section 12.6 triple integrals in cylindrical coordinates
TRANSCRIPT
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SECTION 12.6
TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
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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
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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
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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.
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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.
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P612.6
CYLINDRICAL COORDINATES
To convert from cylindrical to rectangular coordinates, we use:
x = r cos y = r sin
z = z
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P712.6
CYLINDRICAL COORDINATES
To convert from rectangular to cylindrical coordinates, we use:
r2 = x2 + y2 tan = y/x
z = z
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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).
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P912.6
Example 1(a) SOLUTION
The point with cylindrical coordinates (2, 2/3, 1) is plotted in Figure 3.
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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
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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
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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)
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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.
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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.
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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.
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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.
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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.
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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.
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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()}
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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
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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
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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.
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P2312.6
TRIPLE INTEGN. IN CYL. COORDS.
Figure 7 shows how to remember this.
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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.
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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
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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.
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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}
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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
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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
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P3012.6
Example 4
Evaluate
2
2 2 2
2 4 2 2 2
2 4
x
x x yx y dz dy dx
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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
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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
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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.
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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