vector

Vectors 457
Example 110. Apply Divergence Theorem to evaluate
F . ˆ , where
V
F
4x 3 iˆ x 2 y ˆj x 2 zk ˆ and S is the surface of the cylinder x 2 + y 2 = a 2 bounded by the
planes z = 0 and z = b.
Solution. We have,
(U.P. Ist Semester, Dec. 2006)
F
4x 3 iˆ
x 2 y ˆj x 2 zk ˆ
ˆ
3 2 2 ˆ
div F =
ˆ
ˆ
i j k (4 x iˆ x yj ˆ x zk)
x y z
3 2 2
= (4 x ) ( x y) ( x z)
x y z
= 12x 2 – x 2 + x 2 = 12 x 2
Now,
V
div FdV
=
2 2
12 a a
x b
xay a
2
x
2
z 0
x
2
dzdydx
a a
2
x
2
2 b
xa y a
2
x
2
=
= 12 x ( z)
0 dydx
a
a
a 2 2
2 a
x
a a
2
x
2
12 b x ( y)
dx
2 2 2
2 2 2
= 12 b x .2 a x dx = 24 b
x a x dx
=
=
a
2 2 2
48 b x a x dx
0
/2 2 2
48 b a sin a cos a cos d
0
a
a
[Put x = a sin , dx = a cos d]
3 3
4 /2
2 2
= 48 ba sin cos
d
4
= 48 ba
2 2
0
23
1 1
4 2 2
= 48 ba
= 3 b a 4 Ans.
22
Example 111. Evaluate surface integral Fnds ˆ ,
where F = (x 2 + y 2 + z 2 ) iˆ
ˆj kˆ
( ), S
is the surface of the tetrahedron x = 0, y = 0, z = 0, x + y + z = 2 and n is the unit normal in
the outward direction to the closed surface S.
Solution. By Divergence theorem
Fnds
ˆ = div Fdv
S V
where S is the surface of tetrahedron x = 0, y = 0, z = 0, x + y + z = 2
=
i ˆ
ˆ j k ˆ 2 2 2
( x y z )( i ˆ ˆ j k ˆ)
dv
V
x y z
= (2x 2y 2 z)
dv
V
= 2 ( x y z)
dx dy dz
=
=
V
2 2x 2 x y
2 dx dy ( x y z)
dz
2
0 0 0
2
2 x
y
2 2 x
z
dx dy
xz
yz
2
0 0
0