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

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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).

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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

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