1
Section 9.7/12.8: Triple Integrals in Cylindrical and Spherical
Coordinates
Practice HW from Stewart Textbook (not to hand in)
Section 9.7: p. 689 # 3-23 odd
Section 12.8: p. 887 # 1-11 odd, 13a, 17-21 odd, 23a, 31, 33
Cylindrical Coordinates
Cylindrical coordinates extend polar coordinates to 3D space. In the cylindrical
coordinate system, a point P in 3D space is represented by the ordered triple (r ,  , z ) .
Here, r represents the distance from the origin to the projection of the point P onto the x-y
plane,  is the angle in radians from the x axis to the projection of the point on the x-y
plane, and z is the distance from the x-y plane to the point P.
z
P(r , , z )
r
y

x
As a review, the next page gives a review of the sine, cosine, and tangent functions at
basic angle values and the sign of each in their respective quadrants.
2
Sine and Cosine of Basic Angle Values
 Degrees
 Radians
0
30
0

6

4

3

2

3
2
2
45
60
90
180
270
360
sin 
cos 
cos
sin 
cos 0  1
sin 0  0
1
2
0
2
2
3
2
1
0
1
undefined
-1
0
0
-1
0
undefined
1
0
0
3
2
2
2
1
2
tan  
3
3
3
Signs of Basic Trig Functions in Respective Quadrants
Quadrant
cos
sin 
I
II
III
IV
+
+
+
+
-
tan  
sin 
cos 
+
+
-
The following represent the conversion equations from cylindrical to rectangular
coordinates and vice versa.
Conversion Formulas
To convert from cylindrical coordinates ( r ,  , z ) to rectangular form (x, y, z) and vise
versa, we use the following conversion equations.
From polar to rectangular form: x  r cos , y  r sin  , z = z.
From rectangular to polar form: r 2  x 2  y 2 , tan  
y
, and z = z
x
3
Example 1: Convert the points ( 2 , 2 ,3) and (3, 3 ,1) from rectangular to
cylindrical coordinates.
Solution:
█
4
Example 2: Convert the point (3,

4
,1) from cylindrical to rectangular coordinates.
Solution:
█
Graphing in Cylindrical Coordinates
Cylindrical coordinates are good for graphing surfaces of revolution where the z axis is
the axis of symmetry. One method for graphing a cylindrical equation is to convert the
equation and graph the resulting 3D surface.
5
Example 3: Identify and make a rough sketch of the equation z  r 2 .
Solution:
z
y
x
█
Example 4: Identify and make a rough sketch of the equation  

4
.
Solution:
z
y
█
x
6
Spherical Coordinates
Spherical coordinates represents points from a spherical “global” perspective. They are
good for graphing surfaces in space that have a point or center of symmetry.
Points in spherical coordinates are represented by the ordered triple
(  , ,  )
where  is the distance from the point to the origin O,  , where is the angle in radians
from the x axis to the projection of the point on the x-y plane (same as cylindrical

coordinates), and  is the angle between the positive z axis and the line segment OP
joining the origin and the point P (  ,  ,  ) . Note 0     .
z

P(  , ,  )

y

x
7
Conversion Formulas
To convert from cylindrical coordinates (  , ,  ) to rectangular form (x, y, z) and vise
versa, we use the following conversion equations.
From to rectangular form: x   sin  cos  , y   sin  sin  , z   cos 
y
From rectangular to polar form:  2  x 2  y 2  z 2 , tan   , and
x
z
z
  arccos(
)  arccos( )

x2  y2  z 2
Example 5: Convert the points (1, 1, 1) and (3, 3 ,2 2 ) from rectangular to spherical
coordinates.
Solution:
█
8
Example 6: Convert the point (9,

4
,  ) from rectangular to spherical coordinates.
Solution:
█
Example 7: Convert the equation   2 sec  to rectangular coordinates.
Solution:
█
9
Example 8: Convert the equation  

3
to rectangular coordinates.
Solution: For this problem, we use the equation   arccos(
z
) . If we take
x y z
the cosine of both sides of the this equation, this is equivalent to the equation
cos  
Setting  

3

3

2
2
z
x2  y2  z 2
gives
cos
Since cos
2

3

z
x2  y2  z 2
.
1
, this gives
2
1

2
z
x2  y2  z2
or
x 2  y 2  z 2  2z
Hence, x 2  y 2  z 2  2 z is the equation in rectangular coordinates. Doing some
algebra will help us see what type of graph this gives.
Squaring both sides gives
x 2  y 2  z 2  (2 z ) 2
x 2  y 2  z 2  4z 2
x 2  y 2  3z 2  0
The graph of x 2  y 2  3z 2  0 is a cone shape half whose two parts be found by
graphing the two equations  x 2  y 2  z 2  2 z . The graph of the top part,
x 2  y 2  z 2  2 z , is displayed as follows on the next page.
(continued on next page)
10
█
Example 9: Convert the equation x 2  y 2  z to cylindrical coordinates and spherical
coordinates.
Solution: For cylindrical coordinates, we know that r 2  x 2  y 2 . Hence, we have
r 2  z or
r z
For spherical coordinates, we let x   sin  cos  , y   sin  sin  , and z   cos 
to obtain
(  sin  cos ) 2  (  sin  sin  ) 2   cos 
We solve for  using the following steps:
 2 sin 2  cos 2    2 sin 2  sin 2    cos 
 2 sin 2  (cos 2   sin 2  )   cos 
 2 sin 2  (1)   cos   0
 (  sin 2   cos  )  0
  0,  sin 2   cos   0
  0,  
(Square terms)
(Factor  2 sin 2  )
(Use identity cos 2   sin 2   1)
(Factor  )
(Set each factor equal to zero and solve)
cos 
sin 2 
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11
Triple Integrals in Cylindrical Coordinates
Suppose we are given a continuous function of three variables f ( r ,  , z ) expressed over a
solid region E in 3D where we use the cylindrical coordinate system.
z
z  h2 (r, )
E
1
y
r  g 2 ( )
z  h1 (r, )
2
r  g1 ( )
x
Then
  2 r  g 2 ( ) z  h2 ( r , )
 f (r, , z) dV  
E
Volume of E 


f (r , , z ) r dz dr d
 1 r  g1 ( ) z  h1 ( r , )
 dV 
E
  2 r  g 2 ( ) z  h2 ( r , )



 1 r  g1 ( ) z  h1 ( r , )
r dz dr d
12
Example 10: Use cylindrical coordinates to evaluate
 ( x
3
 xy 2 ) dV , where E is the
E
solid in the first octant that lies beneath the paraboloid z  1  x 2  y 2 .
Solution:
█
13
Example 11: Use cylindrical coordinates to find the volume of the solid that lies both
within the cylinder x 2  y 2  4 and the sphere x 2  y 2  z 2  9 .
Solution: Using Maple, we can produce the following graph that represents this solid:
In this graph, the shaft of the solid is represented by the cylinder equation x 2  y 2  4 . It is
capped on the top and bottom by the sphere x 2  y 2  z 2  9 . Solving for z, the upper and
bottom portions of the sphere can be represented by the equations z   9  x 2  y 2 .
Thus, z ranges from z   9  x 2  y 2 to z  9  x 2  y 2 . Since x 2  y 2  r 2 in
cylindrical coordinates, these limits become z   9  r 2 to z  9  r 2 .When this surface
is projected onto the x-y plane, it is represented by the circle x 2  y 2  4 . The graph is
(Continued on next page)
14
This is a circle of radius 2. Thus, in cylindrical coordinates, this circle can be represented
from r = 0 to r = 2 and from   0 to   2 . Thus, the volume can be represented by the
following integral:
Volume 
  2 r  g 2 ( ) z  h2 ( r , )
 dV  


r dz dr d 
 1 r  g1 ( ) z  h1 ( r , )
E
  2 r  2


 0
z  9 r 2

r  0 z   9 r
r dz dr d
2
We evaluate this integral as follows:
  2 r  2

 0

z  9 r 2

r  0 z   9 r
r dz dr d 

 0
2
  2 r  2


2r 9  r 2 dr d

  2

 0
  2

 0
  2

z   9 r

 0

r 0
2
r ( 9  r 2 )  r ( 9  r 2 ) dr d
  2

dr d
rz

 0



 0 r 0
  2 r  2

z  9 r 2
  2 r  2

 0
r 0
3
2
2
 (9  r 2 )
3
r2
d
(Use u - du sub let u  9 - r 2 )
r 0
3
2
3
2
2
2
[ (9  2 2 )   (9  0 2 ) ] d
3
3
3
3
2
2
2
2
[ (5)  (9) ] d
3
3
10
[18 
5 ] d
3
3
2
3
2
(Note (9)  27 and (5)  5 5 )
  2
10
 [18 
5 ]
3
 0
10
5 )2  0
3
20
 36 
5
3
 (18 
Thus, the volume is 36 
20
3
5.
█
15
Triple Integrals in Spherical Coordinates
Suppose we have a continuous function f (  ,  ,  ) defined on a bounded solid region E.
z

P(  , ,  )

y

x
Then
  2  2   h2 ( , )
 f (  , , ) dV  
E
Volume of E 


f (  ,  , )  2 sin  d d d
 1  1   h1 ( , )
  2  2   h2 ( , )
 dV  
E


 1  1   h1 ( , )
 2 sin  d d d
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Example 12: Use spherical coordinates to evaluate
 e
x2  y 2  z 2
dV , where E is
E
enclosed by the sphere x  y  z  9 in the first octant.
2
2
2
Solution:
█
17
Example 13: Convert
2
4 x 2
16 x 2  y 2
2
0
0



x 2  y 2 dz dy d from rectangular to
spherical coordinates and evaluate.
Solution: Using the identities x   sin  cos  and y   sin  sin  , the integrand becomes
x2  y2 
 2 sin 2  cos 2    2 sin 2  sin 2 
  2 sin 2  (cos 2   sin 2  )
  2 sin 2  (1)   sin 
The limits with respect to z range from z = 0 to z  16  x 2  y 2 . Note that
z  16  x 2  y 2 is a hemisphere and is the upper half of the sphere x 2  y 2  z 2  16 .
The limits with respect to y range from y = 0 to y  4  x 2 , which is the semicircle
located on the positive part of the y axis on the x-y plane of the circle x 2  y 2  4 as x
ranges from x  2 to x  2 . Hence, the region described by these limits is given by
the following graph
Thus, we can see that  ranges from   0 to   4 ,  ranges from   0 to  

and
2
 ranges from   0 to    . Using these results, the integral can be evaluated in polar
coordinates as follows:
(continued on next page)
18
2
4 x 2
2
0


 


16 x 2  y 2

x 2  y 2 dz dy d
0



2  4

 sin  (  2 sin  ) d d d
 0  0  0
 





2  4

 3 sin 2  d d d
 0  0  0
 





2
 0  0
 









d d
(Integrate with respect to  )
 0
[
44
sin 2   0]d d  
4
 0
2
1  cos 2 
64 
 d d
2




 0
 




2
32(1  cos 2 ) d d 


2
64 sin 2  d d
(Sub in limits and simplify)
 0
 


1  cos 2u
)
2


2
( 32  32 cos 2 )d d
(Simplify and dist 32)
 0  0
1
( 32  32( ) sin 2 )
2

( 32  16 sin 2 )
 0


(Use trig identity sin 2 u 
 0  0
 




sin 
 
2
 0  0
 
4
 4
2

 0  0
 
4


2
 0

2
 0
(Integrate with respect to  , use u - du sub for cos2 )
d 
 

 0
 
 
 0
 
 0

 
[32    16 sin 2( ) ]  (32(0)  16 sin 0) d
2
 2
 (16  16 sin   0)d   (16  16(0) )d
 
 16 d  16  0
(Integrate with respect to  )
 0
 16 ( )  0  16 2
█