SECTION 12.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES.

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SECTION 12.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES

Transcript of SECTION 12.6 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES.

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SECTION 12.6

TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES

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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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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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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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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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CYLINDRICAL COORDINATES

To convert from cylindrical to rectangular coordinates, we use:

x = r cos y = r sin

z = z

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CYLINDRICAL COORDINATES

To convert from rectangular to cylindrical coordinates, we use:

r2 = x2 + y2 tan = y/x

z = z

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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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Example 1(a) SOLUTION

The point with cylindrical coordinates (2, 2/3, 1) is plotted in Figure 3.

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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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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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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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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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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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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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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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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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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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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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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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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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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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TRIPLE INTEGN. IN CYL. COORDS.

Figure 7 shows how to remember this.

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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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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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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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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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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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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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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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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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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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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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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