View
219
Download
1
Category
Preview:
Citation preview
Multiple Integration14
Copyright © Cengage Learning. All rights reserved.
Triple Integrals in Cylindrical and Spherical Coordinates
Copyright © Cengage Learning. All rights reserved.
14.7
3
Write and evaluate a triple integral in cylindrical coordinates.
Write and evaluate a triple integral in spherical coordinates.
Objectives
4
Triple Integrals in Cylindrical Coordinates
5
Triple Integrals in Cylindrical Coordinates
The rectangular conversion equations for cylindrical
coordinates are
x = r cos θ
y = r sin θ
z = z.
In this coordinate system, the simplest
solid region is a cylindrical block
determined by
r1 ≤ r ≤ r2, θ1 ≤ θ ≤ θ2, z1 ≤ z ≤ z2
as shown in Figure 14.63.Figure 14.63
6
7
To obtain the cylindrical coordinate form of a triple integral, suppose that Q is a solid region whose projection R onto the xy-plane can be described in polar coordinates.
That is,
Q = {(x, y, z): (x, y) is in R, h1(x, y) ≤ z ≤ h2(x, y)}
and
R = {(r, θ): θ1 ≤ θ ≤ θ2, g1(θ) ≤ r ≤ g2(θ)}.
Triple Integrals in Cylindrical Coordinatesskip
8
If f is a continuous function on the solid Q, you can write
the triple integral of f over Q as
where the double integral over R is evaluated in polar
coordinates. That is, R is a plane region that is either
r-simple or θ-simple. If R is r-simple, the iterated form of
the triple integral in cylindrical form is
Triple Integrals in Cylindrical Coordinatesskip
9
To visualize a particular order of integration, it helps to view the iterated integral in terms of three sweeping motions—each adding another dimension to the solid.
For instance, in the order dr dθ dz, the first integration occurs in the r-direction as a point sweeps out a ray.
Then, as θ increases, the line sweeps out a sector.
Finally, as z increases, the sector sweeps out a solid wedge.
Triple Integrals in Cylindrical Coordinates
10
Example 1 – Finding Volume in Cylindrical Coordinates
Find the volume of the solid region Q cut from the sphere
x2 + y2 + z2 = 4 by the cylinder r = 2 sin θ, as shown in
Figure 14.65.
Figure 14.65
1 radius (0,1),center
circle 1)1(
0112
2
)sin(2
)sin(2
22
22
22
2
yx
yyx
yyx
rr
r
11
Because x2 + y2 + z2 = r2 + z2 = 4, the bounds on z are
Let R be the circular projection of the solid onto the rθ-plane.
Then the bounds on R are 0 ≤ r ≤ 2 sin θ and 0 ≤ θ ≤ π.
1 radius (0,1),center
circle 1)1( 22
yx
439
1643
63
32
3
2
23
32
3
11
23
32
3
)(sin)sin(
23
32
)(sin)cos()sin(23
32
))(sin1)(cos(23
32))(cos1(8
3
4
2/32
)2(4
)(cos4)(sin44)sin(2
24
444
2/
0
32/
0
2/
0
22/
0
2/
0
232/
0
2/
0
4
)(cos4
2/32/
0
)(cos4
4
22
2
2/
0
)sin(2
0
22/
0
)sin(2
0
4
0
2
2
2
d
dd
du
ddu
u
ur
rdrduru
rdrdrdzrdrdr
My solution, more detailed
12
Note!Because of square roots and sin/cos, in this caseit is better to just plug in limits into indefinite integrals using at( )
Different approach to drawing using cylindrical coordinates
13
Note the more complex set-upfor opacity of the surfaces to see what is inside.
14
15
Center of Mass: Moment of Inertia about z - axis:
16
17
18
19
35
512
7
1
5
122
7542)4(2
24on Intersecti
:usejust wouldI n,integratio oforder
thechange todecidedbook why surenot am I
7
2
0
7542
2
0
2
0
2
0
42
2
0
2
0
42
2222
22
kk
rrkdrrrk
dzrdrdkrrdzrdrdr
rryxz
rr
skip
20
Center of Mass: Moment of Inertia about z - axis:
21
Mathematica Implementation
22
Triple Integrals in Spherical Coordinates
Triple integrals over spheres or cones are much easier to evaluate by converting to spherical coordinates.
23
The rectangular conversion equations
for spherical coordinates are
x = ρ sin cos θ
y = ρ sin sin θ
z = ρ sin .
In this coordinate system, the simplest
region is a spherical block determined by
{(ρ, θ, ): ρ1 ≤ ρ ≤ ρ2, θ1 ≤ θ ≤ θ2, 1 ≤ ≤ 2}
where ρ1 ≥ 0, θ2 – θ1 ≤ 2π, and 0 ≤ 1 ≤ 2 ≤ π,
as shown in Figure 14.68.
If (ρ, θ, ) is a point in the interior of such a block,
then the volume of the block can be approximated by
V ≈ ρ2 sin ρ θ
Figure 14.68
Triple Integrals in Spherical Coordinates
24
Using the usual process involving an inner partition,
summation, and a limit, you can develop the following
version of a triple integral in spherical coordinates for a
continuous function f defined on the solid region Q.
Triple Integrals in Spherical Coordinates
As in rectangular and cylindrical coordinates, triple integrals in spherical coordinates are evaluated with iterated integrals.
You can visualize a particular order of integration by viewing the iterated integral in terms of three sweeping motions—each adding another dimension to the solid.
25
Triple Integrals in Spherical Coordinates
For instance, the iterated integral
is illustrated in Figure 14.69.
Figure 14.69
26
Example 4 – Finding Volume in Spherical Coordinates
Find the volume of the solid region Q bounded below by
the upper nappe of the cone z2 = x2 + y2 and above
by the sphere x2 + y2 + z2 = 9, as shown in Figure 14.70.
Figure 14.70
27
Example 4 – Solution
In spherical coordinates, the equation of the sphere is
ρ2 = x2 + y2 + z2 = 9
Furthermore, the sphere and cone intersect when
(x2 + y2) + z2 = (z2) + z2 = 9
and, because z = ρ cos , it follows that
cone z2 = x2 + y2
2
3
3
28
Example 4Consequently, you can use the
integration order dρ d dθ,
where 0 ≤ ρ ≤ 3, 0 ≤ ≤ π/4,
and 0 ≤ θ ≤ 2π.
The volume is
cont’d
2292
2118))cos((18
)sin(3
2)sin(
4/
0
4/
0
3
0
32
0
4/
0
3
0
2
dddd
29
30
31
32
33
34
2
1,
2
1,0,0
:
4
1
2
1
04
1
4
1
222
222
222
rcenter
Sphere
zyx
zzyx
zzyx
35
Recommended