as
In the notations above, f(x,y) is called the integrand , D is called the region of integration and dA is called the area element ; it indicates that this is an integral over a two-dimensional region.

11. Hence, for the rectangular region
and a function f of two variables
Example.
For

12. Example.
For
Example.
For

13. is called a regular x-region.
It is natural to think of double integrals over nonrectangular regions. The first type of regions that come to mind in this respect are regular regions.
Regular Regions.
Let m and n be two functions of one variable, a, b  ℝ, a < b . Assume that m(x) ≤ n(x) for all x  [a,b]. The region x y
(0,0)
y=m(x)
y=n(x) a b D
is called a regular x-region.
A regular x-region looks like the figure on the right.
Clearly, every rectangular region is a regular x-region. We shall have examples of nonrectangular regular x-regions.

14. Example. is a regular x-region. Example.y
(0,0) 1 3
(3,2 )
(3,3)
(1,2) D
(1,1) y x
(0,0) 2
y= x
(4,4)
Example.
The region below the line y = x, above the parabola y = (x-2)2 is a regular x-region (see the figure on the right). 4 D
(1,1)
y= (x-2)2 1
(x-2)2 = x  x2 -5x + 4 = 0
 (x-1)(x-4) = 0
 x = 1 veya x = 4.
In set notations,

15. is called a regular y-region.
Let r and s be two functions of one variable, let c, d  ℝ, c < d . Assume that r(y) ≤ s(y) for all y  [c,d]. The region x y
(0,0)
x=r(y)
x=s(y) c d D
is called a regular y-region.
The figure on the right shows how a regular y-region looks like.
Clearly, every rectangular region is a regular y-region. We shall have examples of nonrectangular regular y-regions.

16. Example. is a regular y-region. x y (0,0) x= -y/2+7 (y=-2x+14) x= y2/4
(4,4)
(5,4) 4 D 2
(1,2)
(6,2)

17. Example.
The region enclosed by the parabola x = y2 and the line y = x-2 is a regular y-region. It is the shaded region below.
y = x-2 , x = y2  y = y2 -2  y2 – y -2 = 0
 (y+1)(y-2) = 0
 y = -1 veya y = 2. y x
(0,0)
(4,2) 2
x= y2
x= y+2 D -1
(1,-1)

18. Consider the region D enclosed by the parabola y=x2 and Example.
Any rectangular region can be considered as both a regular x-region and also as a regular y-region. There are nonrectangular regiopns thet can be considered as both regular x-region and regular y-region.
Consider the region D enclosed by the parabola y=x2 and
Example.
x=0 veya x=1 y x
(0,0)
D is a regular x-region: 1 1
D is also a regular y-region: D

19. Double Integrals over Regular Regions.
Just as in the case of rectangular regions, double integrals over regular regions are defined as iterated (simple) integrals.
If f is a function of two variables defined on the regular x-region x y
(0,0)
y=m(x)
y=n(x) a b D
the double integral of f over D is defined as
In the above definition, the limits of the integral inside the square bracket are expressions in the variable x or constants and the computation of the integral gives an expression in x or a constant.
Integral of that expression with respect to x over the interval [a,b] is the integral of the function f over the region D.
Recall that every rectangular region is a regular x-region. If D is a rectangular region, the previşous definition of double integral over D agrees with the above definition.

20. From now on we will omit the square brackets and write
When computing a double integral over a regular x-region, first we integrate with respect to the variable y and then with respect to x.
Example.
Let us find
where x y
(0,0) 1 3 D
(1,1)
(3,3)
(1,2)
(3,2 )

21. Example. Let us find where y= x y x (0,0) y= (x-2)2 1 4 D 2 (1,1)
(4,4)

22. If f is a function of two variables defined on the regular y-region
Double integrals over regular regular y-regions are defined as follows.
If f is a function of two variables defined on the regular y-region x y
(0,0)
x=r(y)
x=s(y) c d D
the double integral of f over D is defined as
In the above definition, the limits of the integral inside the square bracket are expressions in the variable y or constants and the computation of the integral gives an expression in y or a constant.
Integral of that expression with respect to y over the interval [c,d] is the integral of the function f over the region D.
Recall that every rectangular region is a regular y-region. If D is a rectangular region, the previşous definition of double integral over D agrees with the above definition.

23. From now on we will omit the square brackets and write
for double integrals.
When computing a double integral over a regular y-region, first we integrate with respect to the variable x and then with respect to y.
Example.

24. As we have seen before, D is a regular y-bölgesidir.
Example.
Let D be the region enclosed by the parabola x = y2 and the line y = x – 2. Let us calculate the following double integral over the region D :
As we have seen before, D is a regular y-bölgesidir. y
x= y+2 x
(0,0)
x= y2 -1 2 D
(4,2)
(1,-1)

25. As we have seen in some examples before, there are regions which can be considered as both a regular x-region and as a regular y-region.
To compute the integral of a function over such a region, it might become very important whether the region is considered as a regular x-region or a regular n y-region. For some functions it might be impossible to evaluate the integral analytically when the region is considered as a regular x-region while it might be very easy to evaluate the same integral when the region is considered as a regular y-region. Similarly, for some functions it might be impossible to evaluate the integral analytically when the region is considered as a regular y-region while it might be very easy to evaluate the same integral when the region is considered as a regular x-region.
To shift from one choice to another, one should reverse the order of integration. Now we will see some examples for such situations.

26. If a region can be considered as both a regular x-region and a regular y-region, then the double integral of a function over such a region can be evaluated by using both considerations.
Reversing the Order of Integration.
It can be proved that if
then
In the first iterated integral D is considered as a regular x-region and in the second iterated integral it is considered as a regular y-region .
Shifting from one consideration to the other amounts to reversing the order of integration .

27. with both considerations
Example.
The region D below see the figure) is both a regular x-region and a regular y-region. y x
(0,0) 1 D
Let us calculate
with both considerations
D as a regular x-region
D as a reguar y-region

28. D is both a regular x-region and a regular y-region.
Example.
Let D be the region bounded on the left by the y-axis, on the right by the line y = x and above by the line y = 1 (see the figure below). Let us evaluate the integral y x
(0,0) 1 D
D is both a regular x-region and a regular y-region.
D as a regular x-region
It is impossible to find (analytically) an anti-derivative of
with respect to y. Therefore it is impossible to evaluate this integral with that order of integration.
D as a regular y-region: .

29. The region of integration as it is given is a regular y-region:
Example.
It is impossible to evaluate this integral with the given order of integration.
The region of integration as it is given is a regular y-region: y x
(0,0) 1 D
To see whether D is a regular x-region, it is useful to sketch it. The figure on the right shows that D is a regular x-region.
Thus reversing the order of integration, we evaluate the integral qas follows.

30. Double Integrals over Irregular Regions.
Most of the irregular regions can be expressed as a finite union of regular regions. If a region is a union of finite number of nonoverlapping regular regions, then the double integral of a function over that region is the sum of the double integrals of that function over each regular part.
Double Integrals over Irregular Regions.
For example, if D1 and D2 are two nonoverlapping regular regions and D = D1  D2, then (see the figure) x y
(0,0) c d D1 D2 a b
Note here that D1 is a regular x-region and D2 is a regular y-region.
Clearly, one may think of several unions of regular regions.

31. Let us firs sketch the region of integration.
Example.
Let D be the triangular region bounded below by the two lines y = 2x – 4, y = – x +2 and bounded above by the line y = x . We will evaluate the integral
Let us firs sketch the region of integration. y x
(0,0) 2 D1 4 D2 1 y x 4 4 2 D
The region D can be divided into two nonoverlapping regions (see the figure above) both of which are regular x-regions:
(0,0)

32. y x
(0,0) 2 D1 4 D2 1

33. Area and Double Integrals.x y
(0,0)
y=m(x)
y=n(x) a b D
Area of D x y
(0,0)
The area of the region enclosed by the line y = x and the parabola y = 4x - x2 :
y=4x-x2 4
y=x
Example. D 3
x = 4x - x2  x = 0 or x = 3.

34. Example.
The area of the region bounded above by the line y = x + 1 and the curve xy = 2 , and bounded below by the line y = 1 : x y
(0,0)
x=2/y
y=x+1
(x=y-1) 1 2 1
x(x+1)=2  x=-2 or x=1 D
y=1 2

35. Volume and Double Integrals
Volume and Double Integrals. Consider a function f of two variables defined on a plane region D which takes nonnegative values on D. The volume of the solid formeed by the graph of f over the region D is given by x y
(1,1,0) ( ,0 )
(1,-1,0)
(-1,1,0) z
Volume of the solid defined by the graph of z=4 – x2 – y2 over the region
Example.

36. Find the volume above the triangle with vertices (0,0), (2,0) and (0,1) that lies below the graph of z = 15x3y.
Example. x y
(0,0)
x=-2y+2
(y=-(1/2)x+1) D
(0,1)
(2,0)