SECTION 12.8
tive x-axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects
the equator. Then the latitude of P is 90 and the
longitude is 360 . Find the great-circle distance
from Los Angeles (lat. 34.06 N, long. 118.25 W) to Montréal (lat. 45.50 N, long. 73.60 W). Take the radius of the
Earth to be 3960 mi. (A great circle is the circle of intersection of a sphere and a plane through the center of the
sphere.)
CAS
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
713
40. Show that
y y y
2
sx 2 y 2 z 2 ex y
2
z 2 dx dy dz 2
(The improper triple integral is defined as the limit of a
triple integral over a solid sphere as the radius of the sphere
increases indefinitely.)
41. (a) Use cylindrical coordinates to show that the volume of
the solid bounded above by the sphere r 2 z 2 a 2 and
below by the cone z r cot 0 (or 0 ), where
0 0 2, is
39. The surfaces 1 5 sin m sin n have been used as
1
models for tumors. The “bumpy sphere” with m 6 and
n 5 is shown. Use a computer algebra system to find the
volume it encloses.
V
2a 3
1 cos 0 3
(b) Deduce that the volume of the spherical wedge given by
1 2 , 1 2 , 1 2 is
V 23 13
cos 1 cos 2 2 1 3
(c) Use the Mean Value Theorem to show that the volume
in part (b) can be written as
V 2 sin where lies between 1 and 2 , lies between 1 and
2 , 2 1 , 2 1 , and 2 1 .
12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
In one-dimensional calculus we often use a change of variable (a substitution) to simplify an integral. By reversing the roles of x and u, we can write the Substitution Rule
(5.5.6) as
y
1
f x dx y f tutu du
b
d
a
c
where x tu and a tc, b td. Another way of writing Formula 1 is as
follows:
y
2
b
a
f x dx y f xu
d
c
dx
du
du
A change of variables can also be useful in double integrals. We have already seen
one example of this: conversion to polar coordinates. The new variables r and are
related to the old variables x and y by the equations
x r cos y r sin and the change of variables formula (12.3.2) can be written as
yy f x, y dA yy f r cos , r sin r dr d
R
S
where S is the region in the r -plane that corresponds to the region R in the xy-plane.
714
■
CHAPTER 12
MULTIPLE INTEGRALS
More generally, we consider a change of variables that is given by a transformation T from the uv-plane to the xy-plane:
Tu, v x, y
where x and y are related to u and v by the equations
3
x tu, v
y hu, v
x xu, v
y yu, v
or, as we sometimes write,
We usually assume that T is a C 1 transformation, which means that t and h have continuous first-order partial derivatives.
A transformation T is really just a function whose domain and range are both subsets of ⺢ 2. If Tu1, v1 x 1, y1, then the point x 1, y1 is called the image of the point
u1, v1. If no two points have the same image, T is called one-to-one. Figure 1 shows
the effect of a transformation T on a region S in the uv-plane. T transforms S into a
region R in the xy-plane called the image of S, consisting of the images of all points
in S.
√
y
T
S
R
(u¡, √¡)
T –!
u
0
(x¡, y¡)
0
x
FIGURE 1
If T is a one-to-one transformation, then it has an inverse transformation T 1
from the xy-plane to the uv-plane and it may be possible to solve Equations 3 for u
and v in terms of x and y :
u Gx, y
v Hx, y
√
S£
(0, 1)
(1, 1)
S¢
S
0
S™
S¡ (1, 0)
V EXAMPLE 1
u
A transformation is defined by the equations
x u 2 v2
Find the image of the square S u, v 0 u 1, 0 v 1.
T
SOLUTION The transformation maps the boundary of S into the boundary of the
image. So we begin by finding the images of the sides of S. The first side, S1 , is
given by v 0 0 u 1. (See Figure 2.) From the given equations we have
x u 2, y 0, and so 0 x 1. Thus S1 is mapped into the line segment from
0, 0 to 1, 0 in the xy-plane. The second side, S 2, is u 1 0 v 1 and, putting u 1 in the given equations, we get
y
(0, 2)
¥
x= -1
4
¥
x=1- 4
R
(_1, 0)
FIGURE 2
x 1 v2
0
y 2uv
(1, 0)
x
y 2v
Eliminating v, we obtain
4
x1
y2
4
0x1
SECTION 12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
715
which is part of a parabola. Similarly, S 3 is given by v 1 0 u 1, whose
image is the parabolic arc
x
5
y2
1
4
1 x 0
Finally, S4 is given by u 0 0 v 1 whose image is x v 2, y 0, that is,
1 x 0. (Notice that as we move around the square in the counterclockwise
direction, we also move around the parabolic region in the counterclockwise direction.) The image of S is the region R (shown in Figure 2) bounded by the x-axis and
the parabolas given by Equations 4 and 5.
■
Now let’s see how a change of variables affects a double integral. We start with a
small rectangle S in the uv-plane whose lower left corner is the point u0 , v0 and
whose dimensions are u and v. (See Figure 3.)
y
√
u=u ¸
r (u ¸, √)
Î√
S
(u¸, √ ¸)
Îu
T
(x¸, y¸)
√=√ ¸
0
R
r (u, √ ¸)
u
0
x
FIGURE 3
The image of S is a region R in the xy-plane, one of whose boundary points is
x 0 , y0 Tu0 , v0 . The vector
ru, v tu, v i hu, v j
is the position vector of the image of the point u, v. The equation of the lower side
of S is v v0 , whose image curve is given by the vector function ru, v0. The tangent
vector at x 0 , y0 to this image curve is
ru tuu0 , v0 i huu0 , v0 j &x
&y
i
j
&u
&u
Similarly, the tangent vector at x 0 , y0 to the image curve of the left side of S (namely,
u u0 ) is
rv tvu0 , v0 i hvu0 , v0 j r (u¸, √¸+Î√)
&x
&y
i
j
&v
&v
We can approximate the image region R T S by a parallelogram determined by the
secant vectors
r (u ¸, √¸)
a ru0 u, v0 ru0 , v0 b ru0 , v0 v ru0 , v0 a
shown in Figure 4. But
r (u¸+Î u, √¸)
FIGURE 4
ru lim
u l 0
ru0 u, v0 ru0 , v0 u
716
■
CHAPTER 12
MULTIPLE INTEGRALS
and so
ru0 u, v0 ru0 , v0 u ru
Similarly
ru0 , v0 v ru0 , v0 v rv
Î √ r√
r (u ¸, √ ¸)
This means that we can approximate R by a parallelogram determined by the vectors u ru and v rv . (See Figure 5.) Therefore, we can approximate the area of R by
the area of this parallelogram, which, from Section 10.4, is
FIGURE 5
u r v r r
6
u
u
v
rv u v
Computing the cross product, we obtain
i
&x
ru rv &u
&x
&v
j
&y
&u
&y
&v
k
&x
0
&u
&x
0
&v
&y
&x
&u
&u
k
&y
&y
&v
&u
&x
&v
k
&y
&v
The determinant that arises in this calculation is called the Jacobian of the transformation and is given a special notation.
The Jacobian is named after the German
mathematician Carl Gustav Jacob Jacobi
(1804–1851). Although the French mathematician Cauchy first used these special determinants involving partial derivatives, Jacobi
developed them into a method for evaluating
multiple integrals.
■
7 DEFINITION The Jacobian of the transformation T given by x tu, v
and y hu, v is
&x
&x, y
&u
&u, v
&y
&u
&x
&v
&x &y
&x &y
&y
&u &v
&v &u
&v
With this notation we can use Equation 6 to give an approximation to the area A
of R:
A 8
&x, y
u v
&u, v
where the Jacobian is evaluated at u0 , v0 .
Next we divide a region S in the uv-plane into rectangles Sij and call their images
in the xy-plane Rij . (See Figure 6.)
√
y
Sij
R ij
S
Î√
R
Îu
T
(u i , √ j )
0
FIGURE 6
(x i , yj )
u
0
x
SECTION 12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
717
Applying the approximation (8) to each Rij , we approximate the double integral of
f over R as follows:
m
n
yy f x, y dA f x , y A
i
j
i1 j1
R
m
n
f tu , v , hu , v i
j
i
j
i1 j1
&x, y
u v
&u, v
where the Jacobian is evaluated at ui , vj . Notice that this double sum is a Riemann
sum for the integral
yy f tu, v, hu, v
S
&x, y
du dv
&u, v
The foregoing argument suggests that the following theorem is true. (A full proof
is given in books on advanced calculus.)
Suppose that T is a C 1
transformation whose Jacobian is nonzero and that maps a region S in the uvplane onto a region R in the xy-plane. Suppose that f is continuous on R and
that R and S are type I or type II plane regions. Suppose also that T is one-toone, except perhaps on the boundary of S. Then
9 CHANGE OF VARIABLES IN A DOUBLE INTEGRAL
yy f x, y dA yy f xu, v, yu, v
R
¨
r=a
v by expressing x and y in terms of u and v and writing
r=b
S
dA å
¨=å
0
a
r
b
y
r=b
&x, y
du dv
&u, v
Notice the similarity between Theorem 9 and the one-dimensional formula in Equation 2. Instead of the derivative dxdu, we have the absolute value of the Jacobian, that
is, &x, y&u, v .
As a first illustration of Theorem 9, we show that the formula for integration in
polar coordinates is just a special case. Here the transformation T from the r -plane
to the xy-plane is given by
T
¨=∫
&x, y
du dv
&u, v
Theorem 9 says that we change from an integral in x and y to an integral in u and
¨=∫
∫
S
x tr, r cos y hr, r sin R
r=a
∫
0
and the geometry of the transformation is shown in Figure 7. T maps an ordinary rectangle in the r -plane to a polar rectangle in the xy-plane. The Jacobian of T is
¨=å
å
x
FIGURE 7
The polar coordinate transformation
&x
&x, y
&r
&r, &y
&r
&x
&
cos &y
sin &
r sin r cos2 r sin2 r 0
r cos 718
■
CHAPTER 12
MULTIPLE INTEGRALS
Thus Theorem 9 gives
yy f x, y dx dy yy f r cos , r sin R
S
y
y
b
a
&x, y
dr d
&r, f r cos , r sin r dr d
which is the same as Formula 12.3.2.
Use the change of variables x u 2 v 2, y 2uv to evaluate the
integral xxR y dA, where R is the region bounded by the x-axis and the parabolas
y 2 4 4x and y 2 4 4x, y 0.
V EXAMPLE 2
SOLUTION The region R is pictured in Figure 2 (on page 714). In Example 1 we
discovered that TS R, where S is the square 0, 1 0, 1. Indeed, the reason
for making the change of variables to evaluate the integral is that S is a much simpler region than R. First we need to compute the Jacobian:
&x
&x, y
&u
&u, v
&y
&u
Therefore, by Theorem 9,
yy y dA yy 2uv
R
S
8y
1
0
y
1
0
&x
&v
2u
&y
2v
&v
2v
4u 2 4v 2 0
2u
&x, y
1 1
dA y y 2uv4u2 v 2 du dv
0
0
&u, v
u3v uv 3 du dv 8 y
1
0
y 2v 4v 3 dv v 2 v 4
1
[
0
]
1
0
[
1 4
4 v
u
12 u2v 3
u1
]
u0
dv
2
■
NOTE Example 2 was not a very difficult problem to solve because we were given
a suitable change of variables. If we are not supplied with a transformation, then the
first step is to think of an appropriate change of variables. If f x, y is difficult to integrate, then the form of f x, y may suggest a transformation. If the region of integration R is awkward, then the transformation should be chosen so that the corresponding
region S in the uv-plane has a convenient description.
EXAMPLE 3 Evaluate the integral xxR e xyxy dA, where R is the trapezoidal
region with vertices 1, 0, 2, 0, 0, 2, and 0, 1.
SOLUTION Since it isn’t easy to integrate e xyxy, we make a change of variables
suggested by the form of this function:
10
uxy
vxy
These equations define a transformation T 1 from the xy-plane to the uv-plane.
Theorem 9 talks about a transformation T from the uv-plane to the xy-plane. It is
SECTION 12.8
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
■
719
obtained by solving Equations 10 for x and y :
x 12 u v
11
y 12 u v
The Jacobian of T is
&x
&x, y
&u
&u, v
&y
&u
&x
&v
&y
&v
1
2
1
2
1
2
1
2
12
To find the region S in the uv-plane corresponding to R, we note that the sides of R
lie on the lines
√
√=2
(_2, 2)
S
u=_√
y0
(2, 2)
(1, 1)
uv
√=1
0
T
u
v2
u v
v1
S u, v 1 v 2, v u v y
Theorem 9 gives
x-y=1
2
0
_1
xy1
Thus the region S is the trapezoidal region with vertices 1, 1, 2, 2, 2, 2, and
1, 1 shown in Figure 8. Since
T –!
1
x0
and, from either Equations 10 or Equations 11, the image lines in the uv-plane are
u=√
(_1, 1)
xy2
yy e
x
R
xyxy
dA yy e uv
R
x-y=2
_2
S
y
2
y
1
FIGURE 8
1
2
y
v
v
2
1
&x, y
du dv
&u, v
e uv ( 12 ) du dv 12 y [ve uv ]uv dv
2
uv
1
e e1 v dv 34 e e1 ■
TRIPLE INTEGRALS
There is a similar change of variables formula for triple integrals. Let T be a transformation that maps a region S in u vw-space onto a region R in xyz-space by means of
the equations
x tu, v, w
y hu, v, w
z ku, v, w
The Jacobian of T is the following 3 3 determinant:
12
&x
&u
&x, y, z
&y
&u, v, w
&u
&z
&u
&x
&v
&y
&v
&z
&v
&x
&w
&y
&w
&z
&w
720
■
CHAPTER 12
MULTIPLE INTEGRALS
Under hypotheses similar to those in Theorem 9, we have the following formula for
triple integrals:
13
yyy f x, y, z dV
R
yyy f xu, v, w, yu, v, w, zu, v, w
S
&x, y, z
du dv dw
&u, v, w
V EXAMPLE 4 Use Formula 13 to derive the formula for triple integration in
spherical coordinates.
SOLUTION Here the change of variables is given by
x sin cos y sin sin We compute the Jacobian as follows:
sin cos &x, y, z
sin sin & , , cos cos sin sin cos cos sin cos cos sin 0
sin z cos sin sin cos cos sin cos sin sin sin sin cos cos sin sin cos sin sin cos 2 sin cos sin2 2 sin cos cos2 sin sin2 cos2 sin2 sin2 2 sin cos2 2 sin sin2 2 sin Since 0 , we have sin 0. Therefore
&x, y, z
2 sin 2 sin & , , and Formula 13 gives
yyy f x, y, z dV yyy f sin cos , sin sin , cos R
2
sin d d d
S
which is equivalent to Formula 12.7.3.
■
SECTION 12.8
12.8
1–6
■
1. x u 4 v,
y 3u 2v
2. x u 2 v 2,
y u 2 v2
u
,
uv
y
4. x sin ,
5. x u v,
■
■
■
xy 1, xy 2, xy 2 1, xy 2 2; u xy, v xy 2.
Illustrate by using a graphing calculator or computer to
draw R.
v
■
■
z e uvw
■
■
■
■
■
■
■
■ Find the image of the set S under the given
transformation.
7–10
7. S u, v
0 u 3,
x 2u 3v, y u v
10. S is the disk given by u 2 v 2 1;
11–16
11.
■
■
■
19.
■
x au, y bv
■
■
■
■
xxR x 3y dA,
20.
■
where R is the triangular region with
vertices 0, 0, 2, 1, and 1, 2; x 2u v, y u 2v
xxR 4 x 8y dA,
13.
xxR x 2 dA,
15.
■
■
■
■
■
■
■
■
■ Evaluate the integral by making an appropriate change
of variables.
Use the given transformation to evaluate the integral.
12.
14.
■
■
19–23
9. S is the triangular region with vertices 0, 0, 1, 1, 0, 1;
x u2, y v
■
■
18. Evaluate xxxE x 2 y dV, where E is the solid of Exercise 17(a).
0 v 2;
8. S is the square bounded by the lines u 0, u 1, v 0,
v 1; x v, y u1 v 2 ■
■
ellipsoid x 2a 2 y 2b 2 z 2c 2 1. Use the transformation x au, y bv, z cw.
(b) The Earth is not a perfect sphere; rotation has resulted
in flattening at the poles. So the shape can be approximated by an ellipsoid with a b 6378 km and
c 6356 km. Use part (a) to estimate the volume of the
Earth.
z uw
y e uv,
■
17. (a) Evaluate xxxE dV, where E is the solid enclosed by the
y cos ■
721
2
; 16. xxR y dA, where R is the region bounded by the curves
uv
y vw,
6. x e uv,
■
EXERCISES
Find the Jacobian of the transformation.
3. x CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
where R is the parallelogram with
vertices 1, 3, 1, 3, 3, 1, and 1, 5;
x 14 u v, y 14 v 3u
where R is the region bounded by the ellipse
9x 2 4y 2 36; x 2u, y 3v
xxR x 2 xy y 2 dA,
where R is the region bounded
by the ellipse x 2 xy y 2 2;
x s2 u s23 v, y s2 u s23 v
xxR xy dA, where R is the region in the first quadrant
bounded by the lines y x and y 3x and the hyperbolas
xy 1, xy 3; x uv, y v
21.
x 2y
dA, where R is the parallelogram enclosed by
3x y
R
the lines x 2y 0, x 2y 4, 3x y 1, and
3x y 8
yy
xxR x ye x y
2
2
dA , where R is the rectangle enclosed by
the lines x y 0, x y 2, x y 0, and x y 3
yx
dA, where R is the trapezoidal region
yx
R
with vertices 1, 0, 2, 0, 0, 2, and 0, 1
yy cos
22.
xxR sin9x 2 4y 2 dA,
23.
xxR e xy dA,
■
where R is the region in the first
quadrant bounded by the ellipse 9x 2 4y 2 1
where R is given by the inequality
x y 1
■
■
■
■
■
■
■
■
■
■
24. Let f be continuous on 0, 1 and let R be the triangular
region with vertices 0, 0, 1, 0, and 0, 1. Show that
yy f x y dA y
1
0
R
u f u du
■