vector

456 Vectors
Changing to spherical polar coordinates by putting
x = r sin cos , y= r sin sin , z = r cos , dV = r 2 sin dr d d
To cover V, the limits of r will be 0 to 1, those of will be 0 to 2
and those of will be 0 to
2.
2
2
dV =
V zy
=
=
=
2 /2 1 2 2 2 2
2 ( r cos )( r sin sin ) r sin drd d
0 0 0
2 /2 1 5 3 2
2 r sin cos sin
drd d
0 0 0
6
1
2 2
3 2
r
0
0
6 0
2 sin cos sin
d d
2 2
=
2
2
sin
6 0 4.2 d
1
12
2
2
sin d
=
0
12
Example 108. Use Divergence Theorem to evaluate FdS
2
where F 4 xiˆ
–2y ˆj z
2ˆ k
S
and S is the surface bounding the region x 2 + y 2 = 4, z = 0 and z = 3.
(A.M.I.E.T.E., Summer 2003, 2001)
Solution. By Divergence Theorem,
FdS
= div FdV
S V =
V
ˆ
ˆ
ˆ
ˆ 2ˆ
2
i j k (4 xi 2 y j z kˆ
) dV
x y z
= (4 4y
2 z)
dx dy dz
=
V
3
dx dy (4 4y 2 z)
dz =
= (12 12y
9)
dxdy
Let us put x = r cos , y = r sin
= (21 12r sin )
r d
dr
=
0
= (21 12 )
=
2
2
2
2 2
21r
3
d 4r
sin
2
0
=
0 0
0
2 2
0
2 3
dx dy [4z 4 yz z ] 0
y dxdy
d (21r 12r sin )
dr
d (42 32 sin ) (42 32 cos )
2
0
Ans.
= 84 + 32 – 32 = 84 Ans.
^
Example 109. Apply the Divergence Theorem to compute u nds, where s is the surface of
the cylinder x 2 + y 2 = a 2 bounded by the planes z = 0, z = b and where u ix ˆ – ˆjy kz ˆ .
Solution. By Gauss’s Divergence Theorem
u nds
ˆ = ( u)
dv
V
ˆ
ˆ
=
ˆ
ˆ
i j k ( ix ˆ ˆ
jy kzdv )
V
x y z
= x y z
dv
V
x y z
= 111
dv
V
= dv dx dy dz
V = Volume of the cylinder = a 2 b Ans.
=
V