CHAPTER 12:

SPHERICAL COORDINATES

12.1 DEFINING OF SPHERICAL COORDINATES

A location in three dimensions can be defined with spherical coordinates where

 is the same angle defined for polar and cylindrical coordinates. To gain some insight
into this variable in three dimensions, the set of points consistent with some constant
values of are shown below.

 is the angle between the vector that goes from the origin to the location and the vector
(the z axis). To gain some insight into this variable in three dimensions, the set
of points consistent with some constant values of are shown below.

 is the distance from the origin (0, 0, 0) to the location.

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Example Exercise 12.1.1: Find the point .

Solution: The first requirement of the location is that The following diagrams shows
all points in the xy plane associated with .

The second requirement of the location is that . The following diagram shows the set of
points in the xy plane consistent with (these can be visualized as a vector) rotated
upward until the points on this vector achieve an angle of with the z axis

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The set of points where and can be visualized as the vector shown in the
above diagram. If we start at the origin and proceed along this vector until the distance from the
origin is , we will obtain the location or point associated with in
3-space.

Example Exercise 12.1.2: Find the point .

Solution: Starting with we can obtain the points associated with in the xy
plane.

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Finally, of those where and , we can find the point with a distance of = 6
from the origin. Hence the point associated with can be placed in 3-
space.

12.2 IDENTIFYING SOLIDS ASSOCIATED WITH SPHERICAL CUBES

We have previously seen that as cubic coordinates go from constant to constant, the resulting
solid will be a cube. If cylindrical coordinates go from constant to constant, the resulting solid
will be a cylinder or a segment of a cylinder. Hence, our expectation is that spherical

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coordinates have been designed so that if each coordinate goes from constant to constant, the
resulting solid will be a sphere or a segment of a sphere.

Example Exercise 12.2.1: Find the volume associated with

Solution: The region in the xy plane associated with is shown in the diagram below.

If each of the vectors shown in red in the above diagram is rotated upward from the origin to the
z axis, this will reflect the points associated with for each value of . If it is rotated back
to the xy plane, this will reflect a value of for each value of . If it is rotated halfway
between the z axis and the xy plane, this will reflect a value of for each value of . Hence
the set of points consistent with will be the points above the first
quadrant in the xy plane and below the cone shown in the diagram below.

The volume associated with will be the points in the above

solid that reside between the spheres and . This spherical cube is shown in the
diagram below.

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In the above case, when spherical coordinates went from constant to constant, the resulting solid
was a segment of a sphere. If we repeat the above steps with
, the resulting solid will be an entire sphere of radius R centered at (0,0,0).

12.3 TRANSLATING COORDINATE SYSTEMS

We now have three different coordinate systems with which we can represent a point in 3-space.
A point can be represented with in cubic coordinates, with in cylindrical
coordinates and with in spherical coordinates. It is often helpful to translate a problem
from one coordinate system to another depending on the nature of the problem. As a first step,
the geometry of each of the coordinates in these three coordinate systems is presented in the
following diagram.

The right triangle in the xy plane with angle produces the following relationships between
rectangular and polar coordinates that we should already be familiar with:

 ,
 ,
 or and

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  or when (x, y) is in the first or fourth quadrant and
when the point (x, y) is in the second or third quadrant

The right triangle with angle produces the following relationships:

 ,
 ,
 or and
 or

Putting some of these relationships together we obtain:

 and so
 and so ,
 so or
 and so
Organizing these conclusions, the following are the relations between spherical and cubic
coordinates:



 when (x, y) is in the first or fourth quadrant and when
the point (x, y) is in the second or third quadrant




The following are the relations between cylindrical and spherical coordinates:

 ,

 as the coordinate is shared in both coordinate systems
 and


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12.4 APPROXIMATING THE VOLUME OF A SPHERICAL CUBE

To begin our discussion of the volume of a spherical cube, we will consider the solid
shown in the diagram below.

As with cylindrical cubes, we will approximate the volume of this cube with

Volume = length x width x height

As ll corresponding sides will come to resemble parallel lines

and this approximation will become precise.

Beginning with the width of the top and the bottom of the spherical cube, we can see in the
diagram below that goes from 5 to 6 on both the left and right sides of the top and the bottom.
Correspondingly, we will approximate .

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In the following diagram, we can see that the height of the cube on both the left and the right
sides will be approximated with an arc that has angle . The radius associated with the arc
is equal to 5 on the inner side (side closer to the origin) of the cube and is equal to 6 on the outer
side of the cube. Hence we will approximate on the outer side of the
cube and on the inner side of the cube. As, , the
arcs will approach linearity and the outer and inner heights will approach the same length.

We now have approximated the height and the width of the cube however looking at the
diagram, there are four different lengths: The bottom inner side, the bottom outer side, the top
inner side and the top outer side. Starting with the inner sides, we can see in the diagram below
that both the top and bottom lengths are arcs with angle .

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In the following diagram, we can see that the radius associated with the arc that represents the
length of the to and the bottom side of the cube is the value r from cylindrical and polar
coordinates. However we wish to represent the volume in spherical coordinates so we use the
translation from section 12.3 that indicates . The upper inner arc consists of points
where and and hence the radius of the upper inner arc is .
The lower inner arc consists of points where and and hence the radius of the lower
inner arc is .

Hence the lower inner arc has angle , radius and our
approximation for the length will be . The upper inner
arc has angle , radius and our approximation for the length will be
.. The process to find the lengths of the outer arcs on the
top and bottom sides of the cube is the same. The only difference is that the outer arcs use
and the inner arcs use .

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Generalization

When we approximate the volume of a spherical cube, we use

 ,
 and


to obtain,

12.5 USING RIEMANN SUMS AND THE FUNDAMENTAL THEOREM

TO OBTAIN THE MASS OF SPHERICAL CUBES

Example Exercise 12.5.1: A spherical cube 5 ≤ ≤6, ≤ ≤ and ≤ ≤ has density

= . Use Riemann Sums with two divisions in , and to approximate the mass of the
cube using the largest value of each variable to represent a given division. Then express the
Riemann Sum as a triple summation and use the fundamental theorem to find the precise mass of
the cube.

Solution:

Step 1: Divide , and into two parts and identify , , 1, 2, 1, 2 , and :

The following diagram shows the spherical cube associated with 5 ≤ ≤ 6, ≤ ≤ and ≤
≤ .

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The following diagram shows the above spherical cube with two divisions in each variable.

For convenience, we can divide the 8 divisions into the inner 4 divisions where 5 ≤ ≤ 5.5 and
the outer four divisions where 5.5 ≤ ≤ 6 as is shown in the following diagrams

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 Inner Divisions: 5 ≤ ≤ 5.5 Outer Divisions: 5.5 ≤ ≤ 6

In divisions 1, 2, 5 and 6, goes from to and in divisions 3, 4, 7 and 8, goes from to .

In divisions 1, 3, 5 and 7, goes from to and in divisions 2, 4, 6 and 8, goes from to .
In divisions 1, 2, 3 and 4, goes from 5 to 5.5 and in divisions 5, 6, 7 and 8, goes from 5.5 to
6. Hence, , , 1 , 2 , 1 , 2 , and

Step 2: Find the appropriate numeric approximation for the length, width, height, volume,
density and mass for each division:

In Section 12.4, we approximated the volume of a polar cube with volume = length*width*height
where , and . Using the maximum
value of each variable in each division, the following two diagrams show the length, width,
height and density for the 4 inner divisions and the four outer divisions.

Inner Divisions: 5 ≤ ≤ 5.5

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 Outer Divisions: 5.5 ≤ ≤6

The length, width, height, density and mass of each division are summarized in the following
table:

Division Length Width Height Density Mass

(m) (m) (m) ( (kg)

1 5.5 0.5 5.5

2 5.5 0.5 5.5

3 5.5 0.5 5.5

4 5.5 0.5 5.5

5 6 0.5 6

6 6 0.5 6

7 6 0.5 6

8 6 0.5 6

Step 3: Add the masses of the eight divisions to approximate the total mass of the solid:

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Step 4: Repeat Step 2 using , , 1, 2 , 1, 2 , and instead of numerical values
as appropriate.

Division Length Width Height Density Mass

(m) (m) (m) ( (kg)
1
2
3
4
5
6
7
8

Step 5: Add the masses of the eight divisions to approximate the total mass of the solid using r1,
r2, 1, 2 , 1, 2 r, and z:

+ ]+
[
]

Step 6: Express the approximate mass in Step 5 as a triple summation:

+ ]+
[
]

By grouping the [..], we obtain:

}.

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By grouping the {..} (Note: The order within each bracket is followed by , however this
makes no difference with a sum.), we obtain:

Grouping these last two terms we obtain:

Step 7: As , and , this approximation becomes precise and we can apply

the fundamental theorem to find the precise mass of the solid. Note that we change the order of
expressions that are multiplied together so that the sigmas are coupled with the appropriate .

12.6 VOLUMES ASSOCIATED WITH INTEGRALS IN SPHERICAL

COORDINATES

Example Exercise 12.6.1: Find the solid associated with

Solution: From the above section where we use the fundamental theorem and Riemann sums to
find the volume of a spherical solid, we should remember that is necessary for the for
the expression of a volume in spherical coordinates. Hence the integral
is simply a volume and were there a density f, it would be expressed
as .

Working from the outside inward, the first datum from the integrals is indicating that our
volume will contain values of between and

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The second datum from the integral is indicating that for every value of between
and , we will accept values of that reside between and .

The third datum from the integral is indicating that for every value of and in the
defined region, we will accept values of that reside between and . Taking all of the
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data into account, this triple integral will represent the half of an ice cream cone shaped solid
shown in the following diagram.

EXERCISE PROBLEMS:

1) Express the volume of the following solids as a triple integral in (i) spherical coordinates, (ii)
cylindrical coordinates and (iii) cubic coordinates.

A. Above the xy plane and below the sphere .

B. Below the xy plane and above the sphere .
C. Inside the sphere and satisfying that .
D. Inside the sphere and satisfying that and .
E. Inside the sphere and satisfying that and .
F. Inside the sphere , above the xy plane and satisfying that and
.
G. Inside the sphere , above the xy plane and satisfying that and
.

birds
2) The density of birds in a spherical cage is + and we wish to obtain the number of fish in
m3
 3  
the tank described by 2    8,   ,   .
2 2 9 3

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 A. If there are two divisions in each variable and the number of fish is to be approximated using the
minimum value for each variable in each division, find 1 , 2 ,1 ,  2 , 1 and 2 use them to fill in
the following table with numerical values.

Division Length Width Height Density No. of birds

1
2
3
4
5
6
7
8

B. Use the values of 1 , 2 ,1 ,  2 , 1 and 2 to fill in the same table below using 1 , 2 ,1 ,  2 , 1
and 2 ,and,  ,  and  instead of numerical values. (Note, the divisions should not change
between the two tables.)

Division Length Width Height Density No. of birds

1
2
3
4
5
6
7
8

C. Express the approximate no. of birds numerically.

D. Express the number of birds obtained in part C in the form  (...) Take the
appropriate limits to convert the sum in part D to an integral and evaluate the integral.

3) The density of fish in a spherical tank is y and we wish to obtain the number of fish in the tank
  
described by 2    4, 0    ,   .
2 4 2

A. If there are two divisions in each variable and the number of fish is to be approximated using the
maximum value for each variable in each division, find 1 , 2 ,1 ,  2 , 1 and 2 use them to fill
in the following table with numerical values.

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 Division Length Width Height Density No. of fish
1
2
3
4
5
6
7
8

B. Use the values of 1 , 2 ,1 ,  2 , 1 and 2 to fill in the same table below using 1 , 2 ,1 ,  2 , 1
and 2 ,and,  ,  and  instead of numerical values. (Note, the divisions should not change
between the two tables.)

Division Length Width Height Density No. of fish

1
2
3
4
5
6
7
8

C. Express the approximate no. of fish numerically.

D. Express the number of fish obtained in part C in the form  (...) Take the
appropriate limits to convert the sum.

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