vector

Vectors 459
Example 114. Using the divergence theorem, evaluate the surface integral
( yz dy dz zxdz dx xy dy dx)
where S : x 2 + y 2 + z 2 = 4.
S
Solution. ( f1 dydz f2 dx dz f3
dxdy)
S
f1 f2
f3
=
dx dy dz
v
x y z
where S is closed surface bounding a volume V.
( yz dy dz zx dx dz xy dx dy)
S
(AMIETE, Dec. 2010, UP, I Sem., Dec 2008)
( ) ( ) ( )
= yz zx xy
dx dy dz
v
x y z
= (0 0 0) dx dy dz
v
= 0 Ans.
2 2 3 2
Example 115. Evaluate xz dy dz ( x y z ) dzdx (2 xy y z)
dxdy
S
where S is the surface of hemispherical region bounded by
z =
2 2 2
a x y and z = 0.
=
Solution. ( f1 dydz f2 dz dx f3
dxdy)
S
where S is a closed surface bounding a volume V.
=
S
2 2 3 2
xz dy dz ( x y z ) dzdx (2 xy y z)
dxdy
V
V
f1 f2
f3
dx dy dz
x y z
2 2 3
2
( xz ) ( x y z ) (2 xy y z)
dxdydz
x y z
(Here V is the volume of hemisphere)
2 2 2
= ( z x y ) dx dy dz
V
Let x = r sin cos , y = r sin sin , z = r cos
=
=
Example 116. Evaluate
2 2
r ( r sin dr d d )
=
5
a
2 /2
r
( ) 0 ( cos )
0
5
0
=
2 a
2 sin
0 0 0
d d r dr
5
2 ( 0 1) 5
a
=
2a
5
4
5
Ans.
F . n ˆ ds over the entire surface of the region above the xy-plane
S
bounded by the cone z 2 = x 2 + y 2 2
and the plane z = 4, if F = 4xz iˆ
xyz ˆj 3 zkˆ
.
Solution. If V is the volume enclosed by S, then V is bounded by the surfaces z = 0, z = 4,
z 2 = x 2 + y 2 .
By divergence theorem, we have
=
F . n ˆ ds
S
=
=
V
V
V
div Fdxdydz
2
(4 xz) ( xyz ) (3) z dxdydz
x y z
2
(4z xz 3) dxdydz
Limits of z are
2 2
x y and 4.