vector

Vectors 439
On putting the values of curl F,
nˆ
and ds in (1), we get
=
=
=
F
. dr
ˆ ˆ
=
ˆ j k dx dy
(1 2 ) .
C y k
S
2 ˆj
kˆ
. kˆ
2
1
2ydxdy
(1 2 y)
dxdy 2
1
2 1 = (1 2 r sin ) rd d r
0
0
Y
2
2
1 2
( r 2r sin ) d d r
0 0
1
2 3
2
2
r
2r
1 2
d sin sin d
2 3
2 3
0 0
0
2
2 2 2
= – cos – – 0
2 3
0 3 3
= Ans.
Example 88. Apply Stoke’s Theorem to find the value of
c
( ydx z dy xdz)
where c is the curve of intersection of x 2 + y 2 + z 2 = a 2 and x + z = a. (Nagpur, Summer 2001)
Solution. ( ydx z dy xdz)
c
ˆ
ˆ
= ( ˆ ˆ ) ( ˆ ˆ
yi zj xk idx j dy kdz)
c
=
= curl ( yiˆ zj ˆ xkˆ
) nds ˆ
S
C
( yiˆ
zj ˆ xkˆ
) dr
(By Stoke’s Theorem)
ˆ
ˆ
=
ˆ
ˆ
i j k ( yiˆ zj ˆ xk)
nˆ
ds
S
x y z
= ˆ ˆ ˆ
( i j k)
nˆ
ds
S
...(1)
where S is the circle formed by the intersection of x 2 + y 2 + z 2 = a 2 and x + z = a.
ˆ
ˆ
ˆ
i j k ( x z a)
ˆn = | = x y z
|
| |
iˆ
kˆ
ˆn =
2 2
Putting the vlaue of ˆn in (1), we have
ˆ ˆ
= ( ˆ ˆ ˆ
i k
i j k)
ds
S
2 2
1 1
=
ds
S
2 2
2 2
2 2
a a
= ds
2
S
2 2 2
Example 89. Directly or by Stoke’s Theorem, evaluate
=
iˆ
kˆ
1
1
2 2
2 2 2 2 a a
Use
r R p a
2 2
Ans.
ˆ , ˆ ˆ ˆ
curl vnds v iy jz kx,
s is
the surface of the paraboloid z = 1 – x 2 – y 2 , z 3 > 0 and ˆn is the unit vector normal to s.
s
O
r ddr
X