9.17 Change of Variables in Multiple Integrals

INTRODUCTION

In many instances it is either a matter of convenience or of necessity to make a substitution, or change of variable, in a definite integral in order to evaluate it. If f is continuous on [a, b], x = g(u) has a continuous derivative, and dx = g′(u) du, then

f(x) dx = f(g(u)) g′(u) du,(1)

where the u-limits of integration c and d are defined by a = g(c) and b = g(d). There are three things that bear emphasizing in (1). To change the variable in a definite integral we replace x where it appears in the integrand by g(u), we change the interval of integration [a, b] on the x-axis to the corresponding interval [c, d] on the u-axis, and we replace dx by a function multiple (namely, the derivative of g) of du. If we write J(u) = dx/du, then (1) has the form

f(x) dx = f(g(u)) J(u) du.(2)

If the function g is one-to-one, then it has an inverse and so c = g−1(a) and d = g−1(b).

For example, using x = 2 sin θ, –π/2 ≤ θ ≤ π/2, we get

Double Integrals

Although changing variables in a multiple integral is not as straightforward as the procedure in (1), the basic idea illustrated in (2) carries over. To change variables in a double integral we need two equations such as

(3)

To be analogous with (2), we expect that a change of variables in a double integral would take the form

(4)

where S is the region in the uv-plane corresponding to the region R in the xy-plane and J(u, v) is some function that depends on the partial derivatives of the equations in (3). The symbol dA′ on the right side of (4) represents either du dv or dv du.

In Section 9.11 we briefly discussed how to change a double integral F(x, y) dA from rectangular coordinates to polar coordinates. Recall that in Example 2 of that section the substitutions

x = r cos θ, y = r sin θ(5)

led to (6)

As we see in FIGURE 9.17.1, the introduction of polar coordinates changes the original region of integration R in the xy-plane to the more convenient rectangular region of integration S in the rθ-plane. We note, too, that by comparing (4) with (6), we can identify J(r, θ) = r and dA′ = dr dθ.

2 graphs. Graph a. Caption. Region R in x y plane. Graph. A circle, labeled x^2 + y^2 = 8, is graphed on an x y coordinate plane. The center of the circle is at the origin. A line labeled y = x, begins at the origin and ends on the circle at the point (2, 2). The region bound by the y axis, the line y = x, and the circular arc between the point (2, 2) and the y axis, is shaded and labeled R. Graph b. Caption. Region S in r theta plane. Graph. A rectangular region is graphed on a r theta coordinate plane. The vertices of the rectangle are at the following points: (0, pi over 4), (sqrt(8), pi over 4), (sqrt(8), pi over 2), (0, pi over 2). The region inside the rectangle is shaded and labeled S.

FIGURE 9.17.1 Region S is used to evaluate (6)

The change-of-variable equations in (3) define a transformation or mapping T from the uv-plane to the xy-plane. A point (x0, y0) in the xy-plane determined from x0 = f(u0, v0), y0 = g(u0, v0) is said to be an image of (u0, v0).

EXAMPLE 1 Image of a Region

Find the image of the region S shown in FIGURE 9.17.2(a) under the transformation x = u2 + v2, y = u2 – v2.

2 graphs. Graph a. 2 curves and a horizontal line are graphed on a u v coordinate plane. The line, labeled S subscript 1, begins at the point (1, 0), goes horizontally to the right and ends at the point (2, 0). The first curve, labeled u^2 + v^2 = 4, begins on the vertical v axis, goes down to the right with increasing steepness, passes through the point (sqrt(5 over 2), sqrt(3 over 2)), intersects the line at the point (2, 0), and exits the bottom right of the viewing window. The second curve, labeled u^2 minus v^2 = 1, enters the bottom left of the viewing window in the fourth quadrant, goes up to the right with decreasing steepness, intersects the line at the point (1, 0), intersects the first curve at the point (sqrt(5 over 2), sqrt(3 over 2)), and exits the top right of the viewing window. The region bound by the line and 2 curves is shaded and labeled S. The portions of the curves that limit the region S are labeled S subscript 2 and S subscript 3 respectively. 3 arrows placed on the 3 sides of the region S, are oriented counter clockwise. Graph b. A rectangular region is graphed on an x y coordinate plane. The region is bound by 3 lines. The first line, labeled y = x, begins at the point (1, 1), and ends at the point (4, 4). The second line, labeled y = 1, begins at the point (1, 1), and ends at the point (4, 1). The third line, labeled x = 4, begins at the point (4, 1), and ends at the point (4, 4). The region bound by the 3 lines is shaded and labeled R. 3 arrows placed on the 3 sides of the region R, are oriented clockwise.

FIGURE 9.17.2 Region R is the image of region S in Example 1

SOLUTION

We begin by finding the images of the sides of S that we have indicated by S1, S2, and S3.

S1: On this side v = 0 so that x = u2, y = u2. Eliminating u then gives y = x. Now imagine moving along the boundary from (1, 0) to (2, 0) (that is, 1 ≤ u ≤ 2). The equations x = u2, y = u2 then indicate that x ranges from x = 1 to x = 4 and y ranges simultaneously from y = 1 to y = 4. In other words, in the xy-plane the image of S1 is the line segment y = x from (1, 1) to (4, 4).

S2: On this boundary u2 + v2 = 4 and so x = 4. Now as we move from the point (2, 0) to , , the remaining equation y = u2 – v2 indicates that y ranges from y = 22 – 02 = 4 to y = 1. In this case the image of S2 is the vertical line segment x = 4 starting at (4, 4) and going down to (4, 1).

S3: Since u2 – v2 = 1, we get y = 1. But as we move on this boundary from , , to (1, 0), the equation x = u2 + v2 indicates that x ranges from x = 4 to x = 1. The image of S3 is the horizontal line segment y = 1 starting at (4, 1) and ending at (1, 1).

The image of S is the region R given in Figure 9.17.2(b). ≡

Observe in Example 1 that as we traverse the boundary of S in the counterclockwise direction, the boundary of R is traversed in a clockwise manner. We say that the transformation of the boundary of S has induced an orientation on the boundary of R.

While a proof of the formula for changing variables in a multiple integral is beyond the level of this text, we will give some of the underlying assumptions that are made about the equations (3) and the regions R and S. We assume that

  • The functions f and g have continuous first partial derivatives on S.
  • The transformation is one-to-one.
  • Each of the regions R and S consists of a piecewise-smooth simple closed curve and its interior.
  • The determinant

    (7)
    is not zero on S.

A transformation T is said to be one-to-one if each point (x0, y0) in R is the image under T of a unique point (u0, v0) in S. Put another way, no two points in S have the same image in R. With the restrictions that r ≥ 0 and 0 ≤ θ ≤ 2π, the equations in (5) define a one-to-one transformation from the rθ-plane to the xy-plane. The determinant in (7) is called the Jacobian determinant, or simply Jacobian, of the transformation T and is the key to changing variables in a multiple integral. The Jacobian of the transformation defined by the equations in (3) is denoted by the symbol

We will encounter another Jacobian in Section 11.3.

Similar to the notion of a one-to-one function, a one-to-one transformation T has an inverse transformation T−1 such that (u0, v0) is the image under T−1 of (x0, y0). See FIGURE 9.17.3. If it is possible to solve (3) for u and v in terms of x and y, then the inverse transformation is defined by a pair of equations

(8)

2 graphs placed side by side. (graph on the left). A region labeled S is graphed on a u v coordinate plane. The region S contains a point labeled (u subscript 0, v subscript 0). (graph on the right). A region labeled R is graphed on an x y coordinate plane. The region R contains a point labeled (x subscript 0, y subscript 0). A right arrow labeled T connects the points (u subscript 0, v subscript 0) and (x subscript 0, y subscript 0). A left arrow labeled T^negative 1 connects the points (x subscript 0, y subscript 0) and (u subscript 0, v subscript 0).

FIGURE 9.17.3 Transformation T and its inverse

The Jacobian of the inverse transformation T−1 is

(9)

and is related to the Jacobian of the transformation T by

(10)

EXAMPLE 2 Jacobian

The Jacobian of the transformation x = r cos θ, y = r sin θ is

≡

We now turn our attention to the main point of this discussion: how to change variables in a multiple integral. The idea expressed in (4) is valid; the function J(u, v) turns out to be |∂(x, y)/∂(u, v)|. Under the assumptions made above, we have the following result:

THEOREM 9.17.1 Change of Variables in a Double Integral

If F is continuous on R, then

.(11)

Formula (3) of Section 9.11 for changing a double integral to polar coordinates is just a special case of (11) with

since r ≥ 0. In (6) then we have J(r, θ) = |∂(x, y)/∂(r, θ)| = r.

A change of variables in a multiple integral can be used for either a simplification of the integrand or a simplification of the region of integration. The actual change of variables used is often inspired by the structure of the integrand F(x, y) or by equations that define the region R. As a consequence, the transformation is then defined by equations of the form given in (8); that is, we are dealing with the inverse transformation. The next two examples will illustrate these ideas.

EXAMPLE 3 Changing Variables in a Double Integral

Evaluate sin(x + 2y) cos(x – 2y) dA over the region R shown in FIGURE 9.17.4(a).

2 graphs. Graph a. A rectangular region is graphed on an x y coordinate plane. The region is bound by 3 lines. The first line, labeled S subscript 1, begins at the point (0, 0), and ends at the point (2 pi, 0). The second line, labeled S subscript 2, begins at the point (0, 0), and ends at the point (0, pi). The third line, labeled S subscript 3, begins at the point (0, pi), and ends at the point (2 pi, 0). The region bound by the 3 lines is shaded and labeled R. 3 arrows placed on the 3 sides of the region R, are oriented clockwise. Graph b. A rectangular region is graphed on a u v coordinate plane. The region is bound by 3 lines. The first line, labeled v = negative u, begins at the point (0, 0), and ends at the point (2 pi, negative 2 pi). The second line, labeled u = 2 pi, begins at the point (2 pi, negative 2 pi), and ends at the point (2 pi, 2 pi). The third line, labeled v = u, begins at the point (0, 0), and ends at the point (2 pi, 2 pi). The region bound by the 3 lines is shaded and labeled S. 3 arrows placed on the 3 sides of the region R, are oriented counter clockwise.

FIGURE 9.17.4 Region S is the image of region R in Example 3

SOLUTION

The difficulty in evaluating this double integral is clearly the integrand. The presence of the terms x + 2y and x − 2y prompts us to define the change of variables u = x + 2y, v = x − 2y. These equations will map R onto a region S in the uv-plane. As in Example 1, we transform the sides of the region.

S1: y = 0 implies u = x and v = x or v = u. As we move from (2π, 0) to (0, 0), we see that the corresponding image points in the uv-plane lie on the line segment v = u from (2π, 2π) to (0, 0).

S2: x = 0 implies u = 2y and v = –2y, or v = –u. As we move from (0, 0) to (0, π), the corresponding image points in the uv-plane lie on the line segment v = –u from (0, 0) to (2π, –2π).

S3: x + 2y = 2π implies u = 2π. As we move from (0, π) to (2π, 0), the equation v = x − 2y shows that v ranges from v = −2π to v = 2π. Thus, the image of S3 is the vertical line segment u = 2π starting at (−2π, −2π) and going up to (2π, 2π). See Figure 9.17.4(b).

Now, solving for x and y in terms of u and v gives

Therefore,

Hence, from (11) we find that

≡

EXAMPLE 4 Changing Variables in a Double Integral

Evaluate xy dA over the region R shown in FIGURE 9.17.5(a).

2 graphs. Graph a. 4 curves and a region are graphed on an x y coordinate plane. The first curve is labeled x y = 1 and is downward sloping, enters the top left of the viewing window, goes down to the right with decreasing steepness, and exits the bottom right of the viewing window. The second curve is labeled x y = 5 and is downward sloping, enters the top left of the viewing window above the first curve, goes down to the right with decreasing steepness, and exits the bottom right of the viewing window above the first curve. The third curve is labeled y = 4 x^2 and is upward sloping, begins at the origin, goes up to the right with increasing steepness, intersects the first curve, intersects the second curve, and exits the top of the viewing window. The fourth curve is labeled y = x^2 and is upward sloping, begins at the origin, goes up to the right with increasing steepness to the right of the third curve, intersects the first curve, intersects the second curve, and exits the top right of the viewing window. The region bound by the 4 curves is shaded and labeled R. Graph b. A rectangular region is graphed on a u v coordinate plane. The vertices of the rectangle are at the following points: (1, 1), (4, 1), (4, 5), and (1, 5). The region inside the rectangle is shaded and labeled S.

FIGURE 9.17.5 Region S is image of region R in Example 4

SOLUTION

In this case the integrand is fairly simple, but integration over the region R would be tedious since we would have to express xy dA as the sum of three integrals. (Verify this.)

The equations of the boundaries of R suggest the change of variables

(12)

Obtaining the image of R is a straightforward matter in this case, since the images of the curves that make up the four boundaries are simply u = 1, u = 4, v = 1, and v = 5. In other words, the image of the region R is the rectangular region S: 1 ≤ u ≤ 4, 1 ≤ v ≤ 5. See Figure 9.17.5(b).

Now instead of trying to solve the equations in (12) for x and y in terms of u and v, we can compute the Jacobian ∂(x, y)/∂(u, v) by computing ∂(u, v)/∂(x, y) and using (10). We have

and so from (10),

Hence, ≡

Triple Integrals

To change variables in a triple integral, let

be a one-to-one transformation T from a region E in uvw-space to a region D in xyz-space. If F is continuous on D, then

where

We leave it as an exercise for the reader to show that if T is the transformation from spherical to rectangular coordinates defined by

(13)

then

9.17 Exercises Answers to selected odd-numbered problems begin on page ANS-26.

  1. Consider a transformation T defined by x = 4u – v, y = 5u + 4v. Find the images of the points (0, 0), (0, 2), (4, 0), and (4, 2) in the uv-plane under T.
  2. Consider a transformation T defined by x = , y = v + u. Find the images of the points (1, 1), (1, 3), and (, 2) in the xy-plane under T−1.

In Problems 3–6, find the image of the set S under the given transformation.

  1. S: 0 ≤ u ≤ 2, 0 ≤ v ≤ u; x = 2u + v, y = u – 3v
  2. S: –1 ≤ u ≤ 4, 1 ≤ v ≤ 5; u = x – y, v = x + 2y
  3. S: 0 ≤ u ≤ 1, 0 ≤ v ≤ 2; x = u2 – v2, y = uv
  4. S: 1 ≤ u ≤ 2, 1 ≤ v ≤ 2; x = uv, y = v2

In Problems 7–10, find the Jacobian of the transformation T from the uv-plane to the xy-plane.

  1. x = ve−u, y = veu
  2. x = e3u sin v, y = e3u cos v
    1. Find the image of the region S: 0 ≤ u ≤ 1, 0 ≤ v ≤ 1 under the transformation x = u – uv, y = uv.
    2. Explain why the transformation is not one-to-one on the boundary of S.
  6. Determine where the Jacobian ∂(x, y)/∂(u, v) of the transformation in Problem 11 is zero.

In Problems 13–22, evaluate the given integral by means of the indicated change of variables.

  1. (x + y) dA, where R is the region bounded by the graphs of x – 2y = –6, x – 2y = 6, x + y = –1, x + y = 3; u = x – 2y, v = x + y
  2. dA, where R is the region bounded by the graphs of y = x, y = x – π, y = –3x + 3, y = –3x + 6; u = x – y, v = 3x + y
  3. dA, where R is the region bounded by the graphs y = x2, y = x2, x = y2, x = y2; u = , v =
  4. (x2 + y2)−3 dA, where R is the region bounded by the circles x2 + y2 = 2x, x2 + y2 = 4x, x2 + y2 = 2y, x2 + y2 = 6y; u = , v = [Hint: Form u2 + v2.]
  5. (x2 + y2) dA, where R is the region in the first quadrant bounded by the graphs of x2 − y2 = a, x2 − y2 = b, 2xy = c, 2xy = d, 0 < a < b, 0 < c < d; u = x2 – y2, v = 2xy
  6. (x2 + y2) sin xy dA, where R is the region bounded by the graphs of x2 – y2 = 1, x2 – y2 = 9, xy = 2, xy = –2; u = x2 – y2, v = xy
  7. dA, where R is the region in the first quadrant bounded by the graphs of x = 1, y = x2, y = 4 − x2; x = , y = v + u
  8. y dA, where R is the triangular region with vertices (0, 0), (2, 3) and (−4, 1); x = 2u − 4v, y = 3u + v
  9. y4 dA, where R is the region in the first quadrant bounded by the graphs of xy = 1, xy = 4, y = x, y = 4x; u = xy, v = y/x
  10. D (4z + 2x – 2y) dV, where D is the parallelepiped 1 ≤ y + z ≤ 3, −1 ≤ – y + z ≤ 1, 0 ≤ x − y ≤ 3; u = y + z, v = −y + z, w = x − y

In Problems 23–26, evaluate the given double integral by means of an appropriate change of variables.

  3. (6x + 3y) dA, where R is the trapezoidal region in the first quadrant with vertices (1, 0), (4, 0), (2, 4), and (, 1)
  4. (x + y)4 ex−y dA, where R is the square region with vertices (1, 0), (0, 1), (1, 2), and (2, 1)
  5. A problem in thermodynamics is to find the work done by an ideal Carnot engine. This work is defined to be the area of the region R in the first quadrant bounded by the isothermals xy = a, xy = b, 0 < a < b, and the adiabatics xy1.4 = c, xy1.4 = d, 0 < c < d. Use A = dA and an appropriate substitution to find the area shown in FIGURE 9.17.6.
    A graph. 4 downward sloping curves and a region are graphed on an x y coordinate plane. The first curve, labeled x y = a, enters the top left of the viewing window, goes down to the right with decreasing steepness, and exits the bottom of the viewing window. The second curve, labeled x y = b, enters the top left of the viewing window to the right of the first curve, goes down to the right with decreasing steepness, and exits the bottom right of the viewing window to the right of the first curve. The third curve, labeled x y^1.4 = c, enters the left side of the viewing window to the left of the first curve, goes down to the right with decreasing steepness, intersects the first curve, intersects the second curve, and exits the bottom right of the viewing window, above the second curve. The fourth curve, labeled x y^1.4 = d, enters the top left of the viewing window, to the left of the first curve and above the third curve, goes down to the right with decreasing steepness, intersects the first curve, intersects the second curve, and exits the right side of the viewing window, above the third curve. The region bound by the 4 curves is shaded and labeled R.

    FIGURE 9.17.6 Region R for Problem 27

  6. Use V = D dV and the substitutions u = x/a, v = y/b,w = z/c to show that the volume of the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1 is V = πabc.
  7. Evaluate the double integral where R is the elliptical region whose boundary is the graph of x2/25 + y2/9 = 1. Use the substitutions u = x/5, v = y/3, and polar coordinates.
  8. Verify that the Jacobian of the transformation given in (13) is ∂(x, y, z)/∂(ρ, ϕ, θ) = ρ2 sin ϕ.