9.10 Double Integrals

INTRODUCTION

In Section 9.8 we gave the five steps leading to the definition of the definite integral The analogous steps that lead to the concept of the two-dimensional definite integral, known simply as the double integral of a function f of two variables, will be given next.

  1. Let z = f(x, y) be defined in a closed and bounded region R of 2-space.
  2. By means of a grid of vertical and horizontal lines parallel to the coordinate axes, form a partition P of R into n rectangular subregions Rk of areas ΔAk that lie entirely in R.
  3. Let P be the norm of the partition or the length of the longest diagonal of the Rk.
  4. Choose a sample point in each subregion Rk. See FIGURE 9.10.1.
  5. Form the sum f ΔAk.
A graph. A closed curve is graphed in an x y coordinate plane. Several equally spaced out vertical and horizontal lines partition the coordinate plane into several identical rectangles. All the rectangles that lie entirely inside the closed curve are shaded. One of the shaded rectangles contains a sample point (x subscript k asterisk, y subscript k asterisk), and is highlighted.

FIGURE 9.10.1 Sample point in kth rectangle

Thus we have the following definition:

DEFINITION 9.10.1 The Double Integral

Let f be a function of two variables defined on a closed region R of 2-space. Then the double integral of f over R is given by

.(1)

Integrability

If the limit in (1) exists, we say that f is integrable over R and that R is the region of integration. When f is continuous on R, then f is necessarily integrable over R.

Area

When f(x, y) = 1 on R, then simply gives the area A of the region; that is,

(2)

Volume

If f(x, y) ≥ 0 on R, then, as shown in FIGURE 9.10.2, the product f ΔAk can be interpreted as the volume of a rectangular prism of height f and base of area ΔAk. The summation of volumes f ΔAk is an approximation to the volume V of the solid above the region R and below the surface z = f(x, y). The limit of this sum as P → 0, if it exists, gives the exact volume of this solid; that is, if f is nonnegative on R, then

(3)

A graph. A three dimensional surface, a closed curve, and a rectangular prism, are graphed on a three dimensional x y z coordinate system. The surface is labeled z = f(x y). A closed curve is graphed on the surface. The closed curve is projected vertically down on the x y plane. Several equally spaced out lines, parallel to the x and y axes, partitions the region inside the projected curve. All the rectangles in the grid that lie entirely inside the projected curve, are shaded and labeled R. A rectangular prism, with one of the rectangles as its base, and its top face on the surface, connects the surface and the region R. The top face of the rectangular prism on the surface contains a marked point. The base of the rectangular prism on the x y plane, contains a second marked point (x subscript k asterisk, y subscript k asterisk, 0). A dashed vertical line connecting the 2 marked points, passes through the central axis of the rectangular prism. The height of the rectangular prism is labeled f(x subscript k asterisk, y subscript k asterisk).

FIGURE 9.10.2 Volume under a surface

Properties

The double integral possesses the following properties:

THEOREM 9.10.1 Properties of Double Integrals

Let f and g be functions of two variables that are integrable over a region R. Then

  1. kf(x, y) dA = kf(x, y) dA, where k is any constant
  2. [f(x, y) ± g(x, y)] dA = f(x, y) dA ± g(x, y) dA
  3. f(x, y) dA = f(x, y) dA + f(x, y) dA, where R1 and R2 are subregions of R

that do not overlap and R = R1 ∪ R2.

Part (iii) of Theorem 9.10.1 is the two-dimensional equivalent of FIGURE 9.10.3 illustrates the division of a region into subregions R1 and R2 for which R = R1 ∪ R2. R1 and R2 can have no points in common except possibly on their common border. Furthermore, Theorem 9.10.1(iii) extends to any finite number of nonoverlapping subregions whose union is R.

A curve divides a closed curve into 2 regions labeled R subscript 1 and R subscript 2. The following information is given below: R = R subscript 1 union R subscript 2.

FIGURE 9.10.3 R is the union of two regions

Regions of Type I and II

The region shown in FIGURE 9.10.4(a),

R: a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x),

where the boundary functions g1 and g2 are continuous, is called a region of Type I. In Figure 9.10.4(b), the region

R: c ≤ y ≤ d, h1(y) ≤ x ≤ h2(y),

where the functions h1 and h2 are continuous, is called a region of Type II.

2 graphs. Graph a. Caption. Type 1 region. Graph. 2 parallel lines, 2 curves and a region are graphed on a three dimensional x y z coordinate system. The 2 parallel lines begin on the x axis at the values a and b, and go parallel to the y axis. The first curve, labeled y = g subscript 1(x) begins at a marked point on the first line at x = a, and ends at a marked point on the second line at x = b. The second curve, labeled y = g subscript 2(x) begins at a marked point on the first line at x = a, to the right of the first curve with respect to the y axis, and ends at a marked point on the second line at x = b, to the right of the first curve with respect to the y axis. The region on the graph, between the 2 parallel lines and the 2 curves, is shaded and labeled R. Graph b. Caption. Type 2 region. Graph. 2 parallel lines, 2 curves and a region are graphed on a three dimensional x y z coordinate system. The 2 parallel lines begin on the y axis at the values c and d, and go parallel to the x axis. The first curve, labeled x = h subscript 1(y) begins at a marked point on the first line at y = c, and end at a marked point on the second line at y = d. The second curve, labeled x = h subscript 2(y) begins at a marked point on the first line at y = c, to the right of the first curve with respect to the x axis, and ends at a marked point on the second line at y = d, to the right of the first curve with respect to the y axis. The region on the graph, between the 2 parallel lines and the 2 curves, is shaded and labeled R.

FIGURE 9.10.4 Regions of integration

Iterated Integrals

Since the partial integral (x, y) dy is a function of x alone, we may in turn integrate the resulting function now with respect to x. If f is continuous on a region of Type I, we define an iterated integral of f over the region by

.(4)

The basic idea in (4) is to carry out successive integrations. The partial integral with respect to y gives a function of x, which is then integrated in the usual manner from x = a to x = b. The end result of both integrations will be a real number. In a similar manner, we define an iterated integral of a continuous function f over a region of Type II by

.(5)

Evaluation of Double Integrals

Iterated integrals provide the means for evaluating a double integral R f(x, y) dA over a region of Type I or Type II or a region that can be expressed as a union of a finite number of these regions. The following result is due to the Italian mathematician Guido Fubini (1879–1943).

THEOREM 9.10.2 Fubini’s Theorem

Let f be continuous on a region R.

  1. If R is of Type I, then

    .(6)

  2. If R is of Type II, then

    .(7)

Theorem 9.10.2 is the double integral analogue of the Fundamental Theorem of Calculus. While Theorem 9.10.2 is difficult to prove, we can get some intuitive feeling for its significance by considering volumes. Let R be a Type I region and z = f(x, y) be continuous and nonnegative on R. The area A of the vertical plane, as shown in FIGURE 9.10.5, is the area under the trace of the surface z = f(x, y) in the plane x = constant and hence is given by the partial integral

A graph. A surface, a plane and 2 curves are graphed on a three dimensional x y z coordinate system. The surface is labeled z = f(x y). The 2 curves, labeled y = g subscript 1(x) and y = g subscript 2(x), graphed under the surface, on the x y plane, intersect 3 parallel lines to the y axis at x = a, x = constant, and x = b. The curves intersect the line at x = constant at the points (x, g subscript 1(x), 0) and (x, g subscript 2(x), 0). The region between the 2 curves and the parallel lines x = a and x = b, is shaded and labeled R. A vertical rectangular plane, placed on the intersection of the region R and the line at x = constant, connects the region R and the graphed surface above. The area of the vertical plane is labeled A(x). The following information is given about the intersection of the vertical plane and the graphed surface above: trace of surface in plane x = constant.

FIGURE 9.10.5 Geometric interpretation of (6)

By summing all these areas from x = a to x = b, we obtain the volume V of the solid above R and below the surface:

But, as we have already seen in (3), this volume is also given by the double integral

.

EXAMPLE 1 Evaluation of a Double Integral

Evaluate the double integral R ex + 3y dA over the region bounded by the graphs of y = 1, y = 2, y = x, and y = –x + 5. See FIGURE 9.10.6.

A graph. 4 lines and a shaded region are graphed on an x y coordinate plane. The first line is upward sloping and labeled y = x. The second line is downward sloping and labeled y = negative x + 5. The third and fourth lines are dashed, horizontal, and intersect the first 2 lines at y = 1 and y = 2. The region bound by the 4 lines is shaded.

FIGURE 9.10.6 Region of integration in Example 1

SOLUTION

As seen in the figure, the region is of Type II; hence, by (7) we integrate first with respect to x from the left boundary x = y to the right boundary x = 5 – y:

≡

As an aid in reducing a double integral to an iterated integral with correct limits of integration, it is useful to visualize, as suggested in the foregoing discussion, the double integral as a double summation process. Over a Type I region the iterated integral (x, y) dy dx is first a summation in the y-direction. Pictorially, this is indicated by the vertical arrow in FIGURE 9.10.7(a); the typical rectangle in the arrow has area dy dx. The dy placed before the dx signifies that the “volumes” f(x, y) dy dx of prisms built up on the rectangles are summed vertically with respect to y from the lower boundary curve g1 to the upper boundary curve g2. The dx following the dy signifies that the result of each vertical summation is then summed horizontally with respect to x from left (x = a) to right (x = b). Similar remarks hold for double integrals over regions of Type II. See Figure 9.10.7(b). Recall from (2) that when f(x, y) = 1, the double integral A = R dA gives the area of the region. Thus, Figure 9.10.7(a) shows that dx adds the rectangular areas vertically and then horizontally, whereas Figure 9.10.7(b) shows that dy adds the rectangular areas horizontally and then vertically.

2 graphs. Graph a. Caption. Region of Type 1. Graph. 2 curves and 2 vertical lines are graphed on an x y coordinate plane. The vertical lines are graphed at x = a and x = b. The curves, labeled y = g subscript 1(x) and y = g subscript 2(x), each intersect the 2 vertical lines. The region bound by the 2 curves and the 2 vertical lines, is shaded. A vertical arrow begins on the curve y = g subscript 1(x), points up and ends on the second curve y = g subscript 2(x).The arrow is divided into several identical rectangles. The horizontal and vertical sides of the rectangles are labeled d x and d y respectively. Graph b. Caption. Region of Type 2. Graph. 2 curves and 2 horizontal lines are graphed on an x y coordinate plane. The horizontal lines are graphed at y = c and y = d. The curves, labeled x = h subscript 1(y) and x = h subscript 2(y), each intersect the 2 horizontal lines. The region bound by the 2 curves and 2 horizontal lines, is shaded. A horizontal arrow begins on the curve x = h subscript 1(y), points to the right, and ends on the second curve x = h subscript 2(y).The arrow is divided into several identical rectangles. The horizontal and vertical sides of the rectangles are labeled d x and d y respectively.

FIGURE 9.10.7 Summation in y-direction in (a); summation in x-direction in (b)

Reversing the Order of Integration

A problem may become easier when the order of integration is changed or reversed. Also, some iterated integrals that may be impossible to evaluate using one order of integration can perhaps be evaluated using the reverse order of integration.

EXAMPLE 2 Reversing the Order of Integration

Evaluate R x dA over the region R in the first quadrant bounded by the graphs of y = x2, x = 0, y = 4.

SOLUTION

When the region is viewed as Type I, we have from FIGURE 9.10.8 (a), 0 ≤ x ≤ 2, x2 ≤ y ≤ 4, and so

The difficulty here is that the partial integral dy cannot be evaluated, since has no elementary antiderivative with respect to y. However, as we see in Figure 9.10.8(b), we can interpret the same region as a Type II region defined by 0 ≤ y ≤ 4, 0 ≤ x ≤ . Hence, from (7)

≡

2 graphs. Graph a. Caption. Type 1 region. Graph. A curve, a horizontal line and a shaded region are graphed on an x y coordinate plane. The curve, labeled y = x^2, enters the bottom left of the viewing window in the second quadrant, goes down to the right, reaches a low point at the origin, goes up to the right symmetrically, and exits the top right of the viewing window. The horizontal line, labeled y = 4, enters the top left of the viewing window in the second quadrant, goes to the right, intersects the curve at the marked point (2, 4), and exits the top right of the viewing window. The region bound by the vertical y axis, the curve and horizontal the line, is shaded. A vertical arrow begins on the curve, points up and ends on the horizontal line. The arrow is divided into several identical rectangles. Graph b. Caption. Type 2 region. Graph. A curve, a horizontal line and a shaded region are graphed on an x y coordinate plane. The curve, labeled y = x^2, enters the bottom left of the viewing window in the second quadrant, goes down to the right, reaches a low point at the origin, goes up to the right symmetrically, and exits the top right of the viewing window. The horizontal line, labeled y = 4, enters the top left of the viewing window in the second quadrant, goes to the right, intersects the curve at the marked point (2, 4), and exits the top right of the viewing window. The region bound by the vertical y axis, the curve and horizontal the line, is shaded. A horizontal arrow begins on the y axis, points to the right, and ends on the curve. The arrow is divided into several identical rectangles.

FIGURE 9.10.8 Reversing order of integration in Example 2

Laminas with Variable Density—Center of Mass

If ρ is a constant density (mass per unit area), then the mass of the lamina coinciding with a region bounded by the graphs of y = f(x), the x-axis, and the lines x = a and x = b is

(8)

If a lamina corresponding to a region R has a variable density ρ(x, y), where ρ is nonnegative and continuous on R, then analogous to (8) we define its mass m by the double integral

.(9)

The coordinates of the center of mass of the lamina are then

,,(10)

where (11)

are the moments of the lamina about the y- and x-axes, respectively. The center of mass is the point where we consider all the mass of the lamina to be concentrated. If ρ(x, y) is a constant, the center of mass is called the centroid of the lamina.

EXAMPLE 3 Center of Mass

A lamina has the shape of the region in the first quadrant that is bounded by the graphs of y = sin x, y = cos x, between x = 0 and x = π/4. Find its center of mass if the density is ρ(x, y) = y.

SOLUTION

From FIGURE 9.10.9 we see that

A graph. 2 curves and a shaded region are graphed on an x y coordinate plane. The first curve, labeled y = sin x, begins at the origin, goes up and to the right with decreasing steepness, passes through the marked point (pi over 4, sqrt(2) over 2), and exits the top right of the viewing window. The second curve, labeled y = cos x, begins on the y axis above the first curve, goes down and to the right with increasing steepness, intersects the first curve at the marked point (pi over 4, sqrt(2) over 2), and exits the bottom right of the viewing window. The region bound by the y axis and the 2 curves, is shaded. A vertical arrow begins on the curve labeled = sin x, points up and ends on the curve labeled y = cos x. The arrow is divided into several identical rectangles.

FIGURE 9.10.9 Region in Example 3

Now,

Similarly,

Hence, from (10),

Thus the center of mass has the approximate coordinates (0.29, 0.68). ≡

Moments of Inertia

The integrals Mx and My in (11) are also called the first moments of a lamina about the x-axis and y-axis, respectively. The so-called second moments of a lamina or moments of inertia about the x- and y-axes are, in turn, defined by the double integrals

and .(12)

A moment of inertia is the rotational equivalent of mass. For translational motion, kinetic energy is given by K = mv2, where m is mass and v is linear speed. The kinetic energy of a particle of mass m rotating at a distance r from an axis is K = mv2 = m(rω)2 = (mr2)ω2 = Iω2, where I = mr2 is its moment of inertia about the axis of rotation and ω is angular speed.

EXAMPLE 4 Moment of Inertia

Find the moment of inertia about the y-axis of the thin homogeneous disk of mass m shown in FIGURE 9.10.10.

A graph. A circle labeled x^2 + y^2 = r^2, and a shaded region are graphed on an x y coordinate plane. The center of the circle is at the origin. The circle passes through the following points: (negative r, 0), (0, r), (r, 0), and (0, negative r). The region inside the circle is shaded. A vertical arrow begins on the circle in the fourth quadrant at y = negative sqrt(r^2 minus x^2), points up and ends on the circle in the first quadrant at y = sqrt(r^2 minus x^2). The arrow is divided into several identical rectangles.

FIGURE 9.10.10 Disk in Example 4

SOLUTION

Since the disk is homogeneous, its density is the constant ρ(x, y) = m/πr2. Hence, from (12),

≡

Radius of Gyration

The radius of gyration of a lamina of mass m and the moment of inertia I about an axis is defined by

(13)

Since (13) implies that the radius of gyration is interpreted as the radial distance the lamina, considered as a point mass, can rotate about the axis without changing the rotational inertia of the body. In Example 4 the radius of gyration is Rg = = r/2.

9.10 Exercises Answers to selected odd-numbered problems begin on page ANS-25.

In Problems 1–8, evaluate the given partial integral.

  2. tan xy dy
  3. x3exy dy

In Problems 9–12, sketch the region of integration for the given iterated integral.

  1. f(x, y) dy dx

In Problems 13–22, evaluate the double integral over the region R that is bounded by the graphs of the given equations. Choose the most convenient order of integration.

  1. x3y2 dA; y = x, y = 0, x = 1
  2. (x + 1) dA; y = x, x + y = 4, x = 0
  3. (2x + 4y + 1) dA; y = x2, y = x3
  4. xey dA; R the same as in Problem 13
  5. 2xy dA; y = x3, y = 8, x = 0
  6. dA; y = x2 + 1, y = 3 – x2
  7. dA; y = 0, y = 1, x = 0, x = 1
  8. dA; x = y2, x = 0, y = 1, y = 2
  9. dA; x = y, x = –y, x =
  10. dA; y = tan−1x, y = 0, x = 1
  11. Consider the solid bounded by the graphs of x2 + y2 = 4, z = 4 – y, and z = 0 shown in FIGURE 9.10.11. Choose and evaluate the correct integral representing the volume V of the solid.
    1. (4 – y) dy dx
    2. (4 – y) dy dx
    3. (4 – y) dx dy
A graph. A solid, resembling a cylinder with an inclined top base, is graphed on a three dimensional x y z coordinate system. The circular base of the solid, labeled x^2 + y^2 = 4, is at z = 0 on the horizontal x y plane. The center of the horizontal circular base is at the origin. The solid is bonded at the top by a plane slanting to the right and labeled z = 4 minus y. The solid is shaded.

FIGURE 9.10.11 Solid for Problem 23

  1. Consider the solid bounded by the graphs of x2 + y2 = 4 and y2 + z2 = 4. An eighth of the solid is shown in FIGURE 9.10.12. Choose and evaluate the correct integral representing the volume V of the solid.
    1. (4 – y2)1/2 dy dx
    2. (4 – y2)1/2 dx dy
    3. (4 – x2)1/2 dy dx
    A graph. 2 solids in the form of quarter cylinders are graphed on a three dimensional x y z coordinate system. The first solid, labeled x^2 + y^2 = 4, is placed vertically with the center of its quarter circular base on the z axis. The bottom base of the solid is on the x y plane. The 2 lateral flat faces of the solid are on the x z and y z planes. The second solid, labeled y^2 + z^2 = 4, is placed horizontally with the center of its quarter circular base on the x axis. The base of the solid at the back is on the y z plane. The 2 lateral flat faces of the solid are on the x y and x z planes. The intersection of the 2 solids is shaded.

    FIGURE 9.10.12 Solid for Problem 24

In Problems 25–34, find the volume of the solid bounded by the graphs of the given equations.

  1. 2x + y + z = 6, x = 0, y = 0, z = 0, first octant
  2. z = 4 – y2, x = 3, x = 0, y = 0, z = 0, first octant
  3. x2 + y2 = 4, x – y + 2z = 4, x = 0, y = 0, z = 0, first octant
  4. y = x2, y + z = 3, z = 0
  5. z = 1 + x2 + y2, 3x + y = 3, x = 0, y = 0, z = 0, first octant
  6. z = x + y, x2 + y2 = 9, x = 0, y = 0, z = 0, first octant
  7. yz = 6, x = 0, x = 5, y = 1, y = 6, z = 0
  8. z = 4 – x2 – y2, z = 0
  9. z = 4 – y2, x2 + y2 = 2x, z = 0
  10. z = 1 – x2, z = 1 – y2, x = 0, y = 0, z = 0, first octant

In Problems 35–40, evaluate the given iterated integral by reversing the order of integration.

In Problems 41–50, find the center of mass of the lamina that has the given shape and density.

  1. x = 0, x = 4, y = 0, y = 3; ρ(x, y) = xy
  2. x = 0, y = 0, 2x + y = 4; ρ(x, y) = x2
  3. y = x, x + y = 6, y = 0; ρ(x, y) = 2y
  4. y = |x|, y = 3; ρ(x, y) = x2 + y2
  5. y = x2, x = 1, y = 0; ρ(x, y) = x + y
  6. x = y2, x = 4; ρ(x, y) = y + 5
  7. y = 1 – x2, y = 0; density at a point P directly proportional to the distance from the x-axis
  8. y = sin x, 0 ≤ x ≤ π, y = 0; density at a point P directly proportional to the distance from the y-axis
  9. y = ex, x = 0, x = 1, y = 0; ρ(x, y) = y3
  10. y = y = 0; ρ(x, y) = x2

In Problems 51–54, find the moment of inertia about the x-axis of the lamina that has the given shape and density.

  1. x = y – y2, x = 0; ρ(x, y) = 2x
  2. y = x2, y = ; ρ(x, y) = x2
  3. y = cos x, –π/2 ≤ x ≤ π/2, y = 0; ρ(x, y) = k (constant)
  4. y = x = 0, y = 0, first quadrant; ρ(x, y) = y

In Problems 55–58, find the moment of inertia about the y-axis of the lamina that has the given shape and density.

  1. y = x2, x = 0, y = 4, first quadrant; ρ(x, y) = y
  2. y = x2, y = ρ(x, y) = x2
  3. y = x, y = 0, y = 1, x = 3; ρ(x, y) = 4x + 3y
  4. Same R and density as in Problem 47

In Problems 59 and 60, find the radius of gyration about the indicated axis of the lamina that has the given shape and density.

  1. x = x = 0; ρ(x, y) = x; y-axis
  2. x + y = a, a > 0, x = 0, y = 0; ρ(x, y) = k (constant); x-axis
  3. A lamina has the shape of the region bounded by the graph of the ellipse x2/a2 + y2/b2 = 1. If its density is ρ(x, y) = 1, find:
    1. the moment of inertia about the x-axis of the lamina,
    2. the moment of inertia about the y-axis of the lamina,
    3. the radius of gyration about the x-axis [Hint: The area of the ellipse is πab], and
    4. the radius of gyration about the y-axis.
  4. A cross section of an experimental airfoil is the lamina shown in FIGURE 9.10.13. The arc ABC is elliptical, whereas the two arcs AD and CD are parabolic. Find the moment of inertia about the x-axis of the lamina under the assumption that the density is ρ(x, y) = 1.
    A graph. A cross section of an experimental airfoil, represented by a closed curve, is graphed on an x y coordinate plane. The closed curve passes through the following labeled points: A(0, negative b over 2), B(negative a over 3, 0), C(0, b over 2), and D(2a over 3, 0). The arc A B C is elliptical, while the arcs A D and C D are parabolic. The region inside the closed curve is shaded.

    FIGURE 9.10.13 Airfoil in Problem 62

The polar moment of inertia of a lamina with respect to the origin is defined to be

I0 = (x2 + y2)ρ(x, y) dA = Ix + Iy.

In Problems 63–66, find the polar moment of inertia of the lamina that has the given shape and density.

  1. x + y = a, a > 0, x = 0, y = 0; ρ(x, y) = k (constant)
  2. y = x2, y = ; ρ(x, y) = x2 [Hint: See Problems 52 and 56.]
  3. x = y2 + 2, x = 6 – y2; density at a point P inversely proportional to the square of the distance from the origin
  4. y = x, y = 0, y = 3, x = 4; ρ(x, y) = k (constant)
  5. Find the radius of gyration in Problem 63.
  6. Show that the polar moment of inertia about the center of a thin homogeneous rectangular plate of mass m, width w, and length l is I0 = m(l2 + w2)/12.