Example 1: Verify the divergence theorem for the vector …eld
F = x i + y j + z k over the sphere V of radius a centred at the origin x
2
+ y
2
+ z
2
= a
2
:
Solution:
(a) Divergence integral
The divergence of F is
@ r F =
@x
@
( x ) +
@y
@
( y ) +
@z
( z ) = 1 + 1 + 1 = 3 so
I
1
=
ZZZ
V r F dV =3
ZZZ
V dV:
Method 1: Note: the value of the integral
ZZZ
V
1 dV = volume of region V =
4
3 a
3
: use this fact without proof.
Method 2: Relative to spherical polar coordinates the limits are constants and dV = r 2 sin drd d
ZZZ
V dV =
=
=
=
=
Z Z
2
Z a r
2 sin drd d
Z
0
Z
0
2
0
1
3 r
3
0
1
3 a
3
Z
0
0
3
1
1 a
3 a
3
3
[
2 sin cos ]
2 d
0 a
0
Z sin
0
[ ]
2
0
2 d d d
Hence
I
1
= 3
1
3 a
3
4 = 4
(a) Flux integral
Let
F ( x; y; z ) = x
2
+ y
2
+ z
2 a a
3
2
= 0 then r F = 2 x i + 2 y j + 2 z k
1

and jr F j = q
(2 x )
2
+ (2 y )
2
+ (2 z )
2
= 2 a:
Hence the outward unit normal to the sphere is n = r F jr F j
= x i + y j + z k
: a and
F n = ( x i + y j + z k ) x i + y j + z k a
On the surface S of the sphere
F n j
S
= a 2 a
= a:
= x 2 + y 2
Thus,
I
2
=
ZZ
F n dS = a
ZZ dS:
S S
Let S
1
; S
2 denote the upper and lower hemispheres then
I
2
= a
0
@
ZZ dS +
ZZ dS
1
A
= 2 a
ZZ dS
S
1
S
2
S
1 a
+ z 2
(1)
Method 1: Note: the value of the integral
ZZ
S
1
1 dS = area of S
1
=
1
2
4 a 2 : use this fact without proof.
Method 2: Relative to spherical polar coordinates r = a and the element of area is dS = a 2 sin d d
ZZ
S
1
Z
2
Z
1
2 dS = a
2 sin d d
0 0
= a
2
Z
1
2 sin d
= a
2
0
1
[ cos ] 2
0
= a
2
1 2
Z
0
[ ]
2
0
2
= 2 a
2
: d
Hence
I
2
= 2 a 2 a
2
= 4 a
3
:
2

Hence
I
1
= I
2 and the Divergence theorem is veri…ed in this example. Method 2 is the preferred method, but:
Method 3: Returning to equation (1)
From Theorem 4.5, on S
1 q
F 2
;x dS =
+ F 2
;y
+ F 2
;z j F
;z j a
= p a 2 x 2 y 2 dxdy dxdy
= jr F j dxdy = j F
;z j
2 a j 2 z j dxdy = a z dxdy so
I
2
= 2 a
ZZ
A p a 2 where A denotes the disc a x 2 y 2 dxdy z = 0 ; x
2
+ y
2 a
2
Thus, transforming to plane polar coordinates
I
2
= 2 a
2
Z
2
= 2 a
2
Z
0
2
= 2 a
2
Z
0
2
= 4 a
3
:
0
Z h
0 ad a
( a a
2
2 d d
2 )
1 = 2
2 1
= 2 i a d
0 and again
I
1
= I
2
:
Example 2: Verify Stokes’theorem for the vector …eld
F ( r ) = y i x j taken over hemisphere x
2
+ y
2
+ z
2
= 9 ; and its bounding circle x
2
+ y
2
= 9 ;
Solution: (a) Circulation integral z z
= 0 :
0
F ( r ) d r = ( y i x j ) ( dx i + dy j + dz k )
= ydx xdy:
3

The equation of the circle may be written parametrically as x = 3 cos ; y = 3 sin so where 0 dx = 3 sin d ; dy = 3 cos d
I
1
2 : Let
=
I
F ( r ) d r =
I ydx xdy
C
2
= (3 sin ) ( 3 sin d ) (3 cos ) (3 cos d )
0
Z
= 9
0
= 18 :
2 d
(b) Surface integral r F = i j k
@=@x @=@y @=@z y x 0
= 0 i 0 j + ( 1 1) k
= 2 k :
From example 1, with a = 3
Let n = x i + y j + z k
:
3
=
I
2
=
Z Z
S
( r F ) n dS
Z Z x i + y j + z k
( 2 k )
S
3
2
Z Z
= zdS
3
S dS
4

Method 1:
I
2
=
=
=
=
=
=
2
3
2
Z Z
Z
1
2 zdS
S
Z
2
3 cos
3
18
0
Z
1
2
Z
0
18
0
1
2
0 cos sin sin 2 d d
2
Z
3
2 sin d d
0
2 d
1
1
2
36 cos 2
4
0
18 :
Method 2: From Example 1
=
2
3 dS =
Z Z z
3
S
Z Z dxdy z
3 z dxdy
= 2
Z
2
Z
S
3 dxdy
= 2 d d
0
1
0
3
Z
= 2
2
2 d
2
0 0
= 9 2
So
I
1
= I
2 and Stokes’theorem is veri…ed.
Examination:
3 hour paper, two sections, Section A (TS), Section B (DH)
There are three question in each Section.
Answer 5 questions (from 6).
Please use separate booklets for each Section.
There are some formulae at the end of the paper (trig. identities, standard derivatives and integrals).
Make sure you revise enough material to be able to answer 6 questions.
For Section B:
(1) There will be 1 ques. on Coordinate systems/Partial Di¤erentiation
1 ques. on 1st order pdes
5

1 ques. on Vector Calculus
(2) Revision hints:
(a) Understand and learn important de…nitions and theorems.
In stating de…nitions and theorems explain your notation.
(b) Make sure you can do the examples given in the lectures and in the examples sheets.
6