Triple Integrals in Cylindrical Coordinates

Applications in Spherical and Cylindrical Coordinates   

Triple integrals in spherical and cylindrical coordinates occur frequently in applications. For example, it is not common for charge densities and other real-world distributions to have spherical symmetry, which means that the density is a function only of the distance r. ( Note: Scientists and engineers use r both to denote charge density and also to denote distance in spherical coordinates. The context in which r appears will indicate how it is being used).       

EXAMPLE 4    The charge density for a certain charge cloud contained in a sphere of radius 10 cm centered at the origin is given by
r( x,y,z) = 100
x2+y2+z2
      mC
cm3
What is the total charge contained within a sphere? ( mC = micro-coulombs )       

Solution: If W denotes the solid sphere of radius 10 cm centered at the origin, then the total charge is
Q =




W 
100
x2+y2+z2
  dV
However, x2+y2+z2 = r2 leads to
Q = 100
2p

0 

p

0 

10

0 
r   r2sin( f)   drdfdq
: 2500sinf (i.e., charge density is proportional to r ). Evaluation of the integral leads to
Q
=
100
2p

0 

p

0 
 r4
4

10

0 
sin( f)   drdfdq
=
100
2p

0 

p

0 
2500sin( f) dfdq
=
100
2p

0 
5000dq
=
1,000,000p  mC
which is Q = p coulombs.

       

Triple integrals in spherical and cylindrical coordinates are common in the study of electricity and magnetism. In fact, quantities in the fields of electricity and magnetism are often defined in spherical coordinates to begin with.   

EXAMPLE 5    The power emitted by a certain antenna has a power density per unit volume of
p( r,f,q) =  P0
r2
sin4(f) cos2( q)
where P0 is a constant with units in Watts. What is the total power within a sphere of radius 10  m?       

Solution: The total power P will satisfy
P
=





W 
 P0
r2
sin4( f) cos2( q) dV
=

2p

0 

p

0 

10

0 
 P0
r2
sin4( f) cos2( q)    r2sin( f) drdfdq  
=
P0
2p

0 

p

0 

10

0 
sin4( f)   sin( f)   cos2( q)    drdfdq
=
10P0
2p

0 

p

0 
( 1-cos2( f) ) 2  sin( f)   cos2( q)    dfdq
Let us now let u = cos( f) , du = -cos( f)df, u( 0) = 1, and u( p) = -1. Then
P
=
-10P0
2p

0 

-1

1 
( 1-u2) 2  cos2( q)    dudq
=
10P0
2p

0 
 16
15
cos2(q) dq
However, 2cos2( q) = cos( 2q) +1, so that
P =  80
25
P0
2p

0 
( cos( 2q)+1) dq =  32p
5
P0  Watts

       

Check your Reading: What happened to the factor of r2 in the calculations in example 5?