vector

450 Vectors
Line Equ. Lower Upper Fdr .
1 OA
2 AB
3 BC
4 CO
of line limit limit
y 0 dy 0
z 0 dz 0
x = 0 x = 2 2 dx
x 2 dx 0
z 0 dz 0
y = 0 y = 2 4 dy
y 2 dy 0
z 0 dz 0
x = 2 x = 0 4 dx
x 0 dx 0
z 0 dz 0
y = 2 y = 0 4 dy
Fdr =
OA
(2) Along AB, x = 2, dx = 0, z = 0, dz = 0
AB
Fdr =
(3) Along BC, y = 2, dy = 0, z = 0, dz = 0
F dr =
BC
(4) Along CO, x = 0, dx = 0, z = 0, dz = 0
CO
2
2
2 dx [2 x]
0 = 4
0
2
2
4dy
4( y)
0 = 8
0
2
0
(2 0 2) dx (4 x)
2 = – 8
0
Fdr = ( y 0 2) 0 (0 4) dy 0
0
= 4 dy 4( y) 2 = – 8
On putting the values of these integrals in (1), we get
Fdr = 4 + 8 – 8 – 8 = – 4
c
To obtain surface integral
iˆ
ˆj kˆ
F
=
x y z
y z 2 yz 4 xz
= (0 – y) î – (– z + 1) ĵ + (0 – 1) ˆk = – y î + (z – 1) ĵ – ˆk
Here we have to integrate over the five surfaces, ABDE, OCGF, BCGD, OAEF, DEFG.
Over the surface ABDE (x = 2), ˆn = i, ds = dy dz
( F)
nds ˆ = [ yi ( z 1) j k]
idx dz ydydz
2
=
R
( , )
3 ( , , ) z f x y
z f ( x, y)
F x y z dx dy
1