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Calculus Volume 3

5.1 Double Integrals over Rectangular Regions

Calculus Volume 35.1 Double Integrals over Rectangular Regions
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  1. Preface
  2. 1 Parametric Equations and Polar Coordinates
    1. Introduction
    2. 1.1 Parametric Equations
    3. 1.2 Calculus of Parametric Curves
    4. 1.3 Polar Coordinates
    5. 1.4 Area and Arc Length in Polar Coordinates
    6. 1.5 Conic Sections
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Vectors in Space
    1. Introduction
    2. 2.1 Vectors in the Plane
    3. 2.2 Vectors in Three Dimensions
    4. 2.3 The Dot Product
    5. 2.4 The Cross Product
    6. 2.5 Equations of Lines and Planes in Space
    7. 2.6 Quadric Surfaces
    8. 2.7 Cylindrical and Spherical Coordinates
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  4. 3 Vector-Valued Functions
    1. Introduction
    2. 3.1 Vector-Valued Functions and Space Curves
    3. 3.2 Calculus of Vector-Valued Functions
    4. 3.3 Arc Length and Curvature
    5. 3.4 Motion in Space
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  5. 4 Differentiation of Functions of Several Variables
    1. Introduction
    2. 4.1 Functions of Several Variables
    3. 4.2 Limits and Continuity
    4. 4.3 Partial Derivatives
    5. 4.4 Tangent Planes and Linear Approximations
    6. 4.5 The Chain Rule
    7. 4.6 Directional Derivatives and the Gradient
    8. 4.7 Maxima/Minima Problems
    9. 4.8 Lagrange Multipliers
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  6. 5 Multiple Integration
    1. Introduction
    2. 5.1 Double Integrals over Rectangular Regions
    3. 5.2 Double Integrals over General Regions
    4. 5.3 Double Integrals in Polar Coordinates
    5. 5.4 Triple Integrals
    6. 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
    7. 5.6 Calculating Centers of Mass and Moments of Inertia
    8. 5.7 Change of Variables in Multiple Integrals
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Vector Calculus
    1. Introduction
    2. 6.1 Vector Fields
    3. 6.2 Line Integrals
    4. 6.3 Conservative Vector Fields
    5. 6.4 Green’s Theorem
    6. 6.5 Divergence and Curl
    7. 6.6 Surface Integrals
    8. 6.7 Stokes’ Theorem
    9. 6.8 The Divergence Theorem
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Chapter Review Exercises
  8. 7 Second-Order Differential Equations
    1. Introduction
    2. 7.1 Second-Order Linear Equations
    3. 7.2 Nonhomogeneous Linear Equations
    4. 7.3 Applications
    5. 7.4 Series Solutions of Differential Equations
    6. Key Terms
    7. Key Equations
    8. Key Concepts
    9. Chapter Review Exercises
  9. A | Table of Integrals
  10. B | Table of Derivatives
  11. C | Review of Pre-Calculus
  12. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
  13. Index

Learning Objectives

  • 5.1.1. Recognize when a function of two variables is integrable over a rectangular region.
  • 5.1.2. Recognize and use some of the properties of double integrals.
  • 5.1.3. Evaluate a double integral over a rectangular region by writing it as an iterated integral.
  • 5.1.4. Use a double integral to calculate the area of a region, volume under a surface, or average value of a function over a plane region.

In this section we investigate double integrals and show how we can use them to find the volume of a solid over a rectangular region in the xyxy-plane. Many of the properties of double integrals are similar to those we have already discussed for single integrals.

Volumes and Double Integrals

We begin by considering the space above a rectangular region R. Consider a continuous function f(x,y)0f(x,y)0 of two variables defined on the closed rectangle R:

R=[a,b]×[c,d]={(x,y)2|axb,cyd}R=[a,b]×[c,d]={(x,y)2|axb,cyd}

Here [a,b]×[c,d][a,b]×[c,d] denotes the Cartesian product of the two closed intervals [a,b][a,b] and [c,d].[c,d]. It consists of rectangular pairs (x,y)(x,y) such that axbaxb and cyd.cyd. The graph of ff represents a surface above the xyxy-plane with equation z=f(x,y)z=f(x,y) where zz is the height of the surface at the point (x,y).(x,y). Let SS be the solid that lies above RR and under the graph of ff (Figure 5.2). The base of the solid is the rectangle RR in the xyxy-plane. We want to find the volume VV of the solid S.S.

In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d).
Figure 5.2 The graph of f(x,y)f(x,y) over the rectangle RR in the xyxy-plane is a curved surface.

We divide the region RR into small rectangles Rij,Rij, each with area ΔAΔA and with sides ΔxΔx and ΔyΔy (Figure 5.3). We do this by dividing the interval [a,b][a,b] into mm subintervals and dividing the interval [c,d][c,d] into nn subintervals. Hence Δx=bam,Δx=bam, Δy=dcn,Δy=dcn, and ΔA=ΔxΔy.ΔA=ΔxΔy.

In the xy plane, there is a rectangle with corners (a, c), (a, d), (b, c), and (b, d). Between a and b on the x axis lines are drawn from a, x1, x2, …, xi, …, b with distance Delta x between each line; between c and d on the y axis lines are drawn from c, y1, y2, …, yj, …, d with distance Delta y between each line. Among the resulting subrectangles, the one in the second column and third row up has a point marked (x*23, y*23). The rectangle Rij is marked with upper right corner (xi, yj). Within this rectangle the point (x*ij, y*ij) is marked.
Figure 5.3 Rectangle RR is divided into small rectangles Rij,Rij, each with area ΔA.ΔA.

The volume of a thin rectangular box above RijRij is f(xij*,yij*)ΔA,f(xij*,yij*)ΔA, where (xij*,yij*)(xij*,yij*) is an arbitrary sample point in each RijRij as shown in the following figure.

In xyz space, there is a surface z = f(x, y). On the x axis, the lines denoting a and b are drawn; on the y axis the lines for c and d are drawn. When the surface is projected onto the xy plane, it forms a rectangle with corners (a, c), (a, d), (b, c), and (b, d). There are additional squares drawn to correspond to changes of Delta x and Delta y. On the surface, a square is marked and its projection onto the plane is marked as Rij. The average value for this small square is f(x*ij, y*ij).
Figure 5.4 A thin rectangular box above RijRij with height f(xij*,yij*).f(xij*,yij*).

Using the same idea for all the subrectangles, we obtain an approximate volume of the solid SS as Vi=1mj=1nf(xij*,yij*)ΔA.Vi=1mj=1nf(xij*,yij*)ΔA. This sum is known as a double Riemann sum and can be used to approximate the value of the volume of the solid. Here the double sum means that for each subrectangle we evaluate the function at the chosen point, multiply by the area of each rectangle, and then add all the results.

As we have seen in the single-variable case, we obtain a better approximation to the actual volume if m and n become larger.

V=limm,ni=1mj=1nf(xij*,yij*)ΔAorV=limΔx,Δy0i=1mj=1nf(xij*,yij*)ΔA.V=limm,ni=1mj=1nf(xij*,yij*)ΔAorV=limΔx,Δy0i=1mj=1nf(xij*,yij*)ΔA.

Note that the sum approaches a limit in either case and the limit is the volume of the solid with the base R. Now we are ready to define the double integral.

Definition

The double integral of the function f(x,y)f(x,y) over the rectangular region RR in the xyxy-plane is defined as

Rf(x,y)dA=limm,ni=1mj=1nf(xij*,yij*)ΔA.Rf(x,y)dA=limm,ni=1mj=1nf(xij*,yij*)ΔA.
5.1

If f(x,y)0,f(x,y)0, then the volume V of the solid S, which lies above RR in the xyxy-plane and under the graph of f, is the double integral of the function f(x,y)f(x,y) over the rectangle R.R. If the function is ever negative, then the double integral can be considered a “signed” volume in a manner similar to the way we defined net signed area in The Definite Integral.

Example 5.1

Setting up a Double Integral and Approximating It by Double Sums

Consider the function z=f(x,y)=3x2yz=f(x,y)=3x2y over the rectangular region R=[0,2]×[0,2]R=[0,2]×[0,2] (Figure 5.5).

  1. Set up a double integral for finding the value of the signed volume of the solid S that lies above RR and “under” the graph of f.f.
  2. Divide R into four squares with m=n=2,m=n=2, and choose the sample point as the upper right corner point of each square (1,1),(2,1),(1,2),(1,1),(2,1),(1,2), and (2,2)(2,2) (Figure 5.6) to approximate the signed volume of the solid S that lies above RR and “under” the graph of f.f.
  3. Divide R into four squares with m=n=2,m=n=2, and choose the sample point as the midpoint of each square: (1/2,1/2),(3/2,1/2),(1/2,3/2),and(3/2,3/2)(1/2,1/2),(3/2,1/2),(1/2,3/2),and(3/2,3/2) to approximate the signed volume.
    In xyz space, there is a surface z = f(x, y) = 3x2 minus y. The corners of the surface are given as (0, 0, 0), (2, 0, 12), (0, 2, negative 2), and (2, 2, 10). The surface is parabolic along the x axis.
    Figure 5.5 The function z=f(x,y)z=f(x,y) graphed over the rectangular region R=[0,2]×[0,2].R=[0,2]×[0,2].

Solution

  1. As we can see, the function z=f(x,y)=3x2yz=f(x,y)=3x2y is above the plane. To find the signed volume of S, we need to divide the region R into small rectangles Rij,Rij, each with area ΔAΔA and with sides ΔxΔx and Δy,Δy, and choose (xij*,yij*)(xij*,yij*) as sample points in each Rij.Rij. Hence, a double integral is set up as
    V=R(3x2y)dA=limm,ni=1mj=1n[3(xij*)2yij*]ΔA.V=R(3x2y)dA=limm,ni=1mj=1n[3(xij*)2yij*]ΔA.
  2. Approximating the signed volume using a Riemann sum with m=n=2m=n=2 we have ΔA=ΔxΔy=1×1=1.ΔA=ΔxΔy=1×1=1. Also, the sample points are (1, 1), (2, 1), (1, 2), and (2, 2) as shown in the following figure.
    On the xy plane, the points (1, 1), (1, 2), (2, 1), and (2, 2) are marked, and these form the upper right corners of four squares marked R11, R12, R21, and R22, respectively. Each square has area Delta A = 1.
    Figure 5.6 Subrectangles for the rectangular region R=[0,2]×[0,2].R=[0,2]×[0,2].

    Hence,
    V=i=12j=12f(xij*,yij*)ΔA=i=12(f(xi1*,yi1*)+f(xi2*,yi2*))ΔA=f(x11*,y11*)ΔA+f(x21*,y21*)ΔA+f(x12*,y12*)ΔA+f(x22*,y22*)ΔA=f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1)=(31)(1)+(121)(1)+(32)(1)+(122)(1)=2+11+1+10=24.V=i=12j=12f(xij*,yij*)ΔA=i=12(f(xi1*,yi1*)+f(xi2*,yi2*))ΔA=f(x11*,y11*)ΔA+f(x21*,y21*)ΔA+f(x12*,y12*)ΔA+f(x22*,y22*)ΔA=f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1)=(31)(1)+(121)(1)+(32)(1)+(122)(1)=2+11+1+10=24.
  3. Approximating the signed volume using a Riemann sum with m=n=2,m=n=2, we have ΔA=ΔxΔy=1×1=1.ΔA=ΔxΔy=1×1=1. In this case the sample points are (1/2, 1/2), (3/2, 1/2), (1/2, 3/2),
    and (3/2, 3/2).
    Hence
    V=i=12j=12f(xij*,yij*)ΔA=f(x11*,y11*)ΔA+f(x21*,y21*)ΔA+f(x12*,y12*)ΔA+f(x22*,y22*)ΔA=f(1/2,1/2)(1)+f(3/2,1/2)(1)+f(1/2,3/2)(1)+f(3/2,3/2)(1)=(3414)(1)+(27412)(1)+(3432)(1)+(27432)(1)=24+254+(34)+214=454=11.V=i=12j=12f(xij*,yij*)ΔA=f(x11*,y11*)ΔA+f(x21*,y21*)ΔA+f(x12*,y12*)ΔA+f(x22*,y22*)ΔA=f(1/2,1/2)(1)+f(3/2,1/2)(1)+f(1/2,3/2)(1)+f(3/2,3/2)(1)=(3414)(1)+(27412)(1)+(3432)(1)+(27432)(1)=24+254+(34)+214=454=11.

Analysis

Notice that the approximate answers differ due to the choices of the sample points. In either case, we are introducing some error because we are using only a few sample points. Thus, we need to investigate how we can achieve an accurate answer.

Checkpoint 5.1

Use the same function z=f(x,y)=3x2yz=f(x,y)=3x2y over the rectangular region R=[0,2]×[0,2].R=[0,2]×[0,2].

Divide R into the same four squares with m=n=2,m=n=2, and choose the sample points as the upper left corner point of each square (0,1),(1,1),(0,2),(0,1),(1,1),(0,2), and (1,2)(1,2) (Figure 5.6) to approximate the signed volume of the solid S that lies above RR and “under” the graph of f.f.

Note that we developed the concept of double integral using a rectangular region R. This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into R, particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region R. Also, the heights may not be exact if the surface z=f(x,y)z=f(x,y) is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid S approach 0 as m and n approach infinity. Also, the double integral of the function z=f(x,y)z=f(x,y) exists provided that the function ff is not too discontinuous. If the function is bounded and continuous over R except on a finite number of smooth curves, then the double integral exists and we say that ff is integrable over R.

Since ΔA=ΔxΔy=ΔyΔx,ΔA=ΔxΔy=ΔyΔx, we can express dAdA as dxdydxdy or dydx.dydx. This means that, when we are using rectangular coordinates, the double integral over a region RR denoted by Rf(x,y)dARf(x,y)dA can be written as Rf(x,y)dxdyRf(x,y)dxdy or Rf(x,y)dydx.Rf(x,y)dydx.

Now let’s list some of the properties that can be helpful to compute double integrals.

Properties of Double Integrals

The properties of double integrals are very helpful when computing them or otherwise working with them. We list here six properties of double integrals. Properties 1 and 2 are referred to as the linearity of the integral, property 3 is the additivity of the integral, property 4 is the monotonicity of the integral, and property 5 is used to find the bounds of the integral. Property 6 is used if f(x,y)f(x,y) is a product of two functions g(x)g(x) and h(y).h(y).

Theorem 5.1

Properties of Double Integrals

Assume that the functions f(x,y)f(x,y) and g(x,y)g(x,y) are integrable over the rectangular region R; S and T are subregions of R; and assume that m and M are real numbers.

  1. The sum f(x,y)+g(x,y)f(x,y)+g(x,y) is integrable and
    R[f(x,y)+g(x,y)]dA=Rf(x,y)dA+Rg(x,y)dA.R[f(x,y)+g(x,y)]dA=Rf(x,y)dA+Rg(x,y)dA.
  2. If c is a constant, then cf(x,y)cf(x,y) is integrable and
    Rcf(x,y)dA=cRf(x,y)dA.Rcf(x,y)dA=cRf(x,y)dA.
  3. If R=STR=ST and ST=ST= except an overlap on the boundaries, then
    Rf(x,y)dA=Sf(x,y)dA+Tf(x,y)dA.Rf(x,y)dA=Sf(x,y)dA+Tf(x,y)dA.
  4. If f(x,y)g(x,y)f(x,y)g(x,y) for (x,y)(x,y) in R,R, then
    Rf(x,y)dARg(x,y)dA.Rf(x,y)dARg(x,y)dA.
  5. If mf(x,y)M,mf(x,y)M, then
    m×A(R)Rf(x,y)dAM×A(R).m×A(R)Rf(x,y)dAM×A(R).
  6. In the case where f(x,y)f(x,y) can be factored as a product of a function g(x)g(x) of xx only and a function h(y)h(y) of yy only, then over the region R={(x,y)|axb,cyd},R={(x,y)|axb,cyd}, the double integral can be written as
    Rf(x,y)dA=(abg(x)dx)(cdh(y)dy).Rf(x,y)dA=(abg(x)dx)(cdh(y)dy).

These properties are used in the evaluation of double integrals, as we will see later. We will become skilled in using these properties once we become familiar with the computational tools of double integrals. So let’s get to that now.

Iterated Integrals

So far, we have seen how to set up a double integral and how to obtain an approximate value for it. We can also imagine that evaluating double integrals by using the definition can be a very lengthy process if we choose larger values for mm and n.n. Therefore, we need a practical and convenient technique for computing double integrals. In other words, we need to learn how to compute double integrals without employing the definition that uses limits and double sums.

The basic idea is that the evaluation becomes easier if we can break a double integral into single integrals by integrating first with respect to one variable and then with respect to the other. The key tool we need is called an iterated integral.

Definition

Assume a,b,c,a,b,c, and dd are real numbers. We define an iterated integral for a function f(x,y)f(x,y) over the rectangular region RR =[a,b]×[c,d]=[a,b]×[c,d] as


  1. abcdf(x,y)dydx=ab[cdf(x,y)dy]dxabcdf(x,y)dydx=ab[cdf(x,y)dy]dx
    5.2

  2. cdabf(x,y)dxdy=cd[abf(x,y)dx]dy.cdabf(x,y)dxdy=cd[abf(x,y)dx]dy.
    5.3

The notation ab[cdf(x,y)dy]dxab[cdf(x,y)dy]dx means that we integrate f(x,y)f(x,y) with respect to y while holding x constant. Similarly, the notation cd[abf(x,y)dx]dycd[abf(x,y)dx]dy means that we integrate f(x,y)f(x,y) with respect to x while holding y constant. The fact that double integrals can be split into iterated integrals is expressed in Fubini’s theorem. Think of this theorem as an essential tool for evaluating double integrals.

Theorem 5.2

Fubini’s Theorem

Suppose that f(x,y)f(x,y) is a function of two variables that is continuous over a rectangular region R={(x,y)2|axb,cyd}.R={(x,y)2|axb,cyd}. Then we see from Figure 5.7 that the double integral of ff over the region equals an iterated integral,

Rf(x,y)dA=Rf(x,y)dxdy=abcdf(x,y)dydx=cdabf(x,y)dxdy.Rf(x,y)dA=Rf(x,y)dxdy=abcdf(x,y)dydx=cdabf(x,y)dxdy.

More generally, Fubini’s theorem is true if ff is bounded on RR and ff is discontinuous only on a finite number of continuous curves. In other words, ff has to be integrable over R.R.

This figure consists of two figures marked a and b. In figure a, in xyz space, a surface is shown that is given by the function f(x, y). A point x is chosen on the x axis, and at this point, it it written fix x. From this point, a plane is projected perpendicular to the xy plane along the line with value x. This plane is marked Area A(x), and the entire space under the surface is marked V. Similarly, in figure b, in xyz space, a surface is shown that is given by the function f(x, y). A point y is chosen on the y axis, and at this point, it it written fix y. From this point, a plane is projected perpendicular to the xy plane along the line with value y. This plane is marked Area A(y), and the entire space under the surface is marked V.
Figure 5.7 (a) Integrating first with respect to yy and then with respect to xx to find the area A(x)A(x) and then the volume V; (b) integrating first with respect to xx and then with respect to yy to find the area A(y)A(y) and then the volume V.

Example 5.2

Using Fubini’s Theorem

Use Fubini’s theorem to compute the double integral Rf(x,y)dARf(x,y)dA where f(x,y)=xf(x,y)=x and R=[0,2]×[0,1].R=[0,2]×[0,1].

Solution

Fubini’s theorem offers an easier way to evaluate the double integral by the use of an iterated integral. Note how the boundary values of the region R become the upper and lower limits of integration.

Rf(x,y)dA=Rf(x,y)dxdy=y=0y=1x=0x=2xdxdy=y=0y=1[x22|x=0x=2]dy=y=0y=12dy=2y|y=0y=1=2.Rf(x,y)dA=Rf(x,y)dxdy=y=0y=1x=0x=2xdxdy=y=0y=1[x22|x=0x=2]dy=y=0y=12dy=2y|y=0y=1=2.

The double integration in this example is simple enough to use Fubini’s theorem directly, allowing us to convert a double integral into an iterated integral. Consequently, we are now ready to convert all double integrals to iterated integrals and demonstrate how the properties listed earlier can help us evaluate double integrals when the function f(x,y)f(x,y) is more complex. Note that the order of integration can be changed (see Example 5.7).

Example 5.3

Illustrating Properties i and ii

Evaluate the double integral R(xy3xy2)dAR(xy3xy2)dA where R={(x,y)|0x2,1y2}.R={(x,y)|0x2,1y2}.

Solution

This function has two pieces: one piece is xyxy and the other is 3xy2.3xy2. Also, the second piece has a constant 3.3. Notice how we use properties i and ii to help evaluate the double integral.

R(xy3xy2)dA=RxydA+R(−3xy2)dAProperty i: Integral of a sum is the sum of the integrals.=y=1y=2x=0x=2xydxdyy=1y=2x=0x=23xy2dxdyConvert double integrals to iterated integrals.=y=1y=2(x22y)|x=0x=2dy3y=1y=2(x22y2)|x=0x=2dyIntegrate with respect tox,holdingyconstant.=y=1y=22ydyy=1y=26y2dyProperty ii: Placing the constant before the integral.=12ydy612y2dyIntegrate with respect toy.=2y22|126y33|12=y2|122y3|12=(41)2(81)=32(7)=314=−11.R(xy3xy2)dA=RxydA+R(−3xy2)dAProperty i: Integral of a sum is the sum of the integrals.=y=1y=2x=0x=2xydxdyy=1y=2x=0x=23xy2dxdyConvert double integrals to iterated integrals.=y=1y=2(x22y)|x=0x=2dy3y=1y=2(x22y2)|x=0x=2dyIntegrate with respect tox,holdingyconstant.=y=1y=22ydyy=1y=26y2dyProperty ii: Placing the constant before the integral.=12ydy612y2dyIntegrate with respect toy.=2y22|126y33|12=y2|122y3|12=(41)2(81)=32(7)=314=−11.

Example 5.4

Illustrating Property v.

Over the region R={(x,y)|1x3,1y2},R={(x,y)|1x3,1y2}, we have 2x2+y213.2x2+y213. Find a lower and an upper bound for the integral R(x2+y2)dA.R(x2+y2)dA.

Solution

For a lower bound, integrate the constant function 2 over the region R.R. For an upper bound, integrate the constant function 13 over the region R.R.

12132dxdy=12[2x|13]dy=122(2)dy=4y|12=4(21)=4121313dxdy=12[13x|13]dy=1213(2)dy=26y|12=26(21)=26.12132dxdy=12[2x|13]dy=122(2)dy=4y|12=4(21)=4121313dxdy=12[13x|13]dy=1213(2)dy=26y|12=26(21)=26.

Hence, we obtain 4R(x2+y2)dA26.4R(x2+y2)dA26.

Example 5.5

Illustrating Property vi

Evaluate the integral ReycosxdAReycosxdA over the region R={(x,y)|0xπ2,0y1}.R={(x,y)|0xπ2,0y1}.

Solution

This is a great example for property vi because the function f(x,y)f(x,y) is clearly the product of two single-variable functions eyey and cosx.cosx. Thus we can split the integral into two parts and then integrate each one as a single-variable integration problem.

ReycosxdA=010π/2eycosxdxdy=(01eydy)(0π/2cosxdx)=(ey|01)(sinx|0π/2)=e1.ReycosxdA=010π/2eycosxdxdy=(01eydy)(0π/2cosxdx)=(ey|01)(sinx|0π/2)=e1.
Checkpoint 5.2
  1. Use the properties of the double integral and Fubini’s theorem to evaluate the integral
    01−13(3x+4y)dydx.01−13(3x+4y)dydx.
  2. Show that 0RsinπxcosπydA1320RsinπxcosπydA132 where R=(0,14)(14,12).R=(0,14)(14,12).

As we mentioned before, when we are using rectangular coordinates, the double integral over a region RR denoted by Rf(x,y)dARf(x,y)dA can be written as Rf(x,y)dxdyRf(x,y)dxdy or Rf(x,y)dydx.Rf(x,y)dydx. The next example shows that the results are the same regardless of which order of integration we choose.

Example 5.6

Evaluating an Iterated Integral in Two Ways

Let’s return to the function f(x,y)=3x2yf(x,y)=3x2y from Example 5.1, this time over the rectangular region R=[0,2]×[0,3].R=[0,2]×[0,3]. Use Fubini’s theorem to evaluate Rf(x,y)dARf(x,y)dA in two different ways:

  1. First integrate with respect to y and then with respect to x;
  2. First integrate with respect to x and then with respect to y.

Solution

Figure 5.7 shows how the calculation works in two different ways.

  1. First integrate with respect to y and then integrate with respect to x:
    Rf(x,y)dA=x=0x=2y=0y=3(3x2y)dydx=x=0x=2(y=0y=3(3x2y)dy)dx=x=0x=2[3x2yy22|y=0y=3]dx=x=0x=2(9x292)dx=3x392x|x=0x=2=15.Rf(x,y)dA=x=0x=2y=0y=3(3x2y)dydx=x=0x=2(y=0y=3(3x2y)dy)dx=x=0x=2[3x2yy22|y=0y=3]dx=x=0x=2(9x292)dx=3x392x|x=0x=2=15.
  2. First integrate with respect to x and then integrate with respect to y:
    Rf(x,y)dA=y=0y=3x=0x=2(3x2y)dxdy=y=0y=3(x=0x=2(3x2y)dx)dy=y=0y=3[x3xy|x=0x=2]dy=y=0y=3(82y)dy=8yy2|y=0y=3=15.Rf(x,y)dA=y=0y=3x=0x=2(3x2y)dxdy=y=0y=3(x=0x=2(3x2y)dx)dy=y=0y=3[x3xy|x=0x=2]dy=y=0y=3(82y)dy=8yy2|y=0y=3=15.

Analysis

With either order of integration, the double integral gives us an answer of 15. We might wish to interpret this answer as a volume in cubic units of the solid SS below the function f(x,y)=3x2yf(x,y)=3x2y over the region R=[0,2]×[0,3].R=[0,2]×[0,3]. However, remember that the interpretation of a double integral as a (non-signed) volume works only when the integrand ff is a nonnegative function over the base region R.R.

Checkpoint 5.3

Evaluate y=−3y=2x=3x=5(23x2+y2)dxdy.y=−3y=2x=3x=5(23x2+y2)dxdy.

In the next example we see that it can actually be beneficial to switch the order of integration to make the computation easier. We will come back to this idea several times in this chapter.

Example 5.7

Switching the Order of Integration

Consider the double integral Rxsin(xy)dARxsin(xy)dA over the region R={(x,y)|0x3,0y2}R={(x,y)|0x3,0y2} (Figure 5.8).

  1. Express the double integral in two different ways.
  2. Analyze whether evaluating the double integral in one way is easier than the other and why.
  3. Evaluate the integral.
    The function z = f(x, y) = x sin(xy) is shown, which starts with z = 0 along the x axis. Then, the function increases roughly as a normal sin function would, but then skews a bit and decreases as x increases after pi/2.
    Figure 5.8 The function z=f(x,y)=xsin(xy)z=f(x,y)=xsin(xy) over the rectangular region R=[0,π]×[1,2].R=[0,π]×[1,2].

Solution

  1. We can express Rxsin(xy)dARxsin(xy)dA in the following two ways: first by integrating with respect to yy and then with respect to x;x; second by integrating with respect to xx and then with respect to y.y.
    Rxsin(xy)dA=x=0x=πy=1y=2xsin(xy)dydxIntegrate first with respect toy.=y=1y=2x=0x=πxsin(xy)dxdyIntegrate first with respect tox.Rxsin(xy)dA=x=0x=πy=1y=2xsin(xy)dydxIntegrate first with respect toy.=y=1y=2x=0x=πxsin(xy)dxdyIntegrate first with respect tox.
  2. If we want to integrate with respect to y first and then integrate with respect to x,x, we see that we can use the substitution u=xy,u=xy, which gives du=xdy.du=xdy. Hence the inner integral is simply sinudusinudu and we can change the limits to be functions of x,
    Rxsin(xy)dA=x=0x=πy=1y=2xsin(xy)dydx=x=0x=π[u=xu=2xsin(u)du]dx.Rxsin(xy)dA=x=0x=πy=1y=2xsin(xy)dydx=x=0x=π[u=xu=2xsin(u)du]dx.

    However, integrating with respect to xx first and then integrating with respect to yy requires integration by parts for the inner integral, with u=xu=x and dv=sin(xy)dx.dv=sin(xy)dx.
    Then du=dxdu=dx and v=cos(xy)y,v=cos(xy)y, so
    Rxsin(xy)dA=y=1y=2x=0x=πxsin(xy)dxdy=y=1y=2[xcos(xy)y|x=0x=π+1yx=0x=πcos(xy)dx]dy.Rxsin(xy)dA=y=1y=2x=0x=πxsin(xy)dxdy=y=1y=2[xcos(xy)y|x=0x=π+1yx=0x=πcos(xy)dx]dy.

    Since the evaluation is getting complicated, we will only do the computation that is easier to do, which is clearly the first method.
  3. Evaluate the double integral using the easier way.
    Rxsin(xy)dA=x=0x=πy=1y=2xsin(xy)dydx=x=0x=π[u=xu=2xsin(u)du]dx=x=0x=π[cosu|u=xu=2x]dx=x=0x=π(cos2x+cosx)dx=12sin2x+sinx|x=0x=π=0.Rxsin(xy)dA=x=0x=πy=1y=2xsin(xy)dydx=x=0x=π[u=xu=2xsin(u)du]dx=x=0x=π[cosu|u=xu=2x]dx=x=0x=π(cos2x+cosx)dx=12sin2x+sinx|x=0x=π=0.
Checkpoint 5.4

Evaluate the integral RxexydARxexydA where R=[0,1]×[0,ln5].R=[0,1]×[0,ln5].

Applications of Double Integrals

Double integrals are very useful for finding the area of a region bounded by curves of functions. We describe this situation in more detail in the next section. However, if the region is a rectangular shape, we can find its area by integrating the constant function f(x,y)=1f(x,y)=1 over the region R.R.

Definition

The area of the region RR is given by A(R)=R1dA.A(R)=R1dA.

This definition makes sense because using f(x,y)=1f(x,y)=1 and evaluating the integral make it a product of length and width. Let’s check this formula with an example and see how this works.

Example 5.8

Finding Area Using a Double Integral

Find the area of the region R={(x,y)|0x3,0y2}R={(x,y)|0x3,0y2} by using a double integral, that is, by integrating 1 over the region R.R.

Solution

The region is rectangular with length 3 and width 2, so we know that the area is 6. We get the same answer when we use a double integral:

A(R)=02031dxdy=02[x|03]dy=023dy=302dy=3y|02=3(2)=6.A(R)=02031dxdy=02[x|03]dy=023dy=302dy=3y|02=3(2)=6.

We have already seen how double integrals can be used to find the volume of a solid bounded above by a function f(x,y)f(x,y) over a region RR provided f(x,y)0f(x,y)0 for all (x,y)(x,y) in R.R. Here is another example to illustrate this concept.

Example 5.9

Volume of an Elliptic Paraboloid

Find the volume VV of the solid SS that is bounded by the elliptic paraboloid 2x2+y2+z=27,2x2+y2+z=27, the planes x=3x=3 and y=3,y=3, and the three coordinate planes.

Solution

First notice the graph of the surface z=272x2y2z=272x2y2 in Figure 5.9(a) and above the square region R1=[−3,3]×[−3,3].R1=[−3,3]×[−3,3]. However, we need the volume of the solid bounded by the elliptic paraboloid 2x2+y2+z=27,2x2+y2+z=27, the planes x=3x=3 and y=3,y=3, and the three coordinate planes.

This figure consists of two figures marked a and b. In figure a, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from negative 3 to positive 3. The shape looks like a sheet that has been pinned at the corners and forced up gently in the middle. In figure b, in xyz space, the surface z = 20 minus 2x2 minus y2 is shown for x and y from 0 to positive 3. The surface is the upper corner of the figure from part a, and below the surface is marked the solid S.
Figure 5.9 (a) The surface z=272x2y2z=272x2y2 above the square region R1=[−3,3]×[−3,3].R1=[−3,3]×[−3,3]. (b) The solid S lies under the surface z=272x2y2z=272x2y2 above the square region R2=[0,3]×[0,3].R2=[0,3]×[0,3].

Now let’s look at the graph of the surface in Figure 5.9(b). We determine the volume V by evaluating the double integral over R2:R2:

V=RzdA=R(272x2y2)dA=y=0y=3x=0x=3(272x2y2)dxdyConvert to iterated integral.=y=0y=3[27x23x3y2x]|x=0x=3dyIntegrate with respect tox.=y=0y=3(643y2)dy=63yy3|y=0y=3=162.V=RzdA=R(272x2y2)dA=y=0y=3x=0x=3(272x2y2)dxdyConvert to iterated integral.=y=0y=3[27x23x3y2x]|x=0x=3dyIntegrate with respect tox.=y=0y=3(643y2)dy=63yy3|y=0y=3=162.
Checkpoint 5.5

Find the volume of the solid bounded above by the graph of f(x,y)=xysin(x2y)f(x,y)=xysin(x2y) and below by the xyxy-plane on the rectangular region R=[0,1]×[0,π].R=[0,1]×[0,π].

Recall that we defined the average value of a function of one variable on an interval [a,b][a,b] as

fave=1baabf(x)dx.fave=1baabf(x)dx.

Similarly, we can define the average value of a function of two variables over a region R. The main difference is that we divide by an area instead of the width of an interval.

Definition

The average value of a function of two variables over a region RR is

fave=1AreaRRf(x,y)dA.fave=1AreaRRf(x,y)dA.
5.4

In the next example we find the average value of a function over a rectangular region. This is a good example of obtaining useful information for an integration by making individual measurements over a grid, instead of trying to find an algebraic expression for a function.

Example 5.10

Calculating Average Storm Rainfall

The weather map in Figure 5.10 shows an unusually moist storm system associated with the remnants of Hurricane Karl, which dumped 4–8 inches (100–200 mm) of rain in some parts of the Midwest on September 22–23, 2010. The area of rainfall measured 300 miles east to west and 250 miles north to south. Estimate the average rainfall over the entire area in those two days.

A map of Wisconsin and Minnesota that shows many cities with numbers affixed to them. The highest numbers come in a narrow band, and the map is colored accordingly. The map effectively looks like a contour map, but instead of elevations, it uses these numbers.
Figure 5.10 Effects of Hurricane Karl, which dumped 4–8 inches (100–200 mm) of rain in some parts of southwest Wisconsin, southern Minnesota, and southeast South Dakota over a span of 300 miles east to west and 250 miles north to south.

Solution

Place the origin at the southwest corner of the map so that all the values can be considered as being in the first quadrant and hence all are positive. Now divide the entire map into six rectangles (m=2andn=3),(m=2andn=3), as shown in Figure 5.11. Assume f(x,y)f(x,y) denotes the storm rainfall in inches at a point approximately xx miles to the east of the origin and y miles to the north of the origin. Let RR represent the entire area of 250×300=75000250×300=75000 square miles. Then the area of each subrectangle is

ΔA=16(75000)=12500.ΔA=16(75000)=12500.

Assume (xij*,yij*)(xij*,yij*) are approximately the midpoints of each subrectangle Rij.Rij. Note the color-coded region at each of these points, and estimate the rainfall. The rainfall at each of these points can be estimated as:

At (x11,y11)(x11,y11) the rainfall is 0.08.

At (x12,y12)(x12,y12) the rainfall is 0.08.

At (x13,y13)(x13,y13) the rainfall is 0.01.

At (x21,y21)(x21,y21) the rainfall is 1.70.

At (x22,y22)(x22,y22) the rainfall is 1.74.

At (x23,y23)(x23,y23) the rainfall is 3.00.

Another version of the previous storm map, but this time with lines drawn for x = 100, 200, and 300 and for y = 125 and 250. There is a dot in the center of each of the resulting rectangles.
Figure 5.11 Storm rainfall with rectangular axes and showing the midpoints of each subrectangle.

According to our definition, the average storm rainfall in the entire area during those two days was

fave=1AreaRRf(x,y)dxdy=175000Rf(x,y)dxdy175,000i=13j=12f(xij*,yij*)ΔA175,000[f(x11*,y11*)ΔA+f(x12*,y12*)ΔA+f(x13*,y13*)ΔA+f(x21*,y21*)ΔA+f(x22*,y22*)ΔA+f(x23*,y23*)ΔA]175,000[0.08+0.08+0.01+1.70+1.74+3.00]ΔA175,000[0.08+0.08+0.01+1.70+1.74+3.00]12500530[0.08+0.08+0.01+1.70+1.74+3.00]1.10.fave=1AreaRRf(x,y)dxdy=175000Rf(x,y)dxdy175,000i=13j=12f(xij*,yij*)ΔA175,000[f(x11*,y11*)ΔA+f(x12*,y12*)ΔA+f(x13*,y13*)ΔA+f(x21*,y21*)ΔA+f(x22*,y22*)ΔA+f(x23*,y23*)ΔA]175,000[0.08+0.08+0.01+1.70+1.74+3.00]ΔA175,000[0.08+0.08+0.01+1.70+1.74+3.00]12500530[0.08+0.08+0.01+1.70+1.74+3.00]1.10.

During September 22–23, 2010 this area had an average storm rainfall of approximately 1.10 inches.

Checkpoint 5.6

A contour map is shown for a function f(x,y)f(x,y) on the rectangle R=[−3,6]×[−1,4].R=[−3,6]×[−1,4].

A contour map is shown with the highest point being about 18 and centered near (4, negative 1). From this point, the values decrease to 16, 14, 12, 10, 8, and 6 roughly every 0.5 to 1 distance. The lowest point is negative four near (negative 3, 4). There is a local minimum of 2 near (negative 1, 0).
  1. Use the midpoint rule with m=3m=3 and n=2n=2 to estimate the value of Rf(x,y)dA.Rf(x,y)dA.
  2. Estimate the average value of the function f(x,y).f(x,y).

Section 5.1 Exercises

In the following exercises, use the midpoint rule with m=4m=4 and n=2n=2 to estimate the volume of the solid bounded by the surface z=f(x,y),z=f(x,y), the vertical planes x=1,x=1, x=2,x=2, y=1,y=1, and y=2,y=2, and the horizontal plane z=0.z=0.

1.

f(x,y)=4x+2y+8xyf(x,y)=4x+2y+8xy

2.

f(x,y)=16x2+y2f(x,y)=16x2+y2

In the following exercises, estimate the volume of the solid under the surface z=f(x,y)z=f(x,y) and above the rectangular region R by using a Riemann sum with m=n=2m=n=2 and the sample points to be the lower left corners of the subrectangles of the partition.

3.

f(x,y)=sinxcosy,f(x,y)=sinxcosy, R=[0,π]×[0,π]R=[0,π]×[0,π]

4.

f(x,y)=cosx+cosy,f(x,y)=cosx+cosy, R=[0,π]×[0,π2]R=[0,π]×[0,π2]

5.

Use the midpoint rule with m=n=2m=n=2 to estimate Rf(x,y)dA,Rf(x,y)dA, where the values of the function f on R=[8,10]×[9,11]R=[8,10]×[9,11] are given in the following table.

y
x 9 9.5 10 10.5 11
8 9.8 5 6.7 5 5.6
8.5 9.4 4.5 8 5.4 3.4
9 8.7 4.6 6 5.5 3.4
9.5 6.7 6 4.5 5.4 6.7
10 6.8 6.4 5.5 5.7 6.8
6.

The values of the function f on the rectangle R=[0,2]×[7,9]R=[0,2]×[7,9] are given in the following table. Estimate the double integral Rf(x,y)dARf(x,y)dA by using a Riemann sum with m=n=2.m=n=2. Select the sample points to be the upper right corners of the subsquares of R.

y0=7y0=7 y1=8y1=8 y2=9y2=9
x0=0x0=0 10.22 10.21 9.85
x1=1x1=1 6.73 9.75 9.63
x2=2x2=2 5.62 7.83 8.21
7.

The depth of a children’s 4-ft by 4-ft swimming pool, measured at 1-ft intervals, is given in the following table.

  1. Estimate the volume of water in the swimming pool by using a Riemann sum with m=n=2.m=n=2. Select the sample points using the midpoint rule on R=[0,4]×[0,4].R=[0,4]×[0,4].
  2. Find the average depth of the swimming pool.
    y
    x 0 1 2 3 4
    0 1 1.5 2 2.5 3
    1 1 1.5 2 2.5 3
    2 1 1.5 1.5 2.5 3
    3 1 1 1.5 2 2.5
    4 1 1 1 1.5 2
8.

The depth of a 3-ft by 3-ft hole in the ground, measured at 1-ft intervals, is given in the following table.

  1. Estimate the volume of the hole by using a Riemann sum with m=n=3m=n=3 and the sample points to be the upper left corners of the subsquares of R.
  2. Find the average depth of the hole.
    y
    x 0 1 2 3
    0 6 6.5 6.4 6
    1 6.5 7 7.5 6.5
    2 6.5 6.7 6.5 6
    3 6 6.5 5 5.6
9.

The level curves f(x,y)=kf(x,y)=k of the function f are given in the following graph, where k is a constant.

  1. Apply the midpoint rule with m=n=2m=n=2 to estimate the double integral Rf(x,y)dA,Rf(x,y)dA, where R=[0.2,1]×[0,0.8].R=[0.2,1]×[0,0.8].
  2. Estimate the average value of the function f on R.
    A series of curves marked k = negative 1, negative ½, negative ¼, negative 1/8, 0, 1/8, ¼, ½, and 1. The line marked k = 0 serves as an asymptote along the line y = x. The lines originate at (along the y axis) 1, 0.7, 0.5, 0.38, 0, (along the x axis) 0.38, 0.5, 0.7, and 1, with the further out lines curving less dramatically toward the asymptote.
10.

The level curves f(x,y)=kf(x,y)=k of the function f are given in the following graph, where k is a constant.

  1. Apply the midpoint rule with m=n=2m=n=2 to estimate the double integral Rf(x,y)dA,Rf(x,y)dA, where R=[0.1,0.5]×[0.1,0.5].R=[0.1,0.5]×[0.1,0.5].
  2. Estimate the average value of the function f on R.
    A series of quarter circles drawn in the first quadrant marked k = 1/32, 1/16, 1/8, ¼, ½, ¾, and 1. The quarter circles have radii 0. 17, 0.25, 0.35, 0.5, 0.71, 0.87, and 1, respectively.
11.

The solid lying under the surface z=4y2z=4y2 and above the rectangular region R=[0,2]×[0,2]R=[0,2]×[0,2] is illustrated in the following graph. Evaluate the double integral Rf(x,y)dA,Rf(x,y)dA, where f(x,y)=4y2,f(x,y)=4y2, by finding the volume of the corresponding solid.

A quarter cylinder with center along the x axis and with radius 2. It has height 2 as shown.
12.

The solid lying under the plane z=y+4z=y+4 and above the rectangular region R=[0,2]×[0,4]R=[0,2]×[0,4] is illustrated in the following graph. Evaluate the double integral Rf(x,y)dA,Rf(x,y)dA, where f(x,y)=y+4,f(x,y)=y+4, by finding the volume of the corresponding solid.

In xyz space, a shape is created with sides given by y = 0, x = 0, y = 4, x = 2, z = 0, and the plane the runs from z = 4 along the y axis to z = 8 along the plane formed by y = 4.

In the following exercises, calculate the integrals by interchanging the order of integration.

13.

−11(−22(2x+3y+5)dx)dy−11(−22(2x+3y+5)dx)dy

14.

02(01(x+2ey3)dx)dy02(01(x+2ey3)dx)dy

15.

127(12(x3+y3)dy)dx127(12(x3+y3)dy)dx

16.

116(18(x4+2y3)dy)dx116(18(x4+2y3)dy)dx

17.

ln2ln3(0lex+ydy)dxln2ln3(0lex+ydy)dx

18.

02(013x+ydy)dx02(013x+ydy)dx

19.

16(29yx2dy)dx16(29yx2dy)dx

20.

19(42xy2dy)dx19(42xy2dy)dx

In the following exercises, evaluate the iterated integrals by choosing the order of integration.

21.

0π0π/2sin(2x)cos(3y)dxdy0π0π/2sin(2x)cos(3y)dxdy

22.

π/12π/8π/4π/3[cotx+tan(2y)]dxdyπ/12π/8π/4π/3[cotx+tan(2y)]dxdy

23.

1e1e[1xsin(lnx)+1ycos(lny)]dxdy1e1e[1xsin(lnx)+1ycos(lny)]dxdy

24.

1e1esin(lnx)cos(lny)xydxdy1e1esin(lnx)cos(lny)xydxdy

25.

1212(lnyx+x2y+1)dydx1212(lnyx+x2y+1)dydx

26.

1e12x2ln(x)dydx1e12x2ln(x)dydx

27.

1312yarctan(1x)dydx1312yarctan(1x)dydx

28.

0101/2(arcsinx+arcsiny)dydx0101/2(arcsinx+arcsiny)dydx

29.

0112xex+4ydydx0112xex+4ydydx

30.

1201xexydydx1201xexydydx

31.

1e1e(lnyy+lnxx)dydx1e1e(lnyy+lnxx)dydx

32.

1e1e(xlnyy+ylnxx)dydx1e1e(xlnyy+ylnxx)dydx

33.

0112(xx2+y2)dydx0112(xx2+y2)dydx

34.

0112yx+y2dydx0112yx+y2dydx

In the following exercises, find the average value of the function over the given rectangles.

35.

f(x,y)=x+2y,f(x,y)=x+2y, R=[0,1]×[0,1]R=[0,1]×[0,1]

36.

f(x,y)=x4+2y3,f(x,y)=x4+2y3, R=[1,2]×[2,3]R=[1,2]×[2,3]

37.

f(x,y)=sinhx+sinhy,f(x,y)=sinhx+sinhy, R=[0,1]×[0,2]R=[0,1]×[0,2]

38.

f(x,y)=arctan(xy),f(x,y)=arctan(xy), R=[0,1]×[0,1]R=[0,1]×[0,1]

39.

Let f and g be two continuous functions such that 0m1f(x)M10m1f(x)M1 for any x[a,b]x[a,b] and 0m2g(y)M20m2g(y)M2 for any y[c,d].y[c,d]. Show that the following inequality is true:

m1m2(ba)(cd)abcdf(x)g(y)dydxM1M2(ba)(cd).m1m2(ba)(cd)abcdf(x)g(y)dydxM1M2(ba)(cd).

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

40.

1e2Rex2y2dA1,1e2Rex2y2dA1, where R=[0,1]×[0,1]R=[0,1]×[0,1]

41.

π2144RsinxcosydAπ248,π2144RsinxcosydAπ248, where R=[π6,π3]×[π6,π3]R=[π6,π3]×[π6,π3]

42.

0ReycosxdAπ2,0ReycosxdAπ2, where R=[0,π2]×[0,π2]R=[0,π2]×[0,π2]

43.

0R(lnx)(lny)dA(e1)2,0R(lnx)(lny)dA(e1)2, where R=[1,e]×[1,e]R=[1,e]×[1,e]

44.

Let f and g be two continuous functions such that 0m1f(x)M10m1f(x)M1 for any x[a,b]x[a,b] and 0m2g(y)M20m2g(y)M2 for any y[c,d].y[c,d]. Show that the following inequality is true:

(m1+m2)(ba)(cd)abcd[f(x)+g(y)]dydx(M1+M2)(ba)(cd).(m1+m2)(ba)(cd)abcd[f(x)+g(y)]dydx(M1+M2)(ba)(cd).

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

45.

2eR(ex2+ey2)dA2,2eR(ex2+ey2)dA2, where R=[0,1]×[0,1]R=[0,1]×[0,1]

46.

π236R(sinx+cosy)dAπ2336,π236R(sinx+cosy)dAπ2336, where R=[π6,π3]×[π6,π3]R=[π6,π3]×[π6,π3]

47.

π2eπ/2R(cosx+ey)dAπ,π2eπ/2R(cosx+ey)dAπ, where R=[0,π2]×[0,π2]R=[0,π2]×[0,π2]

48.

1eR(eylnx)dA2,1eR(eylnx)dA2, where R=[0,1]×[0,1]R=[0,1]×[0,1]

In the following exercises, the function f is given in terms of double integrals.

  1. Determine the explicit form of the function f.
  2. Find the volume of the solid under the surface z=f(x,y)z=f(x,y) and above the region R.
  3. Find the average value of the function f on R.
  4. Use a computer algebra system (CAS) to plot z=f(x,y)z=f(x,y) and z=favez=fave in the same system of coordinates.
49.

[T] f(x,y)=0y0x(xs+yt)dsdt,f(x,y)=0y0x(xs+yt)dsdt, where (x,y)R=[0,1]×[0,1](x,y)R=[0,1]×[0,1]

50.

[T] f(x,y)=0x0y[cos(s)+cos(t)]dtds,f(x,y)=0x0y[cos(s)+cos(t)]dtds, where (x,y)R=[0,3]×[0,3](x,y)R=[0,3]×[0,3]

51.

Show that if f and g are continuous on [a,b][a,b] and [c,d],[c,d], respectively, then

abcd[f(x)+g(y)]dydx=(dc)abf(x)dxabcd[f(x)+g(y)]dydx=(dc)abf(x)dx

+abcdg(y)dydx=(ba)cdg(y)dy+cdabf(x)dxdy.+abcdg(y)dydx=(ba)cdg(y)dy+cdabf(x)dxdy.

52.

Show that abcdyf(x)+xg(y)dydx=12(d2c2)(abf(x)dx)+12(b2a2)(cdg(y)dy).abcdyf(x)+xg(y)dydx=12(d2c2)(abf(x)dx)+12(b2a2)(cdg(y)dy).

53.

[T] Consider the function f(x,y)=ex2y2,f(x,y)=ex2y2, where (x,y)R=[−1,1]×[−1,1].(x,y)R=[−1,1]×[−1,1].

  1. Use the midpoint rule with m=n=2,4,…,10m=n=2,4,…,10 to estimate the double integral I=Rex2y2dA.I=Rex2y2dA. Round your answers to the nearest hundredths.
  2. For m=n=2,m=n=2, find the average value of f over the region R. Round your answer to the nearest hundredths.
  3. Use a CAS to graph in the same coordinate system the solid whose volume is given by Rex2y2dARex2y2dA and the plane z=fave.z=fave.
54.

[T] Consider the function f(x,y)=sin(x2)cos(y2),f(x,y)=sin(x2)cos(y2), where (x,y)R=[−1,1]×[−1,1].(x,y)R=[−1,1]×[−1,1].

  1. Use the midpoint rule with m=n=2,4,…,10m=n=2,4,…,10 to estimate the double integral I=Rsin(x2)cos(y2)dA.I=Rsin(x2)cos(y2)dA. Round your answers to the nearest hundredths.
  2. For m=n=2,m=n=2, find the average value of f over the region R. Round your answer to the nearest hundredths.
  3. Use a CAS to graph in the same coordinate system the solid whose volume is given by Rsin(x2)cos(y2)dARsin(x2)cos(y2)dA and the plane z=fave.z=fave.

In the following exercises, the functions fnfn are given, where n1n1 is a natural number.

  1. Find the volume of the solids SnSn under the surfaces z=fn(x,y)z=fn(x,y) and above the region R.
  2. Determine the limit of the volumes of the solids SnSn as n increases without bound.
55.

f(x,y)=xn+yn+xy,(x,y)R=[0,1]×[0,1]f(x,y)=xn+yn+xy,(x,y)R=[0,1]×[0,1]

56.

f(x,y)=1xn+1yn,(x,y)R=[1,2]×[1,2]f(x,y)=1xn+1yn,(x,y)R=[1,2]×[1,2]

57.

Show that the average value of a function f on a rectangular region R=[a,b]×[c,d]R=[a,b]×[c,d] is fave1mni=1mj=1nf(xij*,yij*),fave1mni=1mj=1nf(xij*,yij*), where (xij*,yij*)(xij*,yij*) are the sample points of the partition of R, where 1im1im and 1jn.1jn.

58.

Use the midpoint rule with m=nm=n to show that the average value of a function f on a rectangular region R=[a,b]×[c,d]R=[a,b]×[c,d] is approximated by

fave1n2i,j=1nf(12(xi1+xi),12(yj1+yj)).fave1n2i,j=1nf(12(xi1+xi),12(yj1+yj)).
59.

An isotherm map is a chart connecting points having the same temperature at a given time for a given period of time. Use the preceding exercise and apply the midpoint rule with m=n=2m=n=2 to find the average temperature over the region given in the following figure.

A contour map showing surface temperature in degrees Fahrenheit. Given the map, the midpoint rule would give rectangles with values 71, 72, 40, and 43.
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