SECTION 12.6
TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
P212.6
POLAR COORDINATES
In plane geometry, the polar coordinate system is used to give a convenient description of certain curves and regions. See Section 9.3
P312.6
POLAR COORDINATES
Figure 1 enables us to recall the connection between polar and Cartesian coordinates. If the point P has Cartesian coordinates (x, y) and
polar coordinates (r, ), then
x = r cos y = r sin
r2 = x2 + y2 tan = y/x
P412.6
CYLINDRICAL COORDINATES
In three dimensions there is a coordinate system, called cylindrical coordinates, that: Is similar to polar coordinates. Gives a convenient description of commonly
occurring surfaces and solids.
As we will see, some triple integrals are much easier to evaluate in cylindrical coordinates.
P512.6
CYLINDRICAL COORDINATES
In the cylindrical coordinate system, a point P in three-dimensional (3-D) space is represented by the ordered triple (r, , z), where: r and are polar coordinates
of the projection of P onto the xy–plane.
z is the directed distance from the xy-plane to P.
See Figure 2.
P612.6
CYLINDRICAL COORDINATES
To convert from cylindrical to rectangular coordinates, we use:
x = r cos y = r sin
z = z
P712.6
CYLINDRICAL COORDINATES
To convert from rectangular to cylindrical coordinates, we use:
r2 = x2 + y2 tan = y/x
z = z
P812.6
Example 1
a) Plot the point with cylindrical coordinates (2, 2/3, 1) and find its rectangular coordinates.
b) Find cylindrical coordinates of the point with rectangular coordinates (3, –3, –7).
P912.6
Example 1(a) SOLUTION
The point with cylindrical coordinates (2, 2/3, 1) is plotted in Figure 3.
P1012.6
Example 1(a) SOLUTION
From Equations 1, its rectangular coordinates are:
The point is (–1, , 1) in rectangular coordinates.
2 12cos 2 1
3 2
2 32sin 2 3
3 2
1
x
y
z
3
P1112.6
Example 1(b) SOLUTION
From Equations 2, we have:
2 23 ( 3) 3 2
3 7tan 1, so 2
3 47
r
n
z
P1212.6
Example 1(b) SOLUTION
Therefore, one set of cylindrical coordinates is
Another is As with polar coordinates, there are infinitely many
choices.
(3 2, / 4, 7)
(3 2, 7 / 4, 7)
P1312.6
CYLINDRICAL COORDINATES
Cylindrical coordinates are useful in problems that involve symmetry about an axis, and the z-axis is chosen to coincide with this axis of symmetry. For instance, the axis of the circular cylinder with
Cartesian equation x2 + y2 = c2 is the z-axis.
P1412.6
CYLINDRICAL COORDINATES
In cylindrical coordinates, this cylinder has the very simple equation r = c.
This is the reason for the name “cylindrical” coordinates.
P1512.6
Example 2
Describe the surface whose equation in cylindrical coordinates is z = r.
SOLUTION The equation says that the z-value, or height, of each
point on the surface is the same as r, the distance from the point to the z-axis.
Since doesn’t appear, it can vary.
P1612.6
Example 2 SOLUTION
So, any horizontal trace in the plane z = k (k > 0) is a circle of radius k.
These traces suggest the surface is a cone. This prediction can be confirmed by converting the
equation into rectangular coordinates.
P1712.6
Example 2 SOLUTION
From the first equation in Equations 2, we have:
z2 = r2 = x2 + y2
We recognize the equation z2 = x2 + y2
(by comparison with Table 1 in Section 10.6) as being a circular cone whose axis is the z-axis.
See Figure 5.
P1812.6
EVALUATING TRIPLE INTEGRALS WITH CYLINDRICAL COORDINATES
Suppose that E is a type 1 region whose projection D on the xy-plane is conveniently described in polar coordinates.
See Figure 6.
P1912.6
EVALUATING TRIPLE INTEGRALS
In particular, suppose that f is continuous and
E = {(x, y, z) | (x, y) D, u1(x, y) ≤ z ≤ u2(x, y)}
where D is given in polar coordinates by:
D = {(r, ) | ≤ ≤ , h1() ≤ r ≤ h2()}
P2012.6
EVALUATING TRIPLE INTEGRALS
We know from Equation 6 in Section 12.5 that:
However, we also know how to evaluate double integrals in polar coordinates.
In fact, combining Equation 3 with Equation 3 in Section 12.3, we obtain the following formula.
2
1
( , )
( , )( , , ) , ,
u x y
u x yE D
f x y z dV f x y z dz dA
P2112.6
Formula 4
This is the formula for triple integration in cylindrical coordinates.
2 2
1 1
( ) cos , sin
( ) cos , sin
, ,
cos , sin ,
E
h u r r
h u r r
f x y z dV
f r r z r dz dr d
P2212.6
TRIPLE INTEGN. IN CYL. COORDS.
It says that we convert a triple integral from rectangular to cylindrical coordinates by: Writing x = r cos , y = r sin . Leaving z as it is. Using the appropriate limits of integration for z, r,
and . Replacing dV by r dz dr d.
P2312.6
TRIPLE INTEGN. IN CYL. COORDS.
Figure 7 shows how to remember this.
P2412.6
TRIPLE INTEGN. IN CYL. COORDS.
It is worthwhile to use this formula: When E is a solid region easily described in
cylindrical coordinates. Especially when the function f(x, y, z) involves the
expression x2 + y2.
P2512.6
Example 3
A solid lies within: The cylinder x2 + y2 = 1 Below the plane z = 4 Above the paraboloid
z = 1 – x2 – y2
P2612.6
Example 3
The density at any point is proportional to its distance from the axis of the cylinder.
Find the mass of E.
P2712.6
Example 3 SOLUTION
In cylindrical coordinates the cylinder is r = 1 and the paraboloid is z = 1 – r2.
So, we can write:
E ={(r, , z)| 0 ≤ ≤ 2, 0 ≤ r ≤ 1, 1 – r2 ≤ z ≤ 4}
P2812.6
Example 3 SOLUTION
As the density at (x, y, z) is proportional to the distance from the z-axis, the density function is:
where K is the proportionality constant.
2 2, ,f x y z K x y Kr
P2912.6
Example 3 SOLUTION
So, from Formula 13 in Section 12.5, the mass of E is:
2
2 1 42 2
0 0 1
2 1 2 2
0 0
152 1 2 4 3
0 00
( )
4 1
3 25
12
5
rE
m K x y dV Kr r dz dr d
Kr r dr d
rK d r r dr K r
K
P3012.6
Example 4
Evaluate
2
2 2 2
2 4 2 2 2
2 4
x
x x yx y dz dy dx
P3112.6
Example 4 SOLUTION
This iterated integral is a triple integral over the solid region
The projection of E onto the xy-plane is the disk x2 + y2 ≤ 4.
2 2
2 2
{ , , | 2 2, 4 4
, 2}
E x y z x x y x
x y z
P3212.6
Example 4 SOLUTION
The lower surface of E is the cone
Its upper surface is the plane z = 2.
2 2z x y
P3312.6
Example 4 SOLUTION
That region has a much simpler description in cylindrical coordinates:
E = {(r, , z) | 0 ≤ ≤ 2, 0 ≤ r ≤ 2, r ≤ z ≤ 2}
Thus, we have the following result.
P3412.6
Example 4 SOLUTION
2
2 2 2
2 4 2 2 2
2 4
2 2 22 2 2
0 0
2 2 3
0 0
24 51 12 5 0
2
162
5
x
x x y
rE
x y dz dy dx
x y dV r r dz dr d
d r r dr
r r
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