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D
over a region D in space in spherical coordinates, integrating first with respect to p, then with respect to cfJ, and finally with respect to 8, take the following steps. 1.
Sketch. Sketch the region D along with its projection R on the xyplane. Label the surfaces that bound D.
z
x
2.
Find the plimits of integration. Draw a ray M from the origin through D making an angle cfJ with the positive zaxis. Also draw the projection of M on the xyplane (call the projection L). The ray L makes an angle 8 with the positive xaxis. As p increases, M enters D at p == gl (cfJ, 8) and leaves at p == g2( cfJ, 8). These are the plimits of
integration. z
R
    () =
f3
y L
x
3.
e
cfJ that M makes with the cfJmax. These are the cfJ limits of integration.
Find the cfJlimits of integration. For any given 8, the angle
zaxis runs from cfJ
==
cfJmin to cfJ
==
882
Chapter 15: Multiple Integrals
4.
Find the 8limits ofintegration. The ray L sweeps over R as 8 runs from ex to f3. These are the 8limits of integration. The integral is
~~if
()= f31¢=¢max
j(p, cP, 8) dV
SoLution over D.
R (J
FIGURE 15.52
¢=¢rnin
P=g2(¢, ())
1
j(p, cP, 8) p2 sin cP dp dcP d8.
p=gl(¢, ())
EXAMPLE 5 Find the volume of the "ice cream cone" D cut from the solid sphere p :::; 1 by the cone cP == 7T/3.
z
Example 5.
1
()=a
D
x
==
L
Y
The ice cream cone in
The volume is V
==
fffn p2 sin cP dp dcP d8,
the integral of j(p, cP, 8)
==
1
To find the limits of integration for evaluating the integral, we begin by sketching D and its projection R on the xyplane (Figure 15.52). The plimits ofintegration. We draw a ray M from the origin through D making an angle cP with the positive zaxis. We also draw L, the projection of M on the xyplane, along with the angle 8 that L makes with the positive xaxis. Ray M enters D at p == 0 and leaves at p == 1. The cPlimits ofintegration. The cone cP == 7T/3 makes an angle of 7T/3 with the positive zaxis. For any given 8, the angle cP can run from cP == 0 to cP == 7T/3. The 8limits of integration. The ray L sweeps over R as 8 runs from 0 to 27T. The volume is
{27T {7T/3 [p3] 1 {27T {7T/3 1 3 0 sin 4J d4J dO = Jo Jo 3 sin 4J d4J dO
= Jo Jo 27T
==
1 o
[1   cos cP ]7T/3 d8 3 0
1
27T
==
0
(16
+ 1) d8 3
== 1
6
(2 7T)
•
7T. == 3
EXAMPLE 6 A solid of constant density 8 == 1 occupies the region D in Example 5. Find the solid's moment of inertia about the zaxis. SoLution
In rectangular coordinates, the moment is
In spherical coordinates, x 2 + y2 Hence,
==
(p sin cP cos 8)2
+ (p sin cP sin 8)2
==
p2 sin2 cPo
For the region in Example 5, this becomes
1_
{27T {7T/3
= SJo Jo 1_
{27T ( 1
=SJo
2
(1  cos
4J) sin 4J d4J dO
1 1) "2+ 1 +243
1_
{27T [
=
SJo
1_
{27T 5
dO=SJo
cos 4J
24 dO
=
cos 3 cP ]7T/3
+ 3
1 24 (27T)
7T
0
=U·
dO •
15.7 Triple Integrals in Cylindrical and Spherical Coordinates
883
Coordinate Conversion Formulas C'YuNDRICAL TO
SPHERICAL TO
SPHERICAL TO
REcTANGULAR
RECTANGULAR
CYLINDRICAL
rcosO
p sin "'cosO
r=psin'"
y = rsinO
Y = psin"'sinO
z=pcos'"
z=z
z=pcos'"
0=0
X =
X =
Corresponding fonnulas for dV in triple integrals:
dV= dxdydz = dzrdrdO = ;
sin '" dp d'" dO
In the next section we offer a more general procedure for detennining dV in cylindrical and spherical coordinates. The resuits, of course, will he the same.
Exercises 15.7 Evaluating Integrals In CyUndrical Coordinates Evaluate the cylindricsl coordinate integrals in Exercises l{j.
1.
'WJ.1J~ dzrdrdO
J.o
0
2. J.,wJ.'J,vts=;' dzrdrdO rl~
rOO
3. J.'w f6/'w f3+24,.' dz r dr dO 4. J.w {'Iw
Jo Jo J.'WJ.'Jl~3 o
5.
0
f'~ z dz r dr dO
Jo Jv'h'
Jo
Changtng the Order of Integration In Cylindrical Coordinates
9. 10.
0
J.
Jo Jo 0 Jo
I
I 0
(r' cos' 0
J. 'l~J.2w o
r2
(rsinO
b. dr dz dO
Jo
+ z')r dO dr dz + I)rdOdzdr
Finding Iterated Integrals In CyUndrical Coordinates 13. Give the limits of integration for evaluatiog the integral
ff!
f(r, 0, z) dz r dr dO
14. Convert the integral
[J.VH[(x' +
to an equivalent integral in cylindrical coordinates and evaluate
the result In Exercises 1520, set up the iterated integral for evaluatiog fffnf(r, 0, z) dz r dr dO over the given regionD.
15. D is the right circular cylinder whose base is the circle r ~ 2 sin 0 in the xyplane and whose top lies in the plane z = 4  y.
11. Let D be the region bounded below by the plane z = 0, above by the sphere + y' + = 4, nod on the sides by the cylinder + y' ~ I. Set up the triple integrals in cylindricsl coordinates that give the volume of D using the following orders of integration.
x'
a. dz drdO
y') dz dx dy
z
0
x'
c. dO dzdr
as an iterated integral over the region that is bounded below by the plnoe z = 0, on the side by the cylinder r = cos 0, nod on top by the paraboloid z ~ 3r'.
+ z')dzrdrdO
The integrals we have seen so far suggest that there are preferred orders of integration for cylindricsl coordinates, but other orders usually work well and are occasionally easier to evaluate. Evaluate the integrsls in Exercises 710. 1 7. J.'w {' f'P r'drdzdO 8. J.'w f1+~"4rdrdOdz
1J.V, f'w
a. dz dr dO
dzrdrdO
6. J.'w f111/2(r'sin'O o 1/2
o
integrals in cylindricsl coordinates that give the volwne of D using the following orders of integration.
z=4y
z'
b. dr dz dO
c. dO dz dr
12. Let D be the region bounded below by the cone z = y'x' + y' nod above by the paraboloid z = 2  x'  y'. Set up the triple
/
884
Chapter 15: Multiple Integrals
16. D is the right circular cylinder whose base is the circle r = 3 cos 0 and whose top lies in the plane z = 5  x.
z
z
2
z=5x
y
"r =
3 cos ()
x
x
y=x
17. D is the solid right cylinder whose base is the region in the xyplane that lies inside the cardioid r = 1 + cos 0 and outside the circle r = 1 and whose top lies in the plane z = 4.
z
EvaLuating IntegraLs in SphericaL Coordinates Evaluate the spherical coordinate integrals in Exercises 2126.
7T {7T {2 sin4J
21.
1io io 1i i 1io io 1 io io 1 1io io
p2 sin c/J dp dc/J dO
0
27T {7T/4 {2
22.
0
0
27T {7T
~
x
r
=
r
=
1
23.
18. D is the solid right cylinder whose base is the region between the circles r = cos 0 and r = 2 cos 0 and whose top lies in the plane z = 3  y.
{(Icos 4J)/2
24. 25.
z
\
26.
{I
27T17T/312
o
3
5p3 sin c/J dp dc/J dO
0
3p2 sin 4> dp d4> dO
sec4J
0
27T {7T/4
z=3y
p2 sin c/J dp dc/J dO
0
37T/2 {7T
1 + cos ()
(p cos c/J) p2 sin c/J dp dc/J dO
0
{sec 4J
(p cos c/J) p2 sin c/J dp dc/J dO
0
Changing the Order of Integration in SphericaL Coordinates The previous integrals suggest there are preferred orders of integration for spherical coordinates, but other orders give the same value and are occasionally easier to evaluate. Evaluate the integrals in Exercises 2730.
/\ x
\ r
r
=
=
cos 0
2 cos ()
19. D is the prism whose base is the triangle in the xyplane bounded by the xaxis and the lines y = x and x = 1 and whose top lies in theplanez = 2  y.
z
27.1 ro 2
o J7T
28. {1T/3
J7T/6
{2 esc q, rz1Tp2 sin c/J dO dp dc/J Jo
Jcsc4J
29. 1111T11T/412P sin3 c/J dc/J dO dp
2 z=2y
ix I
/
/
30. {7T/2 {
1
7T12
J7T/6 J7T12
2
y=x
20. D is the prism whose base is the triangle in the xyplane bounded by the yaxis and the lines y = x and y = 1 and whose top lies in the plane z = 2  x.
5p4 sin3 c/J dp dO dc/J
esc 4J
31. Let D be the region in Exercise 11. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.
a. dp d4> dO
/ /
x
{1T/2p3 sin 2c/J dc/J dO dp
J7T/4
b. d4> dp dO
32. Let D be the region bounded below by the cone z = v'x 2 + y2 and above by the plane z = 1. Set up the triple integrals in spherical coordinates that give the volume of D using the following orders of integration.
a. dp d4> dO
b. d4> dp dO
Finding Iterated IntegraLs in SphericaL Coordinates In Exercises 3338, (a) find the spherical coordinate limits for the integral that calculates the volume of the given solid and then (b) evaluate the integral. 33. The solid between the sphere p = cos 4> and the hemisphere p = 2,z ~
°
z p = cos
ep
2
885
Triple Integrals in Cylindrical and Spherical Coordinates
15.7
volume of D as an iterated triple integral in (a) cylindrical and (b) spherical coordinates. Then (c) find V. 41. Let D be the smaller cap cut from a solid ball of radius 2 units by a plane 1 unit from the center of the sphere. Express the volume of D as an iterated triple integral in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. Then (d) find the volume by evaluating one of the three triple integrals. 42. Express the moment of inertia /z of the solid hemisphere x 2 + y2 + z2 ~ 1, z ~ 0, as an iterated integral in (a) cylindrical and (b) spherical coordinates. Then (c) find /z.
P= 2
VoLumes Find the volumes of the solids in Exercises 4348. 43. x
44.
y
34. The solid bounded below by the hemisphere p = 1, Z above by the cardioid of revolution p = 1 + cos 4>
z
z ~
0, and
z P= 1
y
x
x
y
\
z=~
z = (x 2 + y2)2  1
45. z =y
f
y
x
46.
z
35. The solid enclosed by the cardioid of revolution p = 1  cos 4> 36. The upper portion cut from the solid in Exercise 35 by the xyplane
r
3 cos ()~
=
37. The solid bounded below by the sphere p = 2 cos 4> and above by 2 the cone z = + y2
y
x
"r y
x
v'x
47.
=
3 cos ()
48.
z
x
y
x
38. The solid bounded below by the xyplane, on the sides by the sphere p = 2, and above by the cone 4> = '1T/3
r
/ sin () =
y x
y
r
z
=
cos ()
49. Sphere and cones Find the volume of the portion of the solid sphere p ~ a that lies between the cones 4> = '1T /3 and 4> = 2'1T/3. p=2 x
50. Sphere and halfplanes Find the volume of the region cut from the solid sphere p ~ a by the halfplanes (J = and (J = '1T /6 in the first octant.
°
51. Sphere and plane Find the volume of the smaller region cut from the solid sphere p ~ 2 by the plane z = 1.
Finding TripLe IntegraLs 39. Set up triple integrals for the volume of the sphere p = 2 in (a) spherical, (b) cylindrical, and (c) rectangular coordinates. 40. Let D be the region in the first octant that is bounded below by the cone 4> = '1T /4 and above by the sphere p = 3. Express the
52. Cone and planes Find the volume of the solid enclosed by the cone z = v'x 2 + y2 between the planes z = 1 and z = 2. 53. Cylinder and paraboloid Find the volume of the region bounded below by the plane z = 0, laterally by the cylinder x 2 + y2 = 1, and above by the paraboloid z = x 2 + y2.
886
Chapter 15: Multiple Integrals
54. CyHnder and paraboloids Find 1he volwne ofthe regioo bouoded below by the paraboloid z ~ ",' + y', laterally by the cylinder x' + y' = I,andabovebytheparaboloidz =x' + y' + I.
73. Moment of inertia of solid cone Find the moment of inertia of a right circular cone of base radius I and height I about an axis through the vertex parallel to the base. (Take 8 = I.)
55. Cylinder and cones Find the volume of the solid cut from the thickwalled cylinder 1:5 + y' :5 2 by the cooes z ~
74. Moment of inertia of solid sphere Find the moment of inertia ofa solid sphere of radius a about a diameter. (Take 8 = I.)
x'
±Vx' + y'.
56. Sphere and cylinder
side the sphere
x'
Find 1he volume of the regioo that lies in+ y' + z' ~ 2 and outside the cylinder
x'+y'=1. 57. Cylinder and planes Find the volume ofthe regioo enclosed by the cylinder ",' + y' ~ 4 and the planes z ~ 0 and y + z ~ 4. 58. Cylinder and planes Find the volume of the regioo enclosed by the cylinder + y' = 4 and the planes z = 0 and x + y + z ~ 4.
x'
75. Moment of inertia of solid cone Find the moment of inertia of a right circular cone of base radius a and height h about its axis. (Hint: Place the cone with its vertex at the origin and its axis along the zaxis.) 76. Variable density A solid is bounded on the top by the paraboloid z = r', 00 the bottom by the plane z = 0, and on the sides by the cylinder r = I. Find the center of mass and the moment of inertia about the zaxis if the density is
a. 8(r, 0, z) = z
b. 8(r, 0, z) = r.
59. Region trapped by paraboloids Find the volume of the regioo y' and below by bounded above by 1he paraboloid z = 5 the paraboloid z ~ 4x' + 4y'.
77. Variable density A solid is bounded below by the cone z= x' + y' and above by the plane z = I. Find the center of mass and the moment of inertia about the zaxis if the density is
60. Paraboloid and cylinder Find the volwne of the regioo y', below by the bounded above by the paraboloid z = 9 xyplane, and lying outside the cylinder ",' + y' ~ I.
78. Variable density A solid ball is bounded by the sphere p Find the moment of inertia about the zaxis if the density is
x' 
x' 
61. Cylinder and sphere Find 1he volume of the regioo cut from the solid cylinder + y' :5 I by the sphere + y' + z' = 4.
x'
x'
62. Sphere and paraboloid Find the volume of the regioo bounded + y' + z' ~ 2 and below by the paraboabove by the sphere loidz = x 2 + y2.
x'
Average Values 63. Find the average value of the functioo f(r, 0, z) = r over the regioo bounded by the cylinder r ~ I between the planes z ~ I andz = I.
64. Find the average value of the function f(r, 0, z) = r over the solid + z' ~ I. (This is 1he sphere ball bounded by the sphere + y' + = I.)
r'
z'
x'
65. Find the average value of the function f(p, q" 0) = p over the solid ball p :5 I. 66. Find the average value of the functioo f(p, q" 0) the solid upper ball p :5 1,0 :5 q, :5 'If/2.
~
p cos q, over
Masses, Moments, and Centroids 67. Center of mass A solid of cooatant density is bounded below by the plane z = 0, above by the cone z = r, r O?:: 0, and on the sides by the cylinder r = I. Find the center of mass. 68. Centroid Find the centroid of thVCgioo in the fll'St octant that is bounded above by the cone z = + y', below by the plane z = 0, and 00 the sides by the cylinder + y' = 4 and the planes x ~ 0 and y ~ O.
x'
69. Centroid
x'
Find the centroid of the solid in Exercise 38.
70. Centroid Find the centroid of the solid bounded above by the spherep ~ a and below by the cooe q, ~ 'If/4.
V
a. 8(r, 0, z) = z
b. 8(r, 0, z) = z'.
a. 8(p, q" 0) = p'
~
a.
b. 8(p, q" 0) = r = p sin q,.
79. Centroid of solid semiellipsoid Show that the centroid of the solid semiellipsoid ofrevolutioo (r'/a') + (z'/h') :5 I,z '" 0, lies on the zaxis threeeighths of the way from the base to the top. The special case h = a gives a solid hemisphere. Thus, the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base to the top. 80. Centroid of solid cone Show that the centroid of a solid right eircular eone is onefourth of the way from the base to the vertex. (10 general, the centroid of a solid cone or pyramid is onefourth of the way from the centroid of the base to the vertex.) 81. Density of eenter of a planet A planet is in the shape of a sphere of radius R and total mass M with spherically symmetric
density distribution that increases linearly as one approaches its center. What is the density at the center of this planet if the density at its edge (surface) is tsken to be zero?
82. Mass of planet's atmosphere A spherical planet of radius R bas an atmosphere whose density is p. = lJ<Je ~', where h is the altitude above the surface of1he planet, IJ<J is the density at sea level, and c is a positive constant Find the mass of1he planet's atmosphere. Theory and Examples 83. Vertical planes in cylindrical coordinates
a. Show that planes perpendicular to the xaxis bave equatiooa of the form r
~
a sec 0 in cylindrical coordinates.
b. Show that planes perpendicular to the yaxis bave equations of the form r = b csc O. 84. (Continuation of Exercise 83.) Find an equation of the form r = f(O) in cylindrical coordinates for 1he plane ax + by = c, c # O.
71. Centroid Find the centroid of 1he regioo that is bounded above by the surface z = 00 the sides by the cylinder r = 4, and below by the xyplane.
85. Symmetry What symmetry will you fmd in a surface that bas an equatioo of the form r = f(z) in cylindrical coordinates? Give
Find the centroid of the region cut from the solid ball z':5 I by the balfplanes 0 = 'If/3,r '" O,andO = 'If/3,
86. Symmetry What symmetry will you fmd in a surface that bas an equation of the form p = f( q,) in spherical coordinates? Give
v;.,
72. Centroid
r' + r O?:: O.
reasons for your answer.
reasons for your answer.
15.8 Substitutions in Multiple Integrals
15.8
I_SU_b_Sti_·t_u_ti_O_nS_i_n_M_U_Lti~·p_Le_In_te....::g~ra_Ls
o
_(u_ v)
G
887

The goal oftbis section is to introduce you to the ideas involved in coordinate transformations. You will see how to evaluate multiple integrals by substitution in order to replace complicated integrals by ones that are easier to evaluate. Substitutions accomplish this by simplifying the integrand, the limits of integration, or both. A thorough discussion ofmultivariable transformations and substitutions, and the Jacobian, is best left to a more advanced course following a study oflinear algebra.
Substitutions in Double Integrals u c+o Cartesian uvplane x
y
~ g(u, 0)
1 y
~
h(u, 0)
The polar coordinate substitution ofSeclion 15.4 is a special case ofa more general substitution method for double integrals, a method that pictures changes in variables as transformations of regions. Suppose that a region G in the uvpiane is transformed onetoone into the region R in the xypiane by equations ofthe form
x = g(u, v),
y = h(u, v),
as suggested in Figure 15.53. We call R the image of G under the transformation, and G the preimage of R. Any function j(x, y) defined on R can be thought of as a function j(g(u, v), h(u, v» dermed on G as well. How is the integral of j(x,y) over R related to the integral of j(g(u, v), h(u, v» over G? The answer is: If g, h, and j have continuous partial derivatives and J(u, v) (to be discussed in a moment) is zero only at isolated points, if at all, then Cartesian xyplane
FIGURE 15.53 The equations x = g(u,v) andy = h(u,v) allow us to change an integral over a region R in the xyplane into an integral over a region G in the uvplane by using Equation (I).
ff
j(x,y)dxdy =
ff
(1)
j(g(u,v),h(u,v»IJ(u,v)ldudv.
G
R
The factor J(u, v), whose absolute value appears in Equation (1), is the Jacobian of the coordinate transformation, named after German mathematician Carl Jacobi. It measures how much the transformation is expanding or contracting the area around a point in G as G is transformed into R.
DEFINmON The Jacobian determinant or Jacobian of the coordinate transformationx = g(u, v),y = h(u, v) is
J(u, v) =
ax au
ax av
ay au
ay av
ax ay auav
ay ax au av'

(2)
The Jacobian can also be denoted by
a(x,y) J(u, v) = a(u, v) HIsTORICAL BIOGRAPHY
Carl Gustav Jacob Jacobi (18041851)
to help us remember how the determinant in Equation (2) is constrocted from the partial derivatives ofx andy. The derivation of Equation (1) is intricate and properly belongs to a course in advanced calculus. We do not give the derivation here.
EXAMPLE 1 Find the Jacobian for the polar coordinate transformation x = r cos 0, y = r sin 0, and use Equation (1) to write the Cartesian integral fJR j(x, y) dx dy as a polar integral.
888
Chapter 15: Multiple Integrals
o "f,,, 2
G
Solution Figure 15.54 shows how the equations x = r cos 6, y = r sin 6 transfonn the rectangle G: 0 :5 r :5 1,0 :5 0 :5 Tr/2, into the quarter circle R bounded by x 2 + y2 = I in the first quadrant of the ~plane. For polar coordinates, we have r and 0 in place of u and v. With x = r cos 6 and y = r sin 0, the Jacobian is
J(r,O)
Cartesian rOplane
=
ax oX or 00 oy
oy
or
00
=
Icoso sinO
11
f(x,y)dxdy
=
R
0=
~
1
r(cos2 0 + sin2 0)
=
Since we assume r ;;,: 0 when integrating in polar coordinates, IJ(r, 0)1 Equation (1) gives
I x=rcos8 ~y=rsin9
y
rSin°1 rcosO
11
=
=
r.
Irl
=
f(rcos6,rsinO)rdrdO.
r, so that
(3)
G
This is the same fonnula we derived independently using a geometric argument for polar area in Section 15.4. Notice that the integral on the righthand side of Equation (3) is not the integral of f(r cos 0, r sin 0) over a region in the polar coordinate plane. It is the integral of the product of f(r cos 0, r sin 0) and r over a region G in the Cartesian rOplane. _ Here is an example of a substitution in which the image of a rectangle under the coordinate transfonnation is a trapezoid. Transformations like this one are called linear transformations.
Cartesian xyplane
EXAMPLE 2
Evaluate
FIGURE 15.54 The equations x = r cos 8, y = rsin 8 transform G into R.
1o
4:LX~(Y/2)+1
xy/2
2x  y dxdy 2
by applying the transfonnation
2x  Y u=2'
y v=2
(4)
and integrating over an appropriate region in the uvplane. Solution We sketch the region R of integration in the xyplane and identify its boundaries (Figure 15.55).
y
v 2
u=o
v=2
4
u
G
=1
:ct;;~u
o
v 0
1
x=u+v y = 2v )
o \
1
y~O
FIGURE 15.55 The equations x = u + v and y = 2v transform G into R. Reversing the transformatioo by the equations u = (2x  y)/2 and v ~ y/2 transforms R into G (Example 2).
15.8 Substitutions in Multiple Integrals
889
To apply Equation (I), we need to fmd the corresponding uvregion G and the Jacobian of the transformation. To find them, we first solve Equations (4) for x andy in terms of u and v. From those equations it is easy to see that
x
u + v,
=
y
=
2v.
(5)
We then find the boundaries of G by substituting these expressions into the equations for the boundaries of R (Figure 15.55). .>yequations for the boundary of R
x = y/2 x = (y/2) + I y=O y=4
Corresponding uvequations for the boundary of G
u + v = 2v/2 = v u + v = (2v/2) + I = v 2v = 0 2v = 4
+
Simplified uvequations
u=o u = I v=O v=2
I
The Jacobian of the transformation (again from Equations (5» is
ax au J(u, v) = ay au
ax av ay av
a au(u + v)
a av(u + v)
a au (2v)
a av (2v)
=
I~ ~I
=
2.
We now have everything we need to apply Equation (I):
:L o 1 4
X
l
(Y/2 +1
2x  y 2dxdy
=
1"21"'
x=yj2
v
v=O
ulJ(u,v)ldudv
u=O
=f[(U)(2) du dv
v=u
I
=
f
[u
2
]:
dv
=f dv =2.
c1(+~U
• EXAMPLE 3
tl=2u
Evaluate
r
f'x
10 '10
2
~ (y
 2x)' dy dx.
Solution We sketch the region R ofintegration in the xyplane and identify its boundaries (Figure 15.56). The integrand suggests the transformation u = x + y and v = y  2x. Routine algebra produces x and y as functions of u and v:
y
x x+y=l
x=o R
0+'\"> X
=
u
v
3  3'
y
2u
=
v
3 + 3'
(6)
From Equations (6), we can find the boundaries of the uvregion G (Figure 15.56). .>yequations for the boundary of R
Corresponding uvequations for the boundary of G
Simplified uvequations
y~O
FIGURE 15.56 The equations x ~ (u/3)  (v/3) andy = (2u/3) + (v/3) transform G into R. Reversing the transformation by the equations u ~ x + y and v = y  2x transforms R into G (Example 3).
x+y=1 x=O y=O
(1¥)+(2;+¥)=1 '!_"=O 3 3 2u+,,=0 3
3
u
=I
v=u
v
=2u
890
Chapter 15: Multiple Integrals The Jacobian of the transfonnation in Equations (6) is
J(u, v) =
ax au
ax av
I 3
I 3
ay au
ay av
2 3
I 3
I 3.
Applying Equation (I), we evaluate the integral:
=
1'1" o
11.'
=
9
u'/2 v 2
2u
(I)
1.£1 [I
dv du = 3

3
ul/2(u 3 +8u 3 )du=
11 0
0
0
u 1/ 2 v3 3
]V"
du
v2u
2]1 29
u7/2du=u912 9
=.
0
•
In the next example we illustrate a nonlinear transfonnation ofcoordinates resulting from simplifying the fonn of the integrand Like the polar coordinatea' transfonnation, nonlinear transfonnations can map a straight line boundary of a region into a curved boundary (or vice versa with the inverse transfonnation). In general, nonlinear transfunnations are more complex to analyze than linear ooes, and a complete treatment is left to a more advanced course. y
EXAMPLE 4
Evaluate the integral
2 f\~'~./
1r ~eVxYdxdY. 2
, }'/Y'/X
1
Solution
The square root terms in the integt'lll!d suggest that we might sinoplify the inteand v = VYf;:. Squaring these equatioos, we readily gratioo by substituting u = have u 2 = xy and v 2 = y/x, which inoply that u 2if = y2 and u 2/if = x 2. So we obtain the transfonnation (in the same ordering ofthe variables as discussed before)
cl'''~.x
o
1
2
'\IxY
u x=v
FIGURE 15.57 The region of integration R in Example 4.
y = uv.
and
Let's firat see what happens to the integrand itself under this transfonnatioo. The Jacobian ofthe transfonnation is
u=1
J
J(u, v) =
xy=l
/
v=l
1
2
ax av
I v
u
ay au
ay av
v
u
if
2u v'
If G is the region of integration in the uvplane, then by Equation (I) the transfonned
I
y=x ,
double integral under the substitution is
I
0
ax au
"
FIGURE 15.58 The boundaries of the region G correspond to those of region R in Figure 15.57. Notice as we move counterclockwise around the region R, we also move counterclockwise around the region G. The inverse transformation equations u ~ v'i,Y. v ~ vY7i prodoce the region G from the region R.
11 ~ R
e VxY dx dy =
11
ve"
~ du dv =
G
11
2ue" du dv.
G
The transfonned integrand function is easier to integrate than the original one, so we proceed to detennine the limits ofintegratioo for the transfonned integral. The regioo of integratioo R of the original integral in the xyplane is shown in Figore and v = VYf;:, we see that the inoage of 15.57. From the substitution equations u = the lefthand boundary xy = I for R is the vertical line segment u = 1,2 ;;,: v ;;,: I, in G (see Figure 15.58). Likewise, the righthand boundary y = x of R maps to the horizontal line segment v = I, I :5 U :5 2, in G. Finally, the horizontal top boundary Y = 2 of R
'\IxY
Substitutions in Multiple Integrals
15.8
891
maps to uv == 2, 1 :::; v :::; 2, in G. As we move counterclockwise around the boundary of the region R, we also move counterclockwise around the boundary of G, as shown in Figure 15.58. Knowing the region of integration G in the uvplane, we can now write equivalent iterated integrals:
11~ 2
Y
1
x e vxy dx dy
11 2
==
l/y
1
2
1
/
U
2ue u dv duo
Note the order of integration.
We now evaluate the transformed integral on the righthand side,
11 2
2
U
/
u
2ue dv du =
=
21 21
2
vue U
]~:7/u du
2
(2e U
ue U ) du

=
21 (2 
==
2[(2  u)e U
2
==
2(e
2

u)e U du
+
(e
+
e U ]::~
e))
==
Integrate by parts.
2e(e  2).
•
Substitutions in Triple Integrals The cylindrical and spherical coordinate substitutions in Section 15.7 are special cases of a substitution method that pictures changes of variables in triple integrals as transformations of threedimensional regions. The method is like the method for double integrals except that now we work in three dimensions instead of two. Suppose that a region G in uvwspace is transformed onetoone into the region D in xyzspace by differentiable equations of the form
x
==
g(u, v, w),
y
==
h(u, v, w),
z == k(u, v, w),
as suggested in Figure 15.59. Then any function F(x, y, z) defined on D can be thought of as a function F(g(u, v, w), h(u, v, w), k(u, v, w))
H(u, v, w)
==
defined on G. If g, h, and k have continuous first partial derivatives, then the integral of F(x, y, z) over D is related to the integral of H(u, v, w) over G by the equation
iff
F(x,y,z)dxdydz =
D
iii
H(u,v,w)IJ(u,v,w)ldudvdw.
G
z
w x = g(u, v, w) y = h(u, v, w) z = k(u, v, w) )
y
v u
Cartesian uvwspace
x
Cartesian xyzspace
FIGURE 15.59 The equations x = g(u, v, w),y = h(u, v, w), and z = k(u, v, w) allow us to change an integral over a region D in Cartesian xyzspace into an integral over a region G in Cartesian uywspace using Equation (7).
(7)
892
Chapter 15: Multiple Integrals The factor J(u, v, w), whose absolute value appears in this equation, is the Jacobian determinant
Cube with sides parallel to the coordinate axes
z
J(u,v,w)
==
8
r
ax au ay au
ax av ay av
ax aw ay aw
az au
az av
az aw
a(x, y, z) a(u, v, w)·
Cartesian r8zspace
This determinant measures how much the volume near a point in G is being expanded or contracted by the transformation from (u, v, w) to (x, y, z) coordinates. As in the twodimensional case, the derivation of the changeofvariable formula in Equation (7) is omitted. For cylindrical coordinates, r, 8 , and z take the place of u, v, and w. The transformation from Cartesian r8zspace to Cartesian xyzspace is given by the equations
x = r cos 8 y = r sin 8
1
z=z
z
x
==
r cos 8,
y
==
z==z
r sin 8,
(Figure 15.60). The Jacobian of the transformation is D
8 = constant x
Y
J(r, 8, z)
Cartesian xyzspace
FIGURE 15.60 The equations x = r cos (), y = r sin (), and z = z transform the cube G into a cylindrical wedgeD.
==
ax ar ay ar
ax a8 ay a8
ax az ay az
az ar
az a8
az az
cos 8 sin 8 0
r sin 8 r cos 8 0
0 0 1
The corresponding version of Equation (7) is
fff
F(x,y,z)dxdydz
fff
=
D
H(r,O,z)lrldrdOdz.
G
We can drop the absolute value signs whenever r ~ o. For spherical coordinates, p, c/J, and 8 take the place of u, v, and w. The transformation from Cartesian pc/J8space to Cartesian xyzspace is given by
x
==
p sin c/J cos 8,
y
==
p sin c/J sin 8,
z
==
p cos c/J
(Figure 15.61). The Jacobian of the transformation (see Exercise 19) is
J(p, c/J, 8)
ax ap ay ==
ax
ac/J ay
ap
ac/J
az
az
ap
ac/J
ax a8 ay a8
==
p2 sin c/J.
az a8
The corresponding version of Equation (7) is
fff D
F(x, y, z) dx dy dz
=
fff G
H(p,
Substitutions in Multiple Integrals
15.8
(J
893
p = constant
Cube with sides parallel to the coordinate axes
(J =
constant
sin ep cos (J sin ep sin (J z = p cos ep
x y
= p = p
)
G
(J
p
Cartesian pep(Jspace
x
Cartesian xyzspace
FIGURE 15.61 The equations x = p sin 4> cos (J, y = p sin 4> sin (J, and z = p cos 4> transform the cube G into the spherical wedge D.
We can drop the absolute value signs because sin cP is never negative for 0 :::; cP :::; 'IT. Note that this is the same result we obtained in Section 15.7. Here is an example of another substitution. Although we could evaluate the integral in this example directly, we have chosen it to illustrate the substitution method in a simple (and fairly intuitive) setting.
w
EXAMPLE 5
Evaluate
1 1 1 4
3
2
)
o
~y
x=u+ y Y = 2y
l
=(Y/2)+1
(2X y  + z) dxdydz 2
x=y/2
u == (2x  y)/2,
z = 3w
z
3
Rear plane:
x D
=
~
2'
==
z/3
(8)
SoLution We sketch the region D of integration in xyzspace and identify its boundaries (Figure 15.62). In this case, the bounding surfaces are planes. To apply Equation (7), we need to find the corresponding uvwregion G and the Jacobian of the transformation. To find them, we first solve Equations (8) for x, y, and z in terms of u, v, and w. Routine algebra gives
ory = 2x
)
x
)
4 /
~ + 1, or y
=
==
u
+ v,
y
==
2v,
z == 3w.
(9)
y
Front plane: =
w
v == y/2,
and integrating over an appropriate region in uvwspace.
3
x
0
by applying the transformation
u
x
X
We then find the boundaries of G by substituting these expressions into the equations for the boundaries of D:
2x  2
FIGURE 15.62 The equations x = u + v,y = 2v, andz = 3w transform G into D. Reversing the transformation by the equations u = (2x  y)/2, v = y/2, and w = z/3 transforms D into G (Example 5).
xyzequations for the boundary of D x
==
Corresponding uvwequations for the boundary of G u
y/2
x == (y/2) y==O
+
1
u
+v
+v ==
==
2v/2 == v
(2v/2)
+
1 == v
+
Simplified uvwequations
u u
==
0
==
1
2v == 0
v==O
y==4
2v == 4
v==2
z==O
3w == 0
w==O
z == 3
3w == 3
w==1
894
Chapter 15: Multiple Integrals The Jacobian of the transfonnation, again from Equations (9), is
J(u, v, w) =
ax au
ax av
ax aw
ay au
ay av
ay aw
az au
az av
az aw
I 0 0
I 2 0
0 0 = 6. 3
We now bave everything we need to apply Equation (7):
1 1
4 :£X(Y/2)+ 1
3
(2xy + z) dxdydz
23 2 = 1 1 1\u + w)lJ(u,v,w)ldudvdw
00x=y/2 1
=
=
=
1J1\u+W)(6)dudvdw=61J[~2 +uwJ>vdw
61J (! 6[w
+ W2]~
+
w) dvdw
=
61' [¥
+ vw
J: dw
=
61'(1
+ 2w) dw
•
= 6(2) = 12.
Exercises 15.8 Jacoblans and Tnnsfonned Regions In the Plane 1. a. Solve the system
u = x  y,
v=2x+y
for x and y in terms of u and v. Then rmd the value of the Jacobian a(x, y)/a(u, v). b. Find the image under the transformatioo u ~ x  y, v = 2x + Y of the triangularregioo with vertices (0, 0), (I, I), and (I, 2) in the xyplane. Sketcb the transformed regioo in the uvplane. 2. a. Solve the system
u = x + 2y,
v=xy
for x and y in terms of u and v. Then rmd the value of the Jacobian a(x, y)/a(u, v). b. Find the image under the transformatioo u ~ x + 2y, v = x  y of the triangular region in the xyplane bounded by the lines y = 0, y = x , and x + 2y = 2. Sketch the transformed region in the uvplane. 3. a. Solve the system
u = 3x
by the xaxis, the yaxis, and the line x transformed region in the uvp1ane.
+y
= I. Sketch the
4. a. Solve the system
u
= 2x
 3y,
v = x + y
for x and y in terms of u and v. Then rmd the value of the Jacobian a(x,y)/a(u, v). b. Find the image uoder the transformation u = 2x  3y, v ~ x + y of the parallelogram R in the xyplane with bouodaties x = 3, x = O,y = x, and y = x + 1. Sketch the transformed regioo in the uvplane.
Substitutions In Double Integrals 5. Evaluate the integral
4:£X(Y/2)+1 2x  Y /.o ;;y/2
dxdy 2
from Example I directly by integration with respect to x and y to
confmn that its value is 2. 6. Use the transformation in Exercise I to evaluate the integral
+ 2y,
v = x
+ 4y
for x andy in terms ofu and v. Then rmd the value of the Jacobian a(x, y)/a(u, v). b. Find the image under the transformatioo u ~ 3x + 2y, v = x + 4y ofthe triangular region in thexyplane bounded
JJ(2x 2  xy  y2) dx dy R
for the regioo R in the rust quadrant bounded by the lines y = 2x + 4,y = 2x + 7,y = x  2, andy = x + 1.
15.8
7. Vse the transfonnation in Exercise 3 to evaluate the integral
Substitutions in Multiple Integrals
16. Vse the transformation x = u 2

895
,}l,y = 2uvto evaluate the in
tegral
jj(3x 2 + 14xy + 8y 2) dx dy
t
r2~ Jo Jo
R
for the region R in the fIrst quadrant bounded by the lines y = (3/2)x + I, y = (3/2)x + 3, y = (1/4)x, and y = (1/4)x + 1.
8. Vse the transformation and parailelogramR in Exercise 4 to evaluate the integral
Finding Jacobians
a. x = ucosv, y = usinv
R
9. Let R be the region in the flIst quadrant of the xyplane bounded by the hyperbolas xy = I, xy = 9 and the lines y = x, y = 4x. Vse the transformation x = u/v,y = uv with u > 0 and v > 0 to rewrite
jj
(Ix +
V;;)dxdy
R
as an integral over an appropriate region G in the uvplane. Then evaluate the uvintegral over G. 10. a. Find the Jacobian of the transfonnation x = u, y = uv and sketch the region G: 1 ::s; u ::s; 2, 1 ::s; uv ::s; 2, in the uvplane. b. Then use Equation (I) to transform the integral
2{2y
1J, x 1
12. The area of an ellipse The area '/Tab of the ellipse x2/a 2 + y2/b 2 ~ I can be found by integrating the function j(x, y) = I over the region bounded by the ellipse in the 'o/"plane. Evaluating the integral directly requires a trigonometric substi1ution. An easier way to evaluate the integral is to use the transfor
au, y
=
+ v2
18. Find the Jacobiana(x,y, z)/a(u, v, w) of the transfonnation
a. x = ucosv, y = usinv, z = w b. x = 2u  I, Y = 3v  4, z = (1/2)(w  4). 19. Evaluate the appropriate determinant to sbow that the Jacobian of the transformation from Cartesian p. 20. Substitutions in single integrals How can substi1utions in single defInite integrals be viewed as transfonnations of regions? What is the Jacobian in such a case? lllus1rate with an example.
Substitutions In Triple Integrals x,y,andz.
11. Polar moment of inertia of an elliptical plate A thin plate of constant density covers the region bounded by the ellipse x 2/a 2 + y2/b 2 = I, a > 0, b > 0, in the xyplane. Find the frrst moment of the plate about the origin. (Hint: V se the transformationx = arcos8,y = brsin8.)
=
b. x = usinv, Y = ucosv.
21. Evaluate the integral in Example 5 by integrating with respect to
dydx
into an integral over G, and evaluate bo1h integrals.
mation x
(Hint Show that the image of the triangular region G with vertices (0, 0), (I, 0), (I, I) in the uvplane is the region of integration R in the xyplane defmed by the limits of integration.)
17. Find the Jacobiana(x,y)/a(u, v) of the transformation
jj2(X  y)dxdy.
the disk G: u 2
Yx 2 + y 2 dydx.
bv and evaluate the transformed integral over I in the uvplane. Find the area this way.
22. Volume of an ellipsoid
Find the volume of the ellipsoid
= au, y = 00, and z = cwo Then find the volume of an appropriate region in uvwspace.)
(Hint: Let x 23. Evaluate
over the solid ellipsoid
'"
x2 y2 z2 2+2+2~1.
13. Vse the transformation in Exercise 2 to evaluate the integral
abc
2/3 [22y
o Jy ].
(x + 2y)e(yr) dx dy
(Hint: Let x = au, y = bv, and z appropriate region in uvwspace.)
by flIst writing it as an integral over a region G in the uvplane.
14. Vse the transformation x
~
u
+ (l/2)v,y
~
24. Let D be the region in xyzspace defmed by the inequalities
v to evaluate the 1
integral 2 {(y+4)/2 y'(2x
_ y)e(2ry )' dx dy
].
o }yf2
by lust writing it as an integral over a region G in the uvplane. 15. Vse the transformation x ~ u/v,y ~ uv to evaluate the integral
sam
11 2
1
Y
l(y
= cw. Then integrate over an
::s;
x ::s; 2,
O::s; xy
::s;
2,
O::s; z
~
1.
Evaluate
ffj(x2y
+ 3xyz) dx dy dz
D
by applying the transfonnation
(x 2 + y2) dx dy
+ 1414(Y (x 2 + y2) dx dy. 2
y(4
u
= x,
v = xy,
w = 3z
and integrating over an appropriate region G in uvwspace.
896
Chapter 15: Multiple Integrals
25. Centroid of a solid semieIIipsoid Assuming the result that the centroid of a solid hemisphere lies on the axis of symmetry threeeighths of the way from the base toward the top, show, by traosforming the appropriate integrals, that the center of mass of a solid semiellipsoid (x 2Ia 2) + (y2Ib 2) + (z 2/e 2) '" I, z '" 0, lies on the zaxis threeeighths of the way from the base toward the top. (You can do this without evaluating any of the integrals.)
Chapter
26. Cylindrieal sheIls In Section 6.2, we learned how to rmd the volume of a solid ofrevolution using the shell method; namely, if the region between the curve y = f(x) and the xaxis from a to b (0 < a < b) is revolved about the yaxis, the volume of the 2'lfxf(x) fix. Prove that rmding volumes by resulting solid is using triple integrals gives the same result. (Hint: Use cylindrical coordinates with the roles ofy andz cbanged.)
1:
Questions to Guide Your Review
1. Derme the double integral of a function of two variables over a bounded region in the coordinate plane. 2. How are double integrals evaluated as iterated integrals? Does the order of integration matter? How are the limits of integration determined? Give examples.
3. How are double integrals used to calculate areas and average values. Give examples. 4. How can you change a double integral in rectangular coordinates into a double integral in polar coordinates? Why might it be worthwhile to do so? Give an example. 5. Derme the triple integral of a function f(x, y, z) over a bounded region in space.
7. How are double and triple integrals in rectangular coordinates used to calculate volumes, average values, masses, moments, and
centers of mass? Give examples. 8. How are triple integrals dermed in cylindrical and sphetical coordinates? Why might one prefer working in one of these coordinate systems to worldng in rectangular coordinates? 9. How are triple integrals in cylindrical and sphetical coordinates evaluated? How are the limits ofintegration found? Give examples.
10. How are substitutions in double integrals pictured as transformations of twodimensional regions? Give a sample calculation.
11. How are substitutions in triple integrals pictured as transformations of threedimensional regions? Give a sample calculation.
6. How are triple integrals in rectangular coordinates evaluated? How are the limits of integration determined? Give an example.
Chapter
Practice Exercises
Evaluating Double Iterated Integrals
Areas and Volumes Using Double Integrals
In Exercises 14, sketch the region of integration and evaluate the double integral.
13. Area between line and parabola Find the area of the region enclosed by the line y ~ 2x + 4 and the parabola y ~ 4  x 2 in the xyplane.
['0 {IIY
1.
3.
JI Jo
/,o
ye" fix dy
,/2:£v9=4t' tdsdt V94t 2
2./,1/," 4.
/,o
eY/r dyfix
IJ2v,; xyfixdy v,;
In Exercises 58, sketch the region of integration and write an equivalent integral with the order of integration reversed. Then evaluate both integrals. 5.
7.
4:£(;;4)/2 /,o V4=Y fixdy /,o
,/2:£V9=4Y' yfixdy V94,,'
6.
{".Ix dy fix Jo lx ['
2
{2
8.
rr'
Jo Jo
2xdyfix
9. [' {'4
h~
11.
/, sh o
2
dyfix ~ y4 + 1
fix dy
10.
12.
{2 (I er' fix dy h~
/,'h'
o"o/Y
angular" region in the xyplane that is bounded on the right by the parabola y = x 2 , on the left by the line x + y = 2, and above by the line y ~ 4. 15. Volume of the region under a paraboloid Find the volume under the paraboloid z = x 2 + y2 above the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xyplane.
16. Volume of the region under parabolic cylinder Find the volume under the parabolic cylinder z = x 2 above the region enclosed by the parabola y ~ 6  x 2 and the line y ~ x in the xyplane.
Average Values
Evaluate the integrals in Exercises 912. cos (x 2 )
14. Area bonnded by lines and parabola Find the area of the ''tri
2'If sm'lfX . 2 fix dy x2
Find the average value of f(x, y) and 18.
= xy over the regions in Exercises 17
17. The square bounded by the lines x quadrant
~
1, Y
~
1 in the first
18. The quarter circle x 2 + y2 '" 1 in the Itrst quadnlnt
Chapter 15 PoLar Coordinates Evaluate the integrals in Exercises 19 and 20 by changing to polar coordinates.
19. 20.
1
11~
1 1
b. First quadrant
The triangle with vertices (0, 0), (1, 0),
The first quadrant of the xyplane.
24.
0
2
l
1
0
32. Rectangular to cylindrical coordinates (a) Convert to cylindrical coordinates. Then (b) evaluate the new integral.
11~ ~
1
VoLumes and Average VaLues Using TripLe IntegraLs 27. Volume Find the volume of the wedgeshaped region enclosed on the side by the cylinder x = cosy, 7r/2 ~ Y ~ 7r/2, on the top by the plane z =  2x, and below by the xyplane.
2
(x
Jv'x 2+y2
dzdydx
34. Rectangular, cylindrical, and spherical coordinates Write an iterated triple integral for the integral of j(x, y, z) = 6 + 4yover the region in the first octant bounded by the cone z = v'x 2 + y2 , the cylinder x 2 + y2 = 1, and the coordinate planes in (a) rectangular coordinates, (b) cylindrical coordinates, and (c) spherical coordinates. Then (d) find the integral of j by evaluating one of the triple integrals. 35. Cylindrical to rectangular coordinates Set up an integral in rectangular coordinates equivalent to the integral
{7T/2 {v'3 (vi4=?
io il il =
2
+y ) 21xy2 dz dy dx _(x 2+y2)
1
Z
x
r ~ 0
to (a) rectangular coordinates with the order of integration dz dx dy and (b) spherical coordinates. Then (c) evaluate one of the integrals.
e ~ 1 11~
e(x+y+z) dz dy dx
In4
eix1z2Y 3dydzdx 1
3 dz r dr dO,
r
33. Rectangular to spherical coordinates (a) Convert to spherical coordinates. Then (b) evaluate the new integral.
cos (x + y + z) dx dy dz
{I {X {x+Y 25. Jo Jo Jo (2x  y  z) dz dy dx 26.
Convert
{27T (Vijvl4=? Jo Jo
o
7T
ln71ln2llns
In6
of j(x, y, z) =
by the coordinate planes and the planes x = 1, y = 3, z = 1.
1
EvaLuating TripLe Iterated Integrals Evaluate the integrals in Exercises 2326.
111 1
value
CyLindricaL and SphericaL Coordinates 31. Cylindrical to rectangular coordinates
+ y2 + 1) dx dy
a. Triangular region and (1, \13).
23.
average
2
21. Integrating over lemniscate Integrate the function j(x, y) = 1/(1 + x 2 + y2)2 over the region enclosed by one loop of the lemniscate (x 2 + y2)2  (x 2  y2) = O. 22. Integrate j(x,y) = 1/(1 + x 2 + y2)2 over
7T
value Find the
w+y over the rectangular solid in the first octant bounded
30. Average value Find the average value of p over the solid sphere p ~ a (spherical coordinates).
1
7T
30xz
897
2dydx
_~ (1 + x 2 + y2)2
l 1V!=.? In (x V!=.?
29. Average
Practice Exercises
3
r (sin 0 cos O)z2 dz dr dO.
Arrange the order of integration to be z first, then y, then x.
cosy
36. Rectangular to cylindrical coordinates The volume of a solid is
1
o
1T
x
2
y
28. Volume Find the volume of the solid that is bounded above by the cylinder z = 4  x 2, on the sides by the cylinder x 2 + y2 = 4, and below by the xyplane. z
21~1v'4~Y2 0
v'4x 2_y2
dz dy dx.
a. Describe the solid by giving equations for the surfaces that form its boundary. b. Convert the integral to cylindrical coordinates but do not evaluate the integral.
37. Spherical versus cylindrical coordinates Triple integrals involving spherical shapes do not always require spherical coordinates for convenient evaluation. Some calculations may be accomplished more easily with cylindrical coordinates. As a case in point, find the volume of the region bounded above by the sphere x 2 + y2 + z2 = 8 and below by the plane z = 2 by using (a) cylindrical coordinates and (b) spherical coordinates. Masses and Moments 38. Finding I z in spherical coordinates Find the moment of inertia about the zaxis of a solid of constant density S = 1 that is bounded above by the sphere p = 2 and below by the cone cf> = 7r/3 (spherical coordinates).
898
Chapter 15: Multiple Integrals
39. Moment orinertia of a "thick" sphere Find the moment of inertia of a solid of constant density 8 bounded by two concentric spheres ofradii a and b (a < b) about a diameter.
40. Moment of inertia of an apple Find the moment of inertia about the zaxis of a solid of density 8 ~ I enclosed by the spberical coordinate surface p = I  cos t/> • The solid is the red curve rotated about the zaxis in the accompanying figure.
z
bounded by the line y = x and the parabola y = x 2 in the xyplane ifthe density is 8(x, y) = x + I . 47. Plate with variable density Find the mass and first moments about the coordinate axes of a thin square plate bounded by the linesx ~ ± I, Y ~ ± I in the xyplane if the density is 8(x, y) ~ x 2 + y2 + 1/3. 48. Triangles with same inertial moment Find the moment of inertia about the xaxis of a thin triangular plate of constant density 8 whose base lies along the interval [0, b] on the xaxis and whose vertex lies on the line y = h above the xaxis. As you will see, it does not matter where on the line this vertex lies. All such triangles have the same moment of inertia about the xaxis. 49. Centroid Find the centroid ofthe region in the polar coordinate plane defined by the inequalities 0 :5 r :5 3, 7f/3 :5 9 :5 7f/3. 50. Centroid Find the centroid of the region in the firat quadrant bounded by the rays 9 = 0 and 9 = 7f/2 and the circles r = I andr ~ 3.
51. a. Centroid Find the centroid of the region in the polar coordinate plane that lies inside the cardioid r = 1 + cos (} and outside the circle r ~ 1. 41. Centroid Find the centroid of the "triangular" region bounded by the lines x = 2, Y = 2 and the hyperbola xy = 2 in the xyplane. 42. Centroid Find the centroid of the region between the parabola x + y2  2y = 0 and the line x + 2y = 0 in the xyplane. 43. Polar moment Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 = 3 bounded by the yaxis and the lines y = 2x and y = 4 in the xyplane. 44. Polar moment Find the polar moment of inertia about the center of a thin rectangular sheet of constant density 8 = I bounded by the lines
a. x = ±2, y = ± I in the xyplane ~ ±a, y ~ ±b in the xyplane.
b. x
(Hint: Find Ix. Then use the formula for Ix to fmd Iy and add the two to find 10). 45. Inertial moment Find the moment of inertia about the xaxis of a thin plate of constant density 8 covering the triangle with vertices (0, 0), (3, 0), and (3, 2) in thexyplane. 46. Plate with variable density Find the center of mass and the moments of inertia about the coordinate axes of a thin plate
Chapter
b. Sketch the region and show the centroid in your sketch.
52. a. Centroid Find the centroid of the plane region defined by the polar coordinate inequalities 0 ~ r ~ a, a ::=:; (J ~ a (0 < a::;;
'7T).Howdoesthecentroidmoveasa~'7T?
b. Sketch the region fora ~ 57f/6 and show the centroid in your sketch.
Substitutions 53. Show that if u = x  y and v = y, then
54. What relationship must hold between the constants a, b, and c to mske
(Hint: l.et s = ax + ,By and t = yx + 8y, where (a8  ,By)' ac  b 2. Thenax2 + 2bxy + cy2 = 8 2 + t 2.)
=
Additional and Advanced Exerdses
Volumes 1. Sand pile: double and triple integrals The base of a sand pile covers the region in the xyplane that is bounded by the parabola x 2 + y = 6 and the line y = x . The height of the sand above the point (x, y) is x 2 • Express the volume of sand as (a) a double integral, (b) a triple integral. Then (c) fmd the volume.
3. Solid cylindrical region between two plane. Find the volume ofthe portion of the solidcylinderx 2 + y2:5 I that lies between the planes z ~ o and x + y + z ~ 2.
2. Water in a hemispherical bowl A hemispherical bowl of radius 5 em is filled with water to within 3 em of the top. Find the volume of water in the bowl.
5. Two paraboloids Find the volume of the region bounded above by the paraboloid z = 3  x 2  y2 and below by the paraboloid z ~ 2x 2 + 2y 2.
4. Spbere and paraboloid Find the volume of the region bounded above by the sphere x 2
loidz
= xl
+ y2 + z2
= 2 and below by the parabo
+ y2,
Chapter 15 Additional and Advanced Exercises 6. Spherical coordinates Find the volume of the region enclosed by the spherical coordinate surface p = 2 sin c/J (see accompanying figure).
p=2sinq,
Similarly, it can be shown that
ll
x
V
u
em(xt) f(t) dt du dv = lx (x  t)2 em(xt) f(t) dt. 2 l0 0 0 0
14. Transforming a double integral to obtain constant limits Sometimes a multiple integral with variable limits can be changed into one with constant limits. By changing the order of integration, show that y
1 1
X
f(x) ( l g(x  y)f(y) dY ) dx
1 (1 1
=
7. Hole in sphere A circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is
=
1
f(y)
g(x  y)f(x) dx) dy
Ire Jo
g( Ix  yl )f(x)f(y) dx dy.
"2Jo
Masses and Moments
[2" [V3 [~
Jo Jo Jl
V= 2
899
r dr dz dO.
a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral.
8. Sphere and cylinder Find the volume of material cut from the solid sphere r 2 + z2 ::::; 9 by the cylinder r = 3 sin o. 9. Two paraboloids Find the volume ofthe region enclosed by the surfaces z = x 2 + y2 and z = (x 2 + y2 + 1)/2.
15. Minimizing polar inertia A thin plate of constant density is to occupy the triangular region in the first quadrant of the xyplane having vertices (0, 0), (a, 0), and (a, l/a). What value of a will minimize the plate's polar moment of inertia about the origin?
16. Polar inertia of triangular plate Find the polar moment of inertia about the origin of a thin triangular plate of constant density 8 = 3 bounded by the yaxis and the lines y = 2x and y = 4 in the xyplane.
10. Cylinder and surface z = xy Find the volume of the region in the first octant that lies between the cylinders r = I and r = 2 and that is bounded below by the xyplane and above by the surfacez = xy.
17. Mass and polar inertia of a counterweight The counterweight of a flywheel of constant density I has the form of the smaller segment cut from a circle of radius a by a chord at a distance b from the center (b < a). Find the mass of the counterweight and its polar moment of inertia about the center of the wheel.
Changing the Order of Integration
18. Centroid of boomerang Find the centroid of the boomerangshaped region between the parabolas y2 = 4(x  I) and y2 = 2(x  2) in thexyplane.
11. Evaluate the integral
1
00
eax 
o
dx. Theory and Examples
(Hint: Use the relation
e ax  e bx x
ebx
x
=
l
19. Evaluate b
eXYdy
a
to form a double integral and evaluate the integral by changing the order of integration.)
12. a. Polar coordinates that
l
o
Show, by changing to polar coordinates,
aSinf3j~ In(x 2 + y2)dxdy
(I) "2 '
= a 2{3 Ina
ycotf3
where a
where a and b are positive numbers and
20. Show that
> 0 and 0 < (3 < 'TT' /2.
b. Rewrite the Cartesian integral with the order of integration reversed.
over the rectangle Xo ::::; x ::::; xl, Yo ::::; Y ::::; y[, is
13. Reducing a double to a single integral By changing the order of integration, show that the following double integral can be reduced to a single integral: l
x l
u
em(xt) f(t) dt du = l
x (x  t)em(xt) f(t) dt.
21. Suppose that f(x,y) can be written as a product f(x, y) = F(x)G(y) of a function of x and a function of y. Then
900
Chapter 15: Multiple Integrals
the integral off over the rectangle R: a
~ x ~ b, c be evaluated as a product as well, by the formula
ff
dx)
f(x,y) dA = ( [ F(x)
(t
~
y
~
d can
a. If you have not yet done Exercise 41 in Section 15.4, do it now to show that
(1)
G(y) dy).
R
The argument is that
ii
f(x,y) dA
~
R
= =
t ([ dx) t (G(Y{ dx) t ([ dx )G(Y) F(x)G(y)
F(x)
b. Substitute y = Vi in Equation (2) to show that f( 1/2) = 2I = Yii.
dy
(i)
dy
F(x)
dy
(ii) (iii)
= ([F(X) dx)tG(y) dy.
(iv)
a. Give reasons for steps (i) through (iv).
24. Total electrical charge over circular plate The electrical charge distribution on a circular plate of radius R meters is O'(r, 0) = kr(1  sinO) coulomb/m2 (k a constant). Integrate 0' over the plate to fmd the total charge Q. 25. A parabolic rain gauge A bowl is in the shape of the graph of z = x 2 + y2 from z = 0 to z = 10 in. You plan to calibrate the bowl to make it into a rain gauge. What height in the bowl would correspond to 1 in. ofrain? 3 in. ofrain? 26. Water in a satemte disb A parabolic satellite dish is 2 m wide and 1/2 m deep. Its axis of symmetry is tilted 30 degrees from the vertical.
When it applies, Equation (I) can be a timesaver. Use it to evaluate the following integrals. b.
{1n2 {~/2 eX cosy dy dx
c.
Jo Jo
hold. (Hint: Put your coordinate system so that the satellite dish is in "standsrd position" and the plane of the water level is slanted) (Caution: The limits of integration are not "nice.")
{2 (I x dx dy
11
iI Y2
22. Let Duf denote the derivative of f(x,y) = (x 2 direction ofthe unit vector u = "1 i + U2j.
+ y2)/2
in the
a. Finding average value Find the average value of Duf over the triangular region cut from the fIrSt quadrant by the line x+y=l. b. Average value and centroid Show in general that the average value of Duf over a region in the >;yplane is the value of D.f at the cenlroid of the region. 23. The value ofr(1/2)
a. Set up, but do not evaluate, a triple integral in rectangular coordinates that gives the amount ofwaler the satellite dish will
b. What would be the antallest tilt ofthe satellite dish so that it holds no water? 27. An infinite balfeylinder Let D be the interior of the infmite right circular halfcylinder ofradius I with its singleend face suspended 1 unit above the origin and its axis the ray from (0, 0, I) to 00. Use cylindrical coordinates to evaluate
iff
The gamma function,
1 ,..1
z(r 2
+ z 2 t'/2 dV.
D
00
f(x) =
et dt,
extends the factorial function from the noonegative integers to other real values. Of particular interest in the theory of differential equations is the number
Chapter
J: dx
28. Hypenolnme We have learned that 1 is the length of the interval [a, b] on the number line (onedimensional space), !fR 1 dA is the area ofregionR in thexyplane (two
Technology Application Projects
Mathematica/Maple Module: 1lIke Your Chances: 1l'y the MOllte Carlo Techllique for Numerical llItegratioll ill Three Dimemioll' Use the Monte Carlo technique to integrate numerically in three dimensions. Means alld MomeIrts and Exploring New Plotting Techlliques, Part II Use the method ofmoments in a form that makes use of geometric symmetry as well as multiple integration.
I1ff
16 INTEGRATION IN VECTOR FIELDS OVERVIEW In this chapter we extend the theory of integration to curves and surfaces in space. The resulting theory of line and surface integrals gives powerful mathematical tools for science and engineering. Line integrals are used to find the work done by a force in moving an object along a path, and to find the mass of a curved wire with variable density. Surface integrals are used to find the rate of flow of a fluid across a surface. We present the fundamental theorems of vector integral calculus, and discuss their mathematical consequences and physical applications. In the final analysis, the key theorems are shown as generalized interpretations of the Fundamental Theorem of Calculus.
16.1
Line Integrals
z
To calculate the total mass of a wire lying along a curve in space, or to find the work done by a variable force acting along such a curve, we need a more general notion of integral than was defined in Chapter 5. We need to integrate over a curve C rather than over an interval [a, b]. These more general integrals are called line integrals (although path integrals might be more descriptive). We make our definitions for space curves, with curves in the xyplane being the special case with zcoordinate identically zero. Suppose that I(x, y, z) is a realvalued function we wish to integrate over the curve C lying within the domain of I and parametrized by r(t) == g(t)i + h(t)j + k(t)k, a :::; t :::; b. The values of I along the curve are given by the composite function I(g(t), h(t), k(t)). We are going to integrate this composite with respect to arc length from t == a to t == b. To begin, we first partition the curve C into a finite number n of subarcs (Figure 16.1). The typical subarc has length J).Sk. In each subarc we choose a point (Xh Yk, Zk) and form the sum n
Sn
== 'LI(Xk, Yk, Zk) J).Sk, k=l
x
FIGURE 16.1 The curve r(t) partitioned into small arcs from t = a to t = b. The length of a typical subarc is aSk.
which is similar to a Riemann sum. Depending on how we partition the curve C and pick (Xh Yh Zk) in the kth subarc, we may get different values for Sn. If I is continuous and the functions g, h, and k have continuous first derivatives, then these sums approach a limit as n increases and the lengths J).Sk approach zero. This limit gives the following definition, similar to that for a single integral. In the definition, we assume that the partition satisfies J).Sk~ 0 as n ~ 00.
If I is defined on a curve C given parametrically by ret)
DEFINITION g(t)i
+
h(t)j
+
==
k(t)k, a :::; t :::; b, then the line integral of f over Cis
Jcr !(x, y, z) ds =
lim ±!(Xk'Yk, Zk) Llsk,
n~OO
(1)
k=l
provided this limit exists.
901
902
Chapter 16: Integration in Vector Fields If the curve C is smooth for a :5 t:5 b (so v = dr/dt is continuous and never 0) and the function f is continuous on C, then the limit in Equation (1) can be shown to exist. We can then apply the Fundamental Theorem of Calculus to differentiate the arc length equation,
l'
s(t) =
Eq. (3) of Section 13.3
IV(T) I dT,
withto=a
to express tis in Equation (1) as tis = Iv(t) I dt and evaluate the integral oflover C as
l
!c1(x,y,z)dS =
b
I(g(t), h(t), k(t» Iv(t) I dt.
(2)
Notice that the integral on the right side of Equation (2) is just an ordinary (single) definite integral, as defined in Chapter 5, where we are integrating with respect to the parameter t. The formula evaluates the line integral on the left side correctly no matter what parametrization is used, as long as the parametrization is smooth. Note that the parameter t defines a direction along the path. The starting point on C is the position r(a) and movement along the path is in the direction of increasing t (see Figure 16.1).
How to Evaluate a Line Integral To integrate a continuous function I(x, y, z) over a curve C: 1. Find a smooth parametrization of C,
r(t) = g(t)i
+ h(t)j + k(t)k,
a
:5
t
:5
b.
2. Evaluate the integral as
!c I(x,y, z) tis =
t
=
l
b
I(g(t), h(t), k(t))lv(t) I dt.
If 1 has the constant value 1, then the integral of 1 over C gives the length of C from = b in Figure 16.1.
a to t
EXAMPLE 1 Integrate I(x,y,z) = x  3y 2 origin to the point (1, 1, 1) (Figure 16.2).
+ z over the line segment C joining the
(I, I, 1) I I I I
: cA~
\ \ \ \ \
X
FIGURE 16.2
r(t) = ti +
)oy
I I I I I I
~
(1, 1,0)
Example I.
We choose the simplest parametrization we can think of:
Solution
ti +
tk,
The components have continuous first derivatives and Iv(t)1 = Ii 2
2
V1 + 1 over Cis
2
+1
=
+ i + kl
=
V3 is never 0, so the parametrization is smooth. The integral of 1
The integration path in
!c I(x,y, z) ds =
=
=
l' l'
I(t, t, t)( V3) dt
(t  3t 2
V3.Jo{' (2t 
Eq. (2)
+ t)V3 dt 3t
2
)
dt =
V3 [t 2 
1
t 3 ]0 =
o.
•
16.1
Line Integrals
903
Additivity Line integrals have the useful property that if a piecewise smooth curve C is made by joining a finite number of smooth curves C" C2 , ••• , Cn end to end (Section 13.1), then the integral of a function over C is the sum ofthe integrals over the curves that make it up:
{Ids= {Ids+ {Ids+ ... + (Ids.
J~
Jc (1, 1, 1)
J~
(3)
J~
EXAMPLE 2 Figure 16.3 shows another path from the origin to (I, I, I), the union of line segments CI and C2. Integrate I{x,y, z) = x  3y 2 + Z over C, U C2. Solution We choose the simplest parametrizations for C, and C2 we can find, calculating the lengths ofthe velocity vectors as we go along:
+ Ij, 0:5 I :5 I; Ivl = VI 2 + 12 = Vi i + j + Ik, 0:5 1:5 I; Ivl = V02 + 02 + 12 = 1.
C,:
r{l) = Ii
C2:
r{l) =
With these parametrizations we fmd that FIGURE 16.3
Example 2.
The path afinlegration in
{
I{x,y,z)ds
=
}C,UC2
=
=
=
{
lei
l' l'
I{x,y,z)ds + ( I(x,y,z)ds
Eq. (3)
JC2
l'
I{I, I, O)Vi dl +
I{I, 1,1)(1) dl
{I  31 2 + O)Vidl +
Vi [1
2
2
l'
2
_ / 3 ]1
+ [1  21]1
0
0
2
(I  3
= _
Eq. (2)
+ 1)(I)dl
Vi _ 1. 2
2
•
Notice three things about the integrations in Examples I and 2. First, as soon as the components ofthe appropriate curve were substituted into the fannula for I, the integration became a standard integration with respect to I. Second, the integral oflover c, U C2 was obtained by integrating I over each section ofthe path and adding the results. Third, the integrals oflover C and C, U C2 had different values.
The value of the line integral along a path joining two points can change if you change the path between them.
We investigate this third observation in Section 16.3.
Mass and Moment Calculations We treat coil springs and wires as masses distributed along smooth curves in space. The distribution is described by a continuous density function Il(x, Y, z) representing mass per unit length. When a curve C is parametrized by r{l) = X{I)i + Y{I}j + Z{I)k, a :5 I :5 b, thenx,y, andz are functions ofthe parameter I, the density is the function Il{X{I), y(1), z{I», and the arc length differential is given by
ds=~(:Y + (itY + (~Ydl.
904
Chapter 16: Integration in Vector Fields (See Section 13.3.) The spring's or wire's mass, center of mass, and moments are then calculated with the formulas in Table 16.1, with the integrations in terms of the parameter I over the interval [a, b]. For example, the formula for mass becomes
M =
t
8(x(I),y(I), z(I))
~ ( : ) ' + (~)' + (:)' dl.
These formulas also apply to thin rods, and their derivations are similar to those in Section 6.6. Notice how alike the formulas are to those in Tables 15.1 and 15.2 for double and triple integrals. The double integrals for planar regions, and the triple integrals for solids, become line integrals for coil springs, wires, and thin rods.
Mass and moment formuLas for coil springs, wires, and thin rods lying aLong a smooth curve Cin space
TABLE 16.1
Mass:
M =
fc 8 tis
B
~ B(x, y, z) is the density ,,(x, y, z)
First moments about the coordinate pIanes:
Myz = fc x 8 tis,
Mxy = fc z 8 tis
Coordinates of the center of mass:
y
= Mxz/M,
Moments of inertia about axes and other lines:
Ix
=
fc (y2
+ z2)8 tis,
h= fcr 28dS
r(x,y, z) = distance from the point (x,y, z) to line L
Notice that the element of mass dm is equal to 8 tis in the table rather than 8 dV as in Table 15.1, and that the integrals are taken over the curve C.
EXAMPLE 3 A slender metal arch, denser at the bottom than top, lies along the semicircle y2 + z2 = I, z ;" 0, in theyzplane (Figure 16.4). Find the center ofthe arch's mass if the density at the point (x, y, z) on the arch is 8(x, y, z) = 2  z.
z
c.m.
1
Solution We know that x = 0 and y = 0 because the arch lies in the yzplane with its mass distributed symmetrically about the zaxis. To find z, we parametrize the circle as r(l) = (cos l)j
FIGURE 16.4 Example 3 shows how to find the center ofmass of a circular arch of
variabledeosity.
+
(sin I)k,
o :5 I
:5 1f.
For this parametrization,
IV(I)I = so tis =
Ivldl =
(:)'
dl.
+
(~)' +
(:)'
=
Y(Of
+
(sinl)2
+
(cos 1)2 = I,
16.1
905
Line Integrals
The formulas in Table 16.1 then give M= fc8ds= fc(ZZ)ds= !."(ZSint)dt=Z1TZ Mxy = fc z8 ds = fc z(Z  z) ds = !."(Sint)(Z  sint) dt =
!."(ZSint  sin2 t) dt
_ Mxy Z= M
=
=
8  1T I Z'Z1T Z
8;
1T
8 41T
1T
4'" 0.57.
With z to the nearest hundredth, the center ofmass is (0, 0, 0.57).
•
Line Integrals in the Plane There is an interesting geometric interpretation for line integrals in the plane. If C is a smooth curve in the :>ryplane parametrized by r(t) = x(t)i + y(t)j, a ,;; t ,;; b, we generate a cylindrical surface by moving a straight line along C orthogonal to the plane, holding the line parallel to the zaxis, as in Section IZ.6. Ifz = f(x,y) is a nonnegative continuous function over a region in the plane containing the curve C, then the graph of f is a surface that lies above the plane. The cylinder cuts through this surface, forming a curve on it that lies above the curve C and follows its winding nature. The part of the cylindrical surface that lies beneath the surface curve and above the :>ryplane is like a "winding wall" or "fence" standing on the curve C and orthogonal to the plane. At any point (x, y) along the curve, the height ofthe wallis f(x, y). We show the wall in Figure 16.5, where the "top" of the wall is the curve lying on the surface z = f(x, y). (We do not display the surface formed by the graph of f in the figure, ouiy the curve on it that is cut out by the cylinder.) From the definition
heightj(x, y)
y
t
=b
FIGURE 16.5 The line integral
Ie f
tis
gives the area ofthe portion ofthe cylindrical surface or "wall" beneath z = f(x,y) '" O. where !isk~ 0 as n ~ shown in the figure.
00,
we see that the line integral Ief ds is the area of the wall
Exercises 16.1 Graphs of Vector Equations
Match the vector equations in Exercises 18 with the graphs (a)(h) given here. a.
c.
b.
z
z
d.
( 2,2,2)
2
1
' r2M z
z
,
1
y
y
2
x
y
2
:~y
~~
I
I
IJ,/
/
/
906
Chapter 16: Integration in Vector Fields
e.
z
f.
z 2 (0,0, 1)
r""
1~ __
I
x
(0,0,0) ~
1
(1, 1, 1)
(0,0,0)
~y
~
y x
y
2
(1,1,1)
(b)
(a)
h.
g.
The paths of integration for Exercises 15 and 16.
z
z
16. Integrate f(x, y, z) = x + vY  z' over the path from (0, 0, 0) to (I, I, I) (see accompanying figure) given by
y
~
+ (I
Ii
 l)j,
+ j + Ik,
2. r(l) = I
3. r(l) = (2 cos 1)1 4. r(l)
~
Ii,
5. r(l)
~
Ii
~
0:5 I :5 I
20. Evaluate
Ie f
11. Evaluate e (xy + Y + z) tis along the curve r(l) ~ 21i Ij + (2  21)k, 0 :5 I :5 1.
+
Ie \Ix' + y' tis along the curve r(l) ~ (4 cos 1)1 + + 31k, 2". :5 I :5 2"..
13. Find the line integral of f(x,y, z) = x + Y line segment from (I, 2, 3) to (0, I, I).
2'17".
21. Find the line integral of f(x,y) ~ ye'" along the curve r(l) = 4ti  3tj, I :5 I :5 2. 22. Find the line integral of f(x, y) = x  y r(l) ~ (cos 1)1 + (sin I)j, 0 :5 I :5 2".. 23.
Evaluatel:~3 tis, eY
+ I'j,
I + j + Ik,
2
0:5 I :5 I
3
,
for
1 :s; t :s; 2.
24. Find the line integral of f(x,y) ~ 0/x along the curve r(l) = 1'1 + I'j, 1/2 :5 I :5 1. 25. Evaluate figure.
f e(x + 0) tis where C is given in the accompanying y
+ y' + z') over
__ x
=='~=~
0:5 I :5 I
+ 3 along the curve
where C is the curve x = t , Y = t
15. Integrate f(x, y, z) ~ x + vY  z' over the path from (0, 0, 0) to (I, I, I) (see accompanying figure) given by
~
~
Ie \Ix + 2y tis, where Cis
+ z over the straight
14. Find the line integral of f(x,y, z) = 'II3/(x' the curve r(l) ~ II + Ij + Ik, I :5 I :5 00.
r(l)
o~ t
h. C1 U C,; C1 is the line segment from (0, 0) to (I, 0) and C, is the line segment from (I, 0) to (I, 2).
".
(x  y + z  2) tis where C is the straightline seg10. Evaluate mentx ~ I,y ~ (I  I),z ~ I, from (0, I, l)to(I,O, I).
C,:
+ (a sin I)k,
.. the straightline segment x = I, Y = 41, from (0, 0) to (I, 4).
x ~ I,y ~ (I  I),z ~ 0, from (0, I, 0) to (I, 0, 0).
r(l) ~ II
+ z'overthecircle
h. the parabolic curve x ~ I,y ~ I', from (0,0) to (2, 4).
Evaluating Line Integrals over Space Curves 9. Evaluate Ie (x + y) tis where C is the straightline segment
Cl:
0:5 I :5 I
Ii
.. the straightline segment x = I, Y = 1/2, from (0, 0) to (4, 2).
0:5 I :5 2
+ 21k, 1:5 1:5 I (2 cos 1)1 + (2 sin I)k, 0:5 I :5
12. Evaluate (4 sin I)j
r(l)
Line Integrals over Plane Curves 19. Evaluate fex tis, where Cis
0:5 I :5 2".
7. r(l) = (I'  I)j 8. r(l) =
r(l) = Ij + k,
r(l) = (a cos I)j
1:5 I :5 I
6. r(l) = Ij + (2  21)k,
C,:
18. Integratef(x,y,z) ~ \Ix'
1:5 I :5 I
+ Ij + Ik,
0:5 I :5 I
+ j + k, 0:5 I :5 I 17. Integratef(x,y,z) = (x + Y + z)/(x' + y' + z') over the path r(l) ~ II + Ij + Ik, 0 < a :5 I :5 b.
0:5 I :5 1
+ (2 sin I)j,
r(l) = Ik,
C,:
y
1. r(l)
C1:
(0,0)
16.2
26. Evaluate
1
2
eX
1
+ y2 + I
tis where C is given in the accompanying
figure.
yzplane. Find the moments of inertia of the rod about the three coordinate axes. 39. Two springs of constant density A spring of constant density 8 lies along the helix
y
o :s t :s 2w.
r(t) = (cos t)1 + (sin t)j + tk, (0, I)'+f~(I, I)
a. Find /z'
:c::;++:04:::~x
(0,0)
(1,0)
In Exercises 2730, integrate! over the given curve.
27. !(x,y) = x 3/y, C: y = x 2/2, 0 '" x'" 2 28. !(x,y) = (x + y2)/vT+7, C: y = x 2/2 from (I, 1/2) to (0,0)
29. !(x,y) = x + y,
C:
907
Vector Fields and Line Integrals: Work, Circulation, and Flux
x 2 + y2 = 4 in the first quadrant from
(2, 0) to (0, 2)
30. !(x,y) = x 2  y, C: x 2 + y2 = 4 in the first quadrant from (0, 2) to (v'2, v'2) 31. Find the area of one side of the "winding wall" standing orthogonally on the curve y = x 2• 0 :s x :s 2. and beneath the curve on the surface !(x, y) = x +
b. Suppose that you have another spring of constant density 8 that is twice as long as the spring in part (a) and lies along the helix for 0 '" t '" 4". Do you expect I. for the longer spring to be the same as that for the shorter one, or should it be different? Check your prediction by calcularing I. for the longer spring.
40. Wire of constant density A wire of constant density 8 along the curve
r(t) ~ (tcost)1 + (tsint)j + (2V2/3)t 312k,
Masses and Moments 33. Mass of a wire Find the mass of a wire that lies along the curve r(t) ~ (12  I)j + 2tk, 0 '" t '" I, if the density is 8 ~ (3/2)t. 34. Center of mass of a euned wire A wire of density 8(x,y, z) = IS'\1'.Y+2 lies along the curve r(t) = (t 2  I)j + 2tk, I '" t '" I. Find its cenlet of mass. Then sketch the curve and cenlet ofmass together.
0'" t '" I.
Find. and I•. 41. The arch in Eilimple 3
Find I. for the arch in Example 3.
42. Center of mass and moments of inertia for wire with variable density Find the cenlet of mass and the moments of inertia about the coordinate axes of a thin wire lying along the curve
r(t) = tl + 2 r t 312j +
v'Y.
32. Find the area of one side of the "wall" standing orthogona1ly on the curve 2x + 3y = 6, 0 '" x '" 6, and beneath the curve on the surface !(x, y) = 4 + 3x + 2y.
= I lies
if the density is 8
= I/(t +
~ k,
0 '" t '" 2,
I).
COMPUTER EXPLORATIONS In Exercises 4346, use a CAS to perfOt03 the following steps to evaluate the line integrals. a. Find tis = Iv(t) I dtforthepathr(t) = g(t)i
+ h(t)j +
k(t)k. b. Express the integrand !(g(t), h(t), k(t)) Iv(t) I as a function of the parameter t.
e. Evaluate
Ie ! tis using Equation (2) in the text
35. Mass of wire with variable density Find the mass of a thin wire lying along the curve r(t) ~ V2ti + V2tj + (4  t 2)k, o'" t '" 1, if the density is (a) 8 = 3t and (b) 8 = I.
43. !(x,y,z) =
36. Center of mass of wire with variable density Find the cenlet of mass of a thin wire 1ying along the curve r(t) ~ ti + 2tj + (2/3)t 312k, 0 '" t '" 2, ifthe density is 8 = 3'V"5+t.
44. !(x,y,z) =
37. Moment of Inertia of wire hoop A circulat wire hoop of constant density 8 lies along the circlex 2 + y2 ~ a 2 in thexyplane. Find the hoop's moment of inertia about the zaxis.
45. !(x,y,z) ~ xyY  3z 2 ; r(t) ~ (cos2t)1 + (sin2tJi + Stk, o :5 t :5 2w
38. Inertia of a slender rod A slender rod of constant density lies along the line segment r(t) ~ tj + (2  2t)k, 0 '" t '" I, in the
16.2
r(t) = ti + t 2j + 3t 2k,
VI
+ 3Ox 2 +
VI
+ x 3 + Sy3; r(t)
lOy;
0:51:52
= ti
+
tt
2
j
+ Vlk,
0:51:52
46. !(x,y,z) t'12k,
~
(I
+ ~zl/3
)'/4;
r(t)
~ (cos2t)1 + (sin2t)j +
0 '" t '" 2"
Vector Fields and Line Integrals: Work, Circulation, and Flux Gravitational and electric forces have both a direction and a magnitude. They are represented by a vector at each point in their domain, producing a vector field. In this section we show how to compute the work done in moving an object through such a field by using a line integral involving the vector field. We also discuss velocity