vector

438 Vectors
= Fdx .
Curl F =
iˆ
ˆj kˆ
x y z
yz zx xy
= curl F.
ds
= (x – x) î + (y – y) ĵ + (z – z) ˆk
= 0 = 0 Ans.
Example 86. Using Stoke’s theorem or otherwise, evaluate
2 2
[(2 x y) dx yz dy y z dz]
c
where c is the circle x 2 + y 2 = 1, corresponding to the surface of sphere of unit radius.
(U.P., I Semester, Winter 2001)
2 2
Solution. [(2 x y) dx yz dy y z dz]
c
2 2 ˆ ˆ
= [(2 x y) iˆ yz ˆj y z k] ( iˆdx ˆjdy k dz)
By Stoke’s theorem
Curl F =
c
F
d r = Curl F n ds
S
...(1)
iˆ
ˆj kˆ
F
=
Putting the value of curl F in (1), we get
= ˆ
ˆ
= kˆ nˆ
ds
x y z
2 2
2x y yz y z
= (– 2 yz 2 yz) iˆ
–(0–0) ˆj (0 1)
kˆ kˆ
dx dy
dx dy
k n = dx dy
nk ˆ ˆ = Area of the circle =
ds
( nk ˆ ˆ
)
Example 87. Evaluate F . d r, where F ( x,
y i xj z kand C is the curve of
C
intersection of the plane y + z = 2 and the cylinder x 2 + y 2 = 1. (Gujarat, I sem. Jan. 2009)
2 2
Solution. F. dr curl F . nds ˆ curl (– y iˆ
x ˆj z kˆ
) nds ˆ
...(1)
Normal vector =
C S S
2 2 ˆ
F (x, y, z) =
Curl F =
F
2 2
, z) – y ˆ ˆ ˆ
y iˆ xj ˆ z k
(By Stoke’s Theorem)
iˆ
ˆj kˆ
x y z
2 2
– y x z
= iˆ(0 –0)– ˆj (0 –0) kˆ(12 y) (1
2 y)
kˆ
= ˆ
ˆ
ˆ
i j k ( y z – 2) ˆj kˆ
x y z
ˆj
kˆ
Unit normal vector ˆn =
2
dx dy
ds =
ˆ . kˆ
O
3y + z = 2
1
x 2 + y 2 = 1