Multiple Integrals 12

Triple Integrals in Cylindrical and Spherical Coordinates12.8

3

Cylindrical Coordinates

4

Cylindrical Coordinates

Recall that the cylindrical coordinates of a point P are

(r, , z), where r, , and z are shown in Figure 1.

Figure 1

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Cylindrical Coordinates

Suppose that E is a type 1 region whose projection D onto

the xy-plane is conveniently described in polar coordinates

(see Figure 2).

Figure 2

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Cylindrical CoordinatesIn 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( )}

We know that

7

Cylindrical CoordinatesBut we also know how to evaluate double integrals in polar coordinates. In fact, combining Equation 1 with the equation below,

we obtain

8

Cylindrical CoordinatesFormula 2 is the formula for triple integration in cylindrical coordinates.

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 ,

and replacing dV by r dz dr d.

(Figure 3 shows how to remember this.)

Volume element in cylindricalcoordinates: dV = r dz dr d

Figure 3

9

Cylindrical Coordinates

It is worthwhile to use this formula when E is a solid region

easily described in cylindrical coordinates, and especially

when the function f (x, y, z) involves the expression x2 + y2.

10

Example 1 – Finding Mass with Cylindrical Coordinates

A solid E lies within the cylinder x2 + y2 = 1, below the plane z = 4, and above the paraboloid z = 1 – x2 – y2. (See Figure 4.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E.

Figure 4

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Example 1 – 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}

Since the density at (x, y, z) is proportional to the distance

from the z-axis, the density function is

f (x, y, z) = K

= Kr

where K is the proportionality constant.

12

Example 1 – Solution

Therefore, from Formula , the mass of E is

cont’d

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Example 1 – Solution cont’d

14

Spherical Coordinates

15

Spherical CoordinatesWe have defined the spherical coordinates (, , ) of a point (see Figure 6) and we demonstrated the following relationships between rectangular coordinates and spherical coordinates:

x = sin cos y = sin sin z = cos

Spherical coordinates of PFigure 6

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Spherical CoordinatesIn this coordinate system the counterpart of a rectangular box is a spherical wedge

E = {(, , ) | a b, , c d }

where a 0 and – 2. Although we defined triple integrals by dividing solids into small boxes, it can be shown that dividing a solid into small spherical wedges always gives the same result.

So we divide E into smaller spherical wedges Eijk by means

of equally spaced spheres = i, half-planes = j, and

half-cones = k.

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Spherical CoordinatesFigure 7 shows that Eijk is approximately a rectangular box

with dimensions , i (arc of a circle with radius i,

angle ), and i sin k (arc of a circle with radius

i sin k, angle ).

Figure 7

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Spherical Coordinates

So an approximation to the volume of Eijk is given by

() (i ) (i sin k ) = i2

sin k

Thus an approximation to a typical triple Riemann sum is

But this sum is a Riemann sum for the function

F(, , ) = f ( sin cos , sin sin , cos ) 2 sin

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Spherical Coordinates

Consequently, the following formula for triple integration

in spherical coordinates is plausible.

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Spherical CoordinatesFormula 4 says that we convert a triple integral from rectangular coordinates to spherical coordinates by writing

x = sin cos y = sin sin z = cos

using the appropriate limits of

integration, and replacing dV

by 2 sin d d d.

This is illustrated in Figure 8.

Figure 8

Volume element in sphericalcoordinates: dV = 2 sin d d d

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Spherical Coordinates

This formula can be extended to include more general

spherical regions such as

E = {(, , ) | , c d, g1(, ) g2(,

)}

In this case the formula is the same as in (4) except that the

limits of integration for are g1(, ) and g2(, ).

Usually, spherical coordinates are used in triple integrals

when surfaces such as cones and spheres form the

boundary of the region of integration.

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Example 3Evaluate where B is the unit ball:

B = {(x, y, z) | x2 + y2 + z2 1}

Solution:

Since the boundary of B is a sphere, we use spherical coordinates:

B = {(, , ) | 0 1, 0 2, 0 }

In addition, spherical coordinates are appropriate because

x2 + y2 + z2 = 2

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Example 3 – SolutionThus (4) gives

cont’d