1. DOUBLE INTEGRALS OVER RECTANGLES
SECTION 12.1
DOUBLE INTEGRALS OVER RECTANGLES

2. MULTIPLE INTEGRALS
In this chapter, we extend the idea of a definite integral to double and triple integrals of functions of two or three variables.
These ideas are then used to compute volumes, masses, and centroids of more general regions than we were able to consider in Chapters 7.
We also use double integrals to calculate probabilities when two random variables are involved.
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3. MULTIPLE INTEGRALS
We will see that polar coordinates are useful in computing double integrals over some types of regions.
Similarly, we will introduce two coordinate systems in three-dimensional space that greatly simplify computing of triple integrals over certain commonly occurring solid regions.
Cylindrical coordinates
Spherical coordinates
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4. DOUBLE INTEGRALS OVER RECTANGLES
In much the same way that our attempt to solve the area problem led to the definition of a definite integral, we now seek to find the volume of a solid and in the process we arrive at the definition of a double integral.
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5. DOUBLE INTEGRALS OVER RECTANGLES
Just as our attempt to solve the area problem led to the definition of a definite integral, we now seek to find the volume of a solid.
In the process, we arrive at the definition of a double integral.
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6. DEFINITE INTEGRAL—REVIEW
First, let’s recall the basic facts concerning definite integrals of functions of a single variable.
If f(x) is defined for a ≤ x ≤ b, we start by dividing the interval [a, b] into n subintervals [xi–1, xi] of equal width ∆x = xi – xi–1.
We choose sample points xi* in these subintervals.
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7. DEFINITE INTEGRAL—REVIEW
Then, we form the Riemann sum
Then, we take the limit of such sums as the largest of the length approach 0 to obtain the definite integral of f from a to b:
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8. DEFINITE INTEGRAL—REVIEW
In the special case where f(x) ≥ 0, the Riemann sum can be interpreted as the sum of the areas of the approximating rectangles in Figure 1.
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9. DEFINITE INTEGRAL—REVIEW
Then, represents the area under the curve y = f(x) from a to b.
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10. VOLUMES
In a similar manner, we consider a function f of two variables defined on a closed rectangle R = [a, b] × [c, d] = {(x, y)  R2 | a ≤ x ≤ b, c ≤ y ≤ d and we first suppose that f(x, y) ≥ 0.
The graph of f is a surface with equation z = f(x, y).
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11. S = {(x, y, z)  R3 | 0 ≤ z ≤ f(x, y), (x, y)  R} See Figure 2.
VOLUMES
Let S be the solid that lies above R and under the graph of f, that is,
S = {(x, y, z)  R3 | 0 ≤ z ≤ f(x, y), (x, y)  R}
See Figure 2.
Our goal is to find the volume of S.
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12. The first step is to take a partition P of R into subrectangles.
VOLUMES
The first step is to take a partition P of R into subrectangles.
This is accomplished by dividing the intervals [a, b] and [c, d] as follows:
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13. VOLUMES
By drawing lines parallel to the coordinate axes through these partition points as in Figure 3,
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14. Thus, we form the subrectangles from i = 1, …, m and j = 1, …, n.
VOLUMES
Thus, we form the subrectangles from i = 1, …, m and j = 1, …, n.
There are mn of these subrectangles, and they cover R.
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15. VOLUMES
If we let Δxi = xi – xi–1 and Δyj = yj – yj–1 then the area of Rij is ∆Aij = Δxi Δyj
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16. Let’s choose a sample point (xij*, yij*) in each Rij.
VOLUMES
Let’s choose a sample point (xij*, yij*) in each Rij.
Then, we can approximate the part of S that lies above each Rij by a thin rectangular box (or “column”) with:
Base Rij
Height f (xij*, yij*)
See Figure 4.
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17. Compare Figure 4 with Figure 1.
VOLUMES
Compare Figure 4 with Figure 1.
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18. VOLUMES
The volume of this box is the height of the box times the area of the base rectangle:
f(xij*, yij*) ∆Aij
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19. Thus, we get an approximation to the total volume of S:
VOLUMES
We follow this procedure for all the rectangles and add the volumes of the corresponding boxes.
Thus, we get an approximation to the total volume of S:
See Figure 5.
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20. This double sum means that:
VOLUMES
This double sum means that:
For each subrectangle, we evaluate f at the chosen point and multiply by the area of the subrectangle.
Then, we add the results.
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21. VOLUMES
Our intuition tells us that the approximation given in Equation 3 becomes better as the sub-rectangles become smaller.
So if we denote by max Δxi, Δyj the largest of the lengths of all the subintervals, we would expect that:
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22. VOLUMES
We use the expression in Equation 4 to define the volume of the solid S that lies under the graph of f and above the rectangle R.
It can be shown that this definition is consistent with our formula for volume in Section 7.2
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23. VOLUMES
Limits of the type that appear in Equation 4 occur frequently—not just in finding volumes but in a variety of other situations as well—as we will see in Section 12.4–even when f is not a positive function.
So we make the following definition.
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24. Definition 5
The double integral of f over the rectangle R is: if this limit exists.
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25. DOUBLE INTEGRAL
The precise meaning of the limit in Definition 5 is that, for every number e > 0, there is an corresponding number d such that
for:
All partitions P of R whose subinterval lengths are less than d.
Any choice of sample points (xij*, yij*) in Rij.
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26. A function f is called integrable if the limit in Definition 5 exists.
INTEGRABLE FUNCTION
A function f is called integrable if the limit in Definition 5 exists.
It is shown in courses on advanced calculus that all continuous functions are integrable.
In fact, the double integral of f exists provided that f is “not too discontinuous.”
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27. INTEGRABLE FUNCTION In particular,
If f is bounded [that is, there is a constant M such that |f(x, y)| ≤ M for all (x, y) in R].
And f is continuous there, except on a finite number of smooth curves.
Then f is integrable over R.
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28. In this case we can simply let m →  and n → .
DOUBLE INTEGRAL
If we know that is f integrable, we can choose the partitions P to be regular, that is, all the subrectangles Rij have the same dimensions and therefore the same area: ∆A = ΔxΔy.
In this case we can simply let m →  and n → .
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29. DOUBLE INTEGRAL
The sample point (xij*, yij*) can be chosen to be any point in the subrectangle Rij*.
However, suppose we choose it to be the upper right-hand corner of Rij [namely (xi, yj)].
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30. DOUBLE INTEGRAL
See Figure 3.
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31. Then, the expression for the double integral looks simpler:
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32. If f(x, y) ≥ 0, then the volume V of the solid that
DOUBLE INTEGRAL
By comparing Definitions 4 and 5, we see that a volume can be written as a double integral, as follows.
If f(x, y) ≥ 0, then the volume V of the solid that
lies above the rectangle R and below the surface
z = f(x, y) is:
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33. is called a double Riemann sum.
DOUBLE REIMANN SUM
The sum in Definition 5
is called a double Riemann sum.
It is used as an approximation to the value of the double integral.
Notice how similar it is to the Riemann sum in Equation 1 for a function of a single variable.
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34. If f happens to be a positive function, the double Riemann sum:
DOUBLE REIMANN SUM
If f happens to be a positive function, the double Riemann sum:
Represents the sum of volumes of columns, as in Figure 5.
Is an approximation to the volume under the graph of f and above the rectangle R.
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35. Example 1
Estimate the volume of the solid that lies above the square R = [0, 2] × [0, 2] and below the elliptic paraboloid z = 16 – x2 – 2y2.
Divide R into four equal squares and choose the sample point to be the upper right corner of each square Rij.
Sketch the solid and the approximating rectangular boxes.
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36. The squares are shown in Figure 6.
Example 1 SOLUTION
The squares are shown in Figure 6.
The paraboloid is the graph of f(x, y) = 16 – x2 – 2y2
The area of each square is 1.
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37. Approximating the volume by the Riemann sum with m = n = 2, we have:
Example 1 SOLUTION
Approximating the volume by the Riemann sum with m = n = 2, we have:
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38. Example 1 SOLUTION
That is the volume of the approximating rectangular boxes shown in Figure 7.
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39. Figure 8 shows how, when we use 16, 64, and 256 squares.
DOUBLE INTEGRALS
We get better approximations to the volume in Example 1 if we increase the number of squares.
Figure 8 shows how, when we use 16, 64, and 256 squares.
In Example 7 we will be able to show that the exact volume is 48.
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40. DOUBLE INTEGRALS
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41. If R = {(x, y)| –1 ≤ x ≤ 1, –2 ≤ y ≤ 2}, evaluate the integral
Example 2
If R = {(x, y)| –1 ≤ x ≤ 1, –2 ≤ y ≤ 2}, evaluate the integral
SOLUTION
It would be very difficult to evaluate this integral directly from Definition 5.
However, since , we can compute it by interpreting it as a volume.
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42. Example 2 SOLUTION
If then x2 + z2 = 1 and z ≥ 0
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43. Example 2 SOLUTION
So, the given double integral represents the volume of the solid S that lies:
Below the circular cylinder x2 + z2 = 1.
Above the rectangle R.
See Figure 9.
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44. Example 2 SOLUTION
The volume of S is the area of a semicircle with radius 1 times the length of the cylinder.
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45. Here, we consider only the Midpoint Rule for double integrals.
The methods that we used for approximating single integrals (Midpoint Rule, Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals.
Here, we consider only the Midpoint Rule for double integrals.
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46. THE MIDPOINT RULE
This means that we use a double Riemann sum to approximate the double integral, where all the subrectangles have area DA and the sample point (xij*, yij*) in Rij is chosen to be the center of Rij.
That is, is the midpoint of [xi–1, xi] and is the midpoint of [yj–1, yj] .
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47. MIDPOINT RULE FOR DOUBLE INTEGRALS
where:
is the midpoint of [xi–1, xi]
is the midpoint of [yj–1, yj].
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48. where R = {(x, y) | 0 ≤ x ≤ 2, 1 ≤ y ≤ 2}.
Example 3
Use the Midpoint Rule with m = n = 2 to estimate the value of the integral
where R = {(x, y) | 0 ≤ x ≤ 2, 1 ≤ y ≤ 2}.
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49. Example 3 SOLUTION
In using the Midpoint Rule with m = n = 2, we evaluate f(x, y) = x – 3y2 at the centers of the four subrectangles shown in Figure 10.
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50. Example 3 SOLUTION
Thus,
The area of each subrectangle is ∆A = ½.
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51. Example 3 SOLUTION
Thus,
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52. Example 3 SOLUTION
Thus we have
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53. NOTE
In Example 5, we will see that the exact value of the double integral in Example 3 is –12.
Remember that the interpretation of a double integral as a volume is valid only when the integrand f is a positive function.
The integrand in Example 3 is not a positive function.
So, its integral is not a volume.
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54. NOTE
Suppose we keep dividing each of these subrectangles in Figure 10 into four smaller ones with similar shape.
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55. Then, we get these Midpoint Rule approximations.
NOTE
Then, we get these Midpoint Rule approximations.
Notice how these approximations approach the exact value of the double integral, –12.
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56. MULTIPLE INTEGRALS
Recall that it is usually difficult to evaluate single integrals directly from the definition of an integral.
However, the Fundamental Theorem of Calculus (FTC) provides a much easier method.
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57. INTRODUCTION
The evaluation of double integrals from first principles is even more difficult.
Once we have expressed a double integral as an iterated integral, we can then evaluate it by calculating two single integrals.
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58. We use the notation to mean:
INTRODUCTION
Suppose that f is a function of two variables that is integrable on the rectangle
R = [a, b] × [c, d]
We use the notation to mean:
x is held fixed.
f(x, y) is integrated with respect to y from y = c to y = d.
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59. This procedure is called partial integration with respect to y.
Notice its similarity to partial differentiation.
Now, is a number that depends on the value of x.
So, it defines a function of x:
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60. PARTIAL INTEGRATION
If we now integrate the function A with respect to x from x = a to x = b, we get:
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61. ITERATED INTEGRAL
The integral on the right side of Equation 7 is called an iterated integral.
Usually, the brackets are omitted.
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62. ITERATED INTEGRALS Thus, means that:
First, we integrate with respect to y from c to d.
Then, we integrate with respect to x from a to b.
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63. Similarly, the iterated integral
ITERATED INTEGRALS
Similarly, the iterated integral
means that:
First, we integrate with respect to x (holding y fixed) from x = a to x = b.
Then, we integrate the resulting function of y with respect to y from y = c to y = d.
Notice that, in both Equations 8 and 9, we work from the inside out.
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64. Evaluate the iterated integrals. (a) (b)
Example 4
Evaluate the iterated integrals. (a) (b)
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65. Regarding x as a constant, we obtain:
Example 4(a) SOLUTION
Regarding x as a constant, we obtain:
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66. Example 4(a) SOLUTION
Thus, the function A in the preceding discussion is given by in this example.
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67. We now integrate this function of x from 0 to 3:
Example 4(a) SOLUTION
We now integrate this function of x from 0 to 3:
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68. Here, we first integrate with respect to x:
Example 4(b) SOLUTION
Here, we first integrate with respect to x:
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69. ITERATED INTEGRALS
Notice that, in Example 4, we obtained the same answer whether we integrated with respect to y or x first.
In general, it turns out (see Theorem 10) that the two iterated integrals in Equations 8 and 9 are always equal.
That is, the order of integration does not matter.
This is similar to Clairaut’s Theorem on the equality of the mixed partial derivatives.
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70. ITERATED INTEGRALS
The following theorem gives a practical method for evaluating a double integral by expressing it as an iterated integral (in either order).
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71. If f is continuous on the rectangle R =｛(x, y) |a ≤ x ≤ b, c ≤ y ≤ d ｝
FUBUNI’S THEOREM
If f is continuous on the rectangle
R =｛(x, y) |a ≤ x ≤ b, c ≤ y ≤ d ｝
then
More generally, this is true if we assume that f is
bounded on R, f is discontinuous only on a finite
number of smooth curves, and the iterated
integrals exist.
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72. FUBUNI’S THEOREM
Theorem 10 is named after the Italian mathematician Guido Fubini (1879–1943), who proved a very general version of this theorem in 1907.
However, the version for continuous functions was known to the French mathematician Augustin-Louis Cauchy almost a century earlier.
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73. FUBUNI’S THEOREM
The proof of Fubini’s Theorem is too difficult to include in this book.
However, we can at least give an intuitive indication of why it is true for the case where f(x, y) ≥ 0.
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74. FUBUNI’S THEOREM
Recall that, if f is positive, then we can interpret the double integral as:
The volume V of the solid S that lies above R and under the surface z = f(x, y).
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75. FUBUNI’S THEOREM
However, we have another formula that we used for volume in Chapter 7, namely, where:
A(x) is the area of a cross-section of S in the plane through x perpendicular to the x-axis.
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76. FUBUNI’S THEOREM
From Figure 11, you can see that A(x) is the area under the curve C whose equation is
z = f(x, y) where:
x is held constant
c ≤ y ≤ d
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77. Therefore, Then, we have:
FUBUNI’S THEOREM
Therefore, Then, we have:
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78. FUBUNI’S THEOREM
A similar argument, using cross-sections perpendicular to the y-axis as in Figure 12, shows that:
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79. Evaluate the double integral where
Example 5
Evaluate the double integral where
R = {(x, y)| 0 ≤ x ≤ 2, 1 ≤ y ≤ 2}
Compare with Example 3.
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80. Fubini’s Theorem gives:
Example 5 SOLUTION 1
Fubini’s Theorem gives:
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81. This time, we first integrate with respect to x:
Example 5 SOLUTION 2
This time, we first integrate with respect to x:
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82. Notice the negative answer in Example 5. Nothing is wrong with that.
FUBUNI’S THEOREM
Notice the negative answer in Example 5.
Nothing is wrong with that.
The function f in the example is not a positive function.
So, its integral doesn’t represent a volume.
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83. From Figure 13, we see that f is always negative on R.
FUBUNI’S THEOREM
From Figure 13, we see that f is always negative on R.
Thus, the value of the integral is the negative of the volume that lies above the graph of f and below R.
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84. Evaluate where R = [1, 2] × [0, p]
Example 6
Evaluate where R = [1, 2] × [0, p]
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85. If we first integrate with respect to x, we get:
Example 6 SOLUTION
If we first integrate with respect to x, we get:
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86. If we first integrate with respect to y in Example 6, we get
NOTE
If we first integrate with respect to y in Example 6, we get
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87. NOTE
However this order of integration is much more difficult than the method given in the example because it involves integration by parts twice.
Therefore, when we evaluate double integrals it is wise to choose the order of integration that gives simpler integrals.
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88. Find the volume of the solid S that is bounded by:
Example 7
Find the volume of the solid S that is bounded by:
The elliptic paraboloid x2 + 2y2 + z = 16
The planes x = 2 and y = 2
The three coordinate planes
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89. We first observe that S is the solid that lies:
Example 7 SOLUTION
We first observe that S is the solid that lies:
Under the surface z = 16 – x2 – 2y2
Above the square R = [0, 2] × [0, 2]
See Figure 15
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90. This solid was considered in Example 1.
Example 7 SOLUTION
This solid was considered in Example 1.
Now, however, we are in a position to evaluate the double integral using Fubini’s Theorem.
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91. Example 7 SOLUTION
Thus,
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92. ITERATED INTEGRALS
Consider the special case where f(x, y) can be factored as the product of a function of x only and a function of y only.
Then, the double integral of f can be written in a particularly simple form.
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93. To be specific, suppose that:
ITERATED INTEGRALS
To be specific, suppose that:
f(x, y) = g(x)h(y)
R = [a, b] × [c, d]
Then, Fubini’s Theorem gives:
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94. In the inner integral, y is a constant.
ITERATED INTEGRALS
In the inner integral, y is a constant.
So, h(y) is a constant and we can write:
since is a constant.
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95. ITERATED INTEGRALS
Hence, in this case, the double integral of f can be written as the product of two single integrals: where R = [a, b] × [c, d].
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96. If R = [0, p/2] × [0, p/2], then, by Equation 11,
Example 8
If R = [0, p/2] × [0, p/2], then, by Equation 11,
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97. The function f(x, y) = sin x cos y in Example 8 is positive on R.
ITERATED INTEGRALS
The function f(x, y) = sin x cos y in Example 8 is positive on R.
So, the integral represents the volume of the solid that lies above R and below the graph of f shown in Figure 16.
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98. PROPERTIES OF DOUBLE INTEGRALS
We list here three properties of double integrals that can be proved in the same man-ner as in Section 5.2.
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99. PROPERTIES OF DOUBLE INTEGRALS
We assume that all of the integrals exist. Properties 12 and 13are referred to as the linearity of the integral.
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100. PROPERTIES OF DOUBLE INTEGRALS
If f(x, y) ≥ g(x, y) for all (x, y) in R, then
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