vector

458 Vectors
=
2
2 2 x
2 2 (2 x
y)
dx dy x x xy y xy y
0
0
2 2 2
3 3
2
2 2 2 y (2 x
y)
= 2
dx 2xy x y xy y
0
3 6
2
2
x
3 3
2
2 2 2 (2 x) (2 x)
= 2
dx 2 x(2 x) x (2 x) x (2 x) (2 x)
0
3 6
3 3
2
2 2 3 2 3 2 (2 x) (2 x)
= 2
4x2x 2x x 4x 4 x x (2 x)
0
3 6
2
3 4 3 4 3 4 4
2 4 2 4 (2 ) (2 ) (2 )
= 2 2 x x 2
x x x x x
x x
3 4 3 4 3 12 24
3 4 4
(2 x) (2 x) (2 x)
8 16 16
= 2 = 2
3 12 24 3 12 24
= 4 Ans.
0
Example 112. Use the Divergence Theorem to evaluate
S
( xdydz y dz dx zdxdy)
where S is the portion of the plane x + 2 y + 3 z = 6 which lies in the first Octant.
(U.P., I Semester, Winter 2003)
Solution. ( f1 dydz f2 dx dz f3
dxdy)
S
f1 f2
f3
=
dxdydz
V
x y z
where S is a closed surface bounding a volume V.
( xdydz y dz dx zdxdy)
S
=
V
x y z
dx dy dz
x y z
= 3 V 2
= (1 11)
dx dy dz dx dy dz
V
= 3 (Volume of tetrahedron OABC)
1
= 3[( Area of the base OAB) height OC]
3
1 1
= 3
63
2
3 2
= 18 Ans.
Example 113. Use Divergence Theorem to evaluate : ( xdydz y dz dx zdxdy)
Solution.
over the surface of a sphere radius a.
Here, we have
(K. University, Dec. 2009)
xdydz y dx dz zdxdy
S
f1 f2
f3
x y z
dx dy dz
V
dx dy dz
x y z
V
x y z
(1 + 1 + 1) dx dy dz = 3 (volume of the sphere)
V
4
3
= 3 a
3 = 4 a3 Ans.
0
0