6.
Multiple Integrals
6.1 Double Integrals
For a function f ( x ) of one variable, we define the definite integral of f from a to b as
b

f ( x )dx  lim
a
N 
N
 f ( x )x
i
i 1
i
which corresponds geometrically, if f ( x ) is positive, to the area between the graph, the x-axis and
the lines x = a and x = b.
f (x )
 xi
f ( xi )
a x1 x 2
xi
xN b x
A function f ( x, y ) defines a surface and, if f ( x, y ) is positive, we may analogously determine
the volume enclosed by this surface and a cylinder erected on a region S of the xy-plane.
z
Surface z  f ( x, y )
f ( xi , y i )
y
x
Region S in the xy-plane
Small area  S i   xi  yi
1
Divide S into N small area elements such that the ith element has area  S i   xi  yi
and contains the point ( xi , yi ) . Then f ( xi , yi ) S i is the volume of the column with base  S i
and height f ( xi , yi ) and we define the double integral of f ( x, y ) over the region S to be

S
f x, y dS  
S

f x, y dxdy  lim
N 
N
 f x , y S
i
i 1
i
i
This definition is taken when f ( x, y ) is any two-dimensional scalar field. Such an integral is
sometimes called a surface integral or a repeated integral (as its evaluation often involves two
definite integrals).
The evaluation of such integrals is best illustrated through examples.
Example
  dxdy = area of the region S if f ( x, y)  1 .
S
Example (Rectangular regions)
If S is the region a  x  b, c  y  d
  f x, y dxdy   
f  x, y dx dy

b
d
    f  x, y dy dx

xa 
 y c
S
d
b
y c
xa
(1)
y
1
2
(2)
y
d
d
y
y
c
c
x
a
x
a
b
x
x
b
x
Example
Evaluate
  (2 xy  y
2
)dxdy where S is the rectangle 1  x  2, 0  y  1 .
S
Sol:
2
1
1
2

2
2
2 2


4 y  2 y 2  y  y 2 dy
2
xy

y
dx
dy

x
y

xy
dy


0 1


1
0
0

 3 y 2 y 3  1 11
 
  
3 0 6
 2

1

or
2
2
1

 2 y3  1
 x 2 x  2 4 2 1 1 11
1

2
       


2
xy

y
dy
dx

xy

dx

x

dx





0
1 0


3
3




 2 31 2 3 2 3 6
1
1

2
Example (Non-rectangular regions)
y
h(x)
y
y
h(x)
d
S
h( y )
c
g (x)
a
g (x)
x
x
b
a  x  b, g ( x )  y  h ( x )
c  y  d , h( y )  x  g ( y )
 

b
h( x)
xa
y g ( x)
S
g ( y)
S
f ( x, y ) dy dx

 g ( y ) f ( x, y )dx dy

y c 
 x h ( y )
d
a
b
a  x  b, g ( x )  y  h ( x )
 
b
h( x)
xa
y g ( x)
f ( x, y ) dy dx

Variable Limits unless a
rectangular region
Constant
Limits
Example
Evaluate
  xy dxdy where S is the region bounded by
2
y  0, y  x 2 and x  1 .
S
Sol:
y
A sketch of the region S
is important, but it is
not necessary to be able to
visualize the surface f ( x, y ) .
1
0
y  x2
S
y0
x 1
x
1
3
x
1
1
x 2 
 y3  x2
x7
S xy dxdy  0  0 xy dy dx  0  x 3  0 dx  0 3 dx


2
1
2
x8 1 1

24 0 24

or
1
1
1 2 
 2 x2  1
 y2 y3 
S xy dxdy  0   xy dx dy  0  y 2  y dx  0  2  2 dx
 y

3
4
y
y 1 1
   
8  0 24
6
1
2
In some cases, one order of integration may be easier than the other, e.g.
1
 
2 y
y 0 x  y
1
fdxdy  

x2
x 0 y 0
fdydx  
2

2 x
x 1 y  0
fdydx
y
1
y  x2
y  -x+2
S
0
1
y0
x
2
Example
Let
 1
 y2

 1
f ( x , y )   2
 x
 0

if 0  x  y  1
if 0  y  x  1
otherwise if 0  x, y  1
Show that the double integral
 f ( x, y)dxdy
, where Q  [0,1]  [0,1] can not exist by pointing
Q
1 1
1 1




out    f ( x, y )dy dx     f ( x, y )dx dy .
0 0
0 0


Proof:
4
1 x
1
1



0 0 f ( x, y)dy dx  0 0 f ( x, y)dy  x f ( x, y)dy dx
1
1
1
 x 1
1 
    2 dy   2 dy dx    1dx  1
0 0 x
x y
0

1
1 y
1

1

f
(
x
,
y
)
dx
dy

f
(
x
,
y
)
dx

f
(
x
,
y
)
dx

dy

0 0
0 0
y



1
1
1
y 1
1 
    2 dx   2 dx dy   1dx  1
0 y

0
y x
0
1
Hence,
1 1
1



0 0 f ( x, y)dy dx  0 0 f ( x, y)dxdy
1
As the two iterated integrals exist but are unequal, the double integral
 f ( x, y)dxdy
, where
Q
Q  [0,1]  [0,1] can not exist.
Example
 x

0  x f ( x, y)dy dx .


1
Change the order of the integration in
Sol:
The line y  x and the parabola y  x cut at (0,0) & (1,1). The domain of integration is the
area bounded by y  x and y  x (the same as x  y 2 ).
1y

 x

f
(
x
,
y
)
dy
dx

f
(
x
,
y
)
dx

dy


0  x


2


0 y


1
So
6.2 Change of Variable in Double Integrals
For the definite integral
I   f  x dx
b
a
We know that the change of variable x = x(u) produces

dx
I   f  xu  du

du
where x( )  a, x(  )  b
5
For the double integral
I    f  x, y dxdy
S
The change of variable x = x(u,v), y = y(u,v) gives
I    f  xu , v , y u , v 
S*
  x, y 
dudv
 u , v 
where
 x

  x, y 
J
 det  u
 u, v 
 y
 u
x 

v 
y 

v 
is called the Jacobian of the transformation. S * is the region in the uv-plane corresponding to the
region S in the xy-plane. Also J must be of one sign in S * .
Example
Find the moment of inertia of a hollow circular cylinder of inner radius a, outer radius b, height h
and constant density  about the axis of the cylinder.
Sol:
y
a

b x
S
Change to polar coordinates
x  r cos
x
  x, y  r
J

 r ,  y
r

MI    h x 2  y 2 dS
y  r sin 
x
  cos
y
sin 

i.e. dxdy  rdrd 
This corresponds to an element of area in the r - plane.
6
 r sin 
r
r cos
The region of integration is now a rectangle in the r - plane, 0    2 , a  r  b
MI  
b

2
r  a  0
 h 2
hr 2 rdrd  h 
2  r 4  b
0  4  a
b4  a 4 M 2


a  b 2  where mass  M  h b 2  a 2 
4
2
Example
yx
y x
 e dxdy , where Q is bounded by x  y  2, x  0, y  0 .
Evaluate the integral
Q
Sol:
The presence of y  x & y  x in the integral suggests the change of variables
vu
vu
,y 
.
2
2
Under the transformations u  y  x & v  y  x , the lines x  y  2, x  0, y  0 map onto the
u  y  x & v  y  x . Solving for x and y we find x 
lines v  2, u  v  0, u  v  0 , respectively. Points inside Q are carried into points inside the
triangular region T bounded by v  2, u  v  0, u  v  0 .
 1

 ( x, y )
The J 
 det  2
(u, v)
 1
 2
1

1
1
2    1 or J   ( x, y )  1



1
 (u , v)  (u , v)
2
  1 1  2


det 
2
 ( x, y )
 1 1
Therefore
 e
Q
yx
y x
2 v u
2
1
1  v 
1 
1
1
dxdy   e  dudv     e du dv   v e  dv  e 
2
2 0 v
20 
e
e
T

u
v
Example
Evaluate

R
1
x
2
y

p
2 2
dxdy over the region R: x 2  y 2  1 .
7
Sol:
1

Consider
x
Rk
y
2

p
2 2
dxdy where Rk : 1  x 2  y 2  k 2 .
Then

R
1
x
2
y

p
2 2

dxdy  lim
k 
Rk
1
x
2
y

p
2 2
dxdy if the limit exists.
Let x  r cos  , y  r sin  , where 1  r  k ,0    2 .
 x

 ( x, y )
 det r
The J 
 ( r , )
 x

 
y 

r   det cos
  r sin 
y 


 
sin  
r
r cos 
Then
k
2 r

1 p
R 2 2 p dxdy  1  0 r p d dr  1 2r dr
k x
 y 2
k
1
r 2 p
 2
2 p
2

k 2 p  1
2 p

k
1
Therefore,

R
1
x
2
y

p
2 2
So, if p  2 ,
k 

R
If p  2 ,

R

dxdy  lim
x
x
1
x
2
y
1
2
Rk
1
y

p
2 2

p
2 2
2
y

p
2 2
dxdy 
dxdy  lim
k 
2
2 
1


k 2 p  1 
 lim p 2  1
k


2 p
2 p
k

2
p2
.
dxdy diverges, that means

R
1
x
2
y

p
2 2
Example
Evaluate

R

log x 2  y 2
x
2
y

1
2 2
dxdy
over the region R: x 2  y 2  1 .
Sol:
8
dxdy does not exist.
We observe that the function f ( x, y ) 

log x 2  y 2
x
2
y


1
2 2
is not defined at the point (0,0).
log x 2  y 2 
dxdy over the region R : 0   2  x 2  y 2  1 .
Consider 
1
R x 2  y 2 2
Then

R

log x 2  y 2
x
2
y

1
2 2
dxdy  lim
 0 

R

log x 2  y 2
x
y
2

1
2 2
dxdy
Let x  r cos  , y  r sin  , where   r  1,0    2 . We have

R

log x 2  y 2
x
2
 y2

1
2
dxdy 
1
2 log r 2

2
  0 r rd dr   2 log( r )dr
1
1
1




1
 2 r log( r 2 ) 1   rd log( r 2 )   2   log(  2 )   r 2 2rdr 
r






2
2
 2   log(  )  (2  2 )  2 2   log(  )  2






As a result,
log x 2  y 2 
log x 2  y 2 
dxdy

lim
dxdy  lim 2 2   log(  2 )  2  4
1
R 2 2 1
  0  
 0
R x 2  y 2 2
x  y 2
Remark: The above two examples are the so called improper integrals.
Example

Evaluate
e
 x2
dx .
0
Sol:

Suppose

x
x
 e dx really exists and let I   e dx .
2
2
0
y
0
  2   2 
Then I 2    e  x dx   e  y dy  .
0
 0


Since I   e  x dx is a number (a constant) and
2
0
x

e
0
9
 y2
x
dy is a definite integral. We learnt the rule
b
b
a
a
that k  f ( x)dx   kf ( x)dx where k is a constant.
   x 2    y 2      x 2   y 2
Therefore we have I    e dx   e dy      e dx e dy
0
 0
 00

2

 2  2
2
2
Also   e  x dx e  y  e  y  e  x dx as both
0
0


x
y
 e dx & e
2
2
are numbers.
0

   x 2    y 2      x 2   y 2
   x2 
 y2
Thus, I    e dx   e dy      e dx e dy   e   e dx dy .
0
0
 0
 00

0

2
b
b
a
a
Using the rule k  f ( x)dx   kf ( x)dx again, we have


 
  2   2     2  2



2 
2
2
2
I 2    e  x dx   e  y dy      e  x dx e  y dy   e  y   e  x dx dy     e  y e  x dx dy
0
00
0
 0
 00

0


 


2
2
    e ( x  y ) dx dy
00

  ( x 2  y 2 ) 
dx dy is equivalent
0  0 e


to integrating the function f ( x, y)  e ( x
2
 y2 )
over the region
R of the first quadrant of xy- plane.
From the transformations x  r cos  , y  r sin  ,which are from r -plane to xy-plane, we have
the inverse transformations r  x 2  y 2 ,  tan 1
And
the
transformations
y
, which are from r -plane to xy-plane.
x
r  x 2  y 2 ,  tan 1
y
x
map
the
quarter
circle x 2  y 2  r02 , x, y  0 to r  r0 , therefore map the region R of the first quadrant of xy- plane
onto the region R bounded by r  0,  0, 

2
So
10
.



 2



 ( x, y )
2
( x 2  y 2 )
r 2
r
I     e
dx dy     e rd dr where J 
 ( r , )

00
00




And
 2


 r 2

 r 2
  r 2
 2
e
r
d

dr

e
r
dr


e d (r 2 )   e  r


0  0


2
40
4

0


 1 
 1   
4   4

1
(Remark:
e
2
 lim
a
1
ea
2

0

 1

  2  1
4 e

)
Finally, we have

   x 2    y 2  
2


I    e dx   e dy     e  x dx 

4
2
0
0
 0
 4
2
Example
Evaluate the double integral

y  x 2 dxdy where Q  [1,1]  [0,2] .
Q
Sol:
 x 2  y for  1  x  1,0  y  x 2
f ( x, y )  
2
2
 y  x for  1  x  1, x  y  1
2
y  x dxdy    
1  0
1

2
Q
1 x
2


2
y  x dy dx     x  y dy   y  x 2 dy dx
0
1 

x2

2
2
2
x

    x 2  y dy   y  x 2 dy dx

1 
x2
0
1
2
For the first integral we make the change of
t  y  x 2 . This gives us
variable t  x 2  y and in the second we put
11
2
1 
0
2 x
x


2
2
y

x
dxdy

x

y
dy

y

x
dy
dx


t
dt

t
dt



Q
  0

  2
0 dx
1 
1 
x2
x


3
1
1
3
 2 3
2 
2

2
2
   t 2 0x 2  t 2 02 x dx    x 3  2  x 2 2 dx
3
3
3
3

1 
1 

2
1
2
2

1



3
4
  2  x 2 2 dx
30
From the integration table we have
1
3
4
1
2 2
2

x
dx   x 2  x 2

30
3




 x  1
 3x 2  x 2  6 arcsin 

 2  0

3
2
4 

3 2
Or let x  2 sin  then dx  2 cosd . When x  0,  0; x  1, 

4
So

1




3
4
44
2 2
2

x
dx

2  2 sin 2 


30
30

4



16
  cos 2 
3 0

3
2


3
2
2 cos d

2
16
16
16 4  cos 2  1 
cos d   cos 4  d   cos 2  d   
 d 
3 0
3 0
3 0
2

4



44 3
2 cos d   2 2 cos 2 
30
3
2
4

2


4
4 4  cos 4  1

2
cos
2


2
cos
2


1
d


 2 cos 2  1d



30
3 0
2

4




4
4 
 2 cos 4 8

1

 
 cos 2  2 d   sin 4  sin 2  2  04  
3
3
3
3 2

6

0
4
6.3 Triple Integrals
Suppose that a scalar field f ( x, y , z ) is defined at all points xi , yi , zi  within a region V of
three-dimensional space. Divide V into N subregious such that the ith subregion has volume  Vi
and contains a point ( xi , yi , z i ) . We define the triple integral of f ( x, y , z ) over the region V to be
 
V
f  x, y, z dV   
V

N
f  x, y, z dxdydz  lim  f  xi , y i , z i VL
N 
Such an integral is often called a volume integral.
12
i 1
Examples
(i)
If f ( x, y, z )  1 , then

V
dxdydz = volume of the region V.
(ii) If f ( x, y, z )   , the scalar field giving the density at a point, then

V
 dxdydz = mass of the region V.
(iii) If f ( x, y, z )   d 2 where d is the perpendicular distance of a point in the region V to a fixed
axis, then

V
 d 2 dxdydz = moment of inertia of the region V about the given axis.
(iv) If f ( x, y, z )   is the charge density at a point, then

V
 dxdydz = total charge within the region V.
Example (Rectangular regions)
The density of a rectangular blocks bounded by the planes x = 1, x = 2, y = 0, y = 3, z = -1 and z = 0
is given by the scalar function  ( x, y.z )  x( y  1)  z . Find the mass of the block.
z
3
0
y
1
2
-1
x
x( y  1)  z dzdydx  x 1  y 0  z 1 x( y  1)  z dz dy dx
x 1 y 0 z  1
Mass  
2
3
0
2
3

0


2  3 
2  3 
z2  0 
1 
     x ( y  1) z   dy dx      x ( y  1   dy dx
x 1
y 0
x

1
y

0
2  1 
2 




2
2  xy
2 9x
1 3
3
3  2 51
15
 
 xy  y  dx     3x  dx   x 2  x  
x 1
x 1
2 0
2
2 1 4
2
4
 2
13

Example (Non-rectangular regions)
z
z   2 ( x, y )
z   1 ( x, y )
y
x  a1
x  a2
y   1 ( x)
y   2 ( x)
x
Suppose V is bounded above by the surface z   2 ( x, y) and below by z   1 ( x, y) , and its
projection onto the xy-plane is described by a1  x  a2 , 1 ( x)  y   2 ( x) .
Then,

f ( x, y, z )dxdydz 
V
CONSTANT
 2 ( x )  2 ( x , y )
 
f
(
x
,
y
,
z
)
dz


dy dx




1 ( x , y )
x a1  y 1 ( x ) 
 
a2
FUNCTIONS
OF x IN
GENERAL
FUNCTIONS OF
x AND y IN
GENERAL
Similarly, it may be more convenient to project V onto the xz- or yz-planes.
Example
Evaluate
 
V
1
dxdydz where V is the region enclosed by the planes x = 0,
( x  y  2 z  1) 3
y = 0, z = 0 and x + y + z = 1.
14
z
y  z 1
1
x  z 1
z 1 x  y
o
1
y
y 1 x
1
x
Sol:
The projection of V on xy-plane is  xy which is bounded by x  0, y  0, y  1  x .
1 1 x 1 x  y
 
1 x  y

 
1
1
dz
dxdy

dz dy  dx
  
 

3
3


( x  y  2 z  1)
 0  0 ( x  y  2 z  1)
 xy  0
0 

 
1 1 x
1 x 
 
1  x  y 
 
1
1
1
  
dy  dx     

dy  dx
2
2
2
0
4( x  y  2 z  1) 
4( x  y  1)
4( 3  x  y )  
0  0 
0  0 

1
1

1  x
 1 1
1
1
1
1 
 

dx      

dx

4( x  y  1) 4(3  x  y )  0
8 8 4( x  1) 4(3  x ) 
0 
0 
1
1
 x 1
1 1
    log  x  1  log 3  x   log 3  1
4
 4 4
0 4
Example
Evaluate
   2 xyzdxdydz
where V is the region bounded by the parabolic cylinder z  2 
V
and the planes x = 0, y = x and y  0 .
15
1 2
x
2
z
y=x
y0
z  2
1 2
x
2
2
The region V
y
2
x
z=0
Sol:
Projecting V onto the xz-plane gives  xz which is bounded z  2 


 

x
0
xz
2 xyzdy  dxdz  


xz
1 2
x , x  0, z  0
2
2  12 x 2

2
x

xy 2 z dxdz   x 3 zdxdz     x 3 zdz  dx
0
0
 xz
 0



2
 3
1 2 
x
2

x
 
 
1 2
3 2
2x z 2
2
4
2  
x


 
dx

dx 
2


0
0


2
3
 2 
0




Alternatively, projecting V onto the xy-plane gives  xy which is bounded
2
 
42 z
z 0 x 0

x
y 0
2 xyzdydxdz 
4
3
Example
Evaluate
 xyzdxdydz
, where V is the region enclosed by
V
x 2  y 2  z 2  1, where x  0, z  0, y  0 & x  0, y  0, z  0
16
Sol:
The
projection
of
x 2  y 2  1, where x  0,
V
onto
xy-plane
is
 xy
which
is
bounded
by
y  0 & x  0, y  0
So


xyzdxdydz




V
 xy

1

dxdy  
xyzdz
0
0 



1 x 2  y 2
1 x 2

0




 
dy dx
xyzdz
0
 
 
1 x 2  y 2
Then
 1 x 2  1 x 2  y 2  
1  1 x 2
z 2 1 x 2  y 2 





xyzdxdydz   
xyzdz dy dx    xy
dy dx
0

 
 

2


V
0
0 0
 
 0  0
1  1 x 2
1  1 x 2
 1 x2  y2  
 xy  x 3 y  xy3  
dy dx     
dy dx
     xy
2
2





 
0 
0
0
0



1

 

1
 ( x  x 3 ) y 2 xy 4  1  x 2
(x  x3 ) 1  x 2
x 1 x2

  

dx  

dx
4
8  0
4
8
0
0
1
1

1  x   x  x
2
4
0


3

x 1 x2

2
dx  1  x   x  x
1



0
2
4


3
2
2
1

x  2x3  x5
dx

dx

0
8

 x2 x4 x6  1
1
  
  0 
48
 16 16 48 
Example
Find the (a) volume V and (b) centroid ( x , y , z ) of the region R bounded by the parabolic cylinder
z  4  x 2 and the planes x  0, y  0, y  6, z  0 assuming the density to be a constant   0
Sol:
(a)
2 6
2
 6 4 x 2  


2
V  1dxdydz       1dz dy dx     4  x dy dx   4  x 2 y 60 dx
R
0 
0 0
0

 0  0  
2
2




  24  6 x 2 dx  32
0
(b)
Total mass m   dxdydz  32
R
The centroid ( x, y, z ) of the region R:
17


 6 4 x 2  
xdxdydz       xdzdy dx

0 
Total moment about yz plane
 
 0  0
x
 R

Total mass
32
32
2 6
2
2


    x 4  x 2 dy dx   6 x 4  x 2 xdx   6 x 4  x 2 xdx
24 3
  0
 0 0
 0


32
32
32
32 4
2




Total moment about xz plane
y

Total mass
z
Total moment about xy plane

Total mass

 ydxdydz
R
32
 zdxdydz
R
32


96
3
32
256

8
 5

32
5
3 8
Thus the centroid ( x, y, z ) of the region R: has coordinates  ,3,  .
 4 5
6.4 Change of Variable in Triple Integrals
If I   f ( x, y , z )dxdydz , the change of variable x  x(u, v, w), y  (u, v, w), z  z (u, v, w) , gives
V
I   F (u, v, w)
V
  x, y , z 
dudvdw , where
 u, v, w 
 x

 u
 ( x, y , z )  y
J

 (u, v, w)  u
 z

 u
x
v
y
v
z
v
x 

w 
y 
w 
z 

w 
is the Jacobian of the transformation. V  is the region in uvw-space corresponding to the region V
in xyz-space. J must be of one sign in V  .
The most popular alternative coordinate systems to Cartesian coordinates are cylindrical polar
coordinates and spherical polar coordinates.
18
Cylindrical Polar Coordinates
z

r , , z 
x  r cos
y  r sin 
z
zz
y
r
r  0,
0    2 ,
  z  
x
r
constant – cylinder

constant – plane
constant – plane
z
cos
  x, y , z 
J
 sin 
 r , , z 
0
 r sin 
r cos
0
0
0 r
1
dV  dxdydz  rdrddz
This corresponds to an element of volume in cylindrical polar coordinates.
z
z
r

r

z
y
r
x
V  rrz
19
Example
Find the volume between the surfaces x 2  y 2  2, z  x 2  y 2  2 and the z = 0 plane.
z
z  x 2  y 2  2 parabolic bowl
x 2  y 2  2 cylinder
y
x
Sol:
 2 x 2  x 2  y 2 2  
In Cartesian coordinates we have      1dz dy dx while in cylindrical polar coordinates:
 
 2 
  2 x 2  0
2
Volume  
2
r 0
  rdzddr    r
2
r2 2
2
0 z 0
r 0
2
0
3
 2r dd r  6
Example
Using cylindrical polar coordinates, evaluate
 x
2

 y 2 dxdydz , where V is the solid enclosed
V
by x 2  y 2  2 z, z  2
Sol:
The projection of V onto xy-plane is  xy which is bounded by x 2  y 2  4 .
So
 
 
2  2
2 2
 



r2
x 2  y 2 dxdydz       r 2 rdz d dr     r 3  2 
2
 

0 0
0 0
 r2
2




 
V

2

r2
  2r 3  2 
2

0
2
 
d dr
 


r 6  2
64 32 16
dr   r 4 
 0  16 


6 
6
6
3


20
Spherical Polar Coordinates
x  r sin  cos 
y  r sin  sin 
z  r cos
z
r , ,  
r  0,
r

0   ,
0    2
y

x
r  constant - sphere
– cone
constant

- plane
constant

sin  cos 
  x, y , z 
J
 sin  sin 
 r , ,  
cos
2
 r sin 
r cos cos 
r cos sin 
 r sin 
 r sin  sin 
r sin  cos 
0
dV  dxdydz  r 2 sin drdd
This corresponds to an element of volume in spherical polar coordinates.
z

r
r sin 
r

x

V  r 2 sin r
r
y
Example
In a sample model of the charge distribution around the positively charged (Q) nucleus of the
hydrogen atom the charge density in the electron cloud at a distance r from the atom is
21
Q a
 (r) 
e where a is the Bohr radius. Determine the total charge in the electron cloud.
 a3
Sol:
2r
 
2r
   2  Q  2 r 2
 


Q a 2
a


q       3 e r sin d d dr     2 3 e r sin  d dr
a
 0  0 a
0 
0 0
 



2r

2r



  2 r 
Q 
4Q 
4Q  a
    2 3 e a r 2 cos   0 dr   3  e a r 2 dr   3    r 2 d  e a 
a
a 0
a  2 0
0




2r
2Q  2  a
r
e

a2 

0

e

2r
a
0

d r 2 

Note: Differentiate with respect to b twice the top and the bottom of
b2
2b
ea
r 2e

2r
a 
0
 lim
b2
b 
e
2b
a
 lim
b 
2b
2
e
a
 lim
2b
a
b 
2
2b
4 a
e
a2
0
Therefore,
2Q  2  a
r e
a2 
2r
q


0

e

2r
a
0

2r

2r

2r


2Q 
2Q
4Q  a    
d r 2    2  e a d r 2    2  2re a dr   2    rd  e a 
a 0
a 0
a 0 2 



  2ar  2Q   2ar
2Q
e  
r
d
re

a 0 
a 



2Q  a 1

a  2 2ar
e


0

0

e

0
2r
a

2r
2r

2Q  a
2Q  a  a

dr   
e
dr



e
a 0
a  2


0





   2Q  0  a   Q



a 
2

Note The vector operators ,   and   may be expressed in terms of other coordinates. For
example, in spherical polar coordinates,
 

1 
1 
ir 
i 
i
r
r 
r sin  
And
 2 
1   2  
1
 
 
1
 2
r

sin






  r 2 sin 2   2
r 2 r  r  r r sin   
22