vector

Vectors 433
Solution. Statement: See Article 24.4 on page 576.
Here the closed curve C consists of straight lines OB, BA and AO, where coordinates of A and
B are (3, 0) and (0, – 2) respectively. Let R be the region bounded by C.
Then by Green’s Theorem in plane, we have
2 2
[(3 x –8 y ) dx (4 y –6 xy) dy ]
=
x
y
2 2
(4 y –6 xy)– (3 x –8 y ) dxdy
R
...(1)
= (– 6y 16 y) dx dy 10 ydxdy
R
2
3 0 3 y
= 10 dx 1 ydy 10 dx
x
2
0 (2 –6) 0
3 1
(2 x –6)
3
=
3
5 (2 x – 6) 5 3
= – – (0 6)
9 3
2 54
0
Now we evaluate L.H.S. of (1) along OB, BA and AO.
Along OB, x = 0, dx = 0 and y varies form 0 to –2.
Along BA, x = 1 (6 3 y),
dx 3 dy and y varies from –2 to 0.
2 2
and along AO, y = 0, dy = 0 and x varies from 3 to 0.
L.H.S. of (1) =
=
OB
2 2
[(3 x – 8 y ) dx (4 y –6 xy) dy]
3
2 2 2 2
BA
R
0
=
5 3
– (2 – 6)
9
dx x
0
5
– (216) –20 ...(2)
54
[(3 x – 8 y ) dx (4 y –6 xy) dy] [(3 x – 8 y ) dx (4 x –6 xy) dy ]
AO
2 2
[(3 x – 8 y ) dx (4 y –6 xy) dy ]
–2 0 3 2 23
0
2
= 4 ydy (6 3 y) –8 y dy [4 y –3(6 3 y) y] dy 3x dx
0 –2
4
2
3
=
2 –2 0 9
2 2 2 3 0
[2 y ]
0
(63 y) –12y 4 y –18 y –9 y dy ( x )
3
–2
8
0 9
2 2
= 2[4]
(6 3 y) –21 y –14 y dy (0 –27)
–2
8
0
3
9(6 3 y) 3 2
216
3 2
=
8 – 7 y – 7 y –27 –19 7(– 2) 7(–2)
8 3 3 8
–2
= – 19 + 27 – 56 + 28 = – 20 ...(3)
With the help of (2) and (3), we find that (1) is true and so Green’s Theorem is verified.
2 2 2 2
Example 82. Apply Green’s Theorem to evaluate [(2 x y ) dx ( x y ) dy],
where C
is the boundary of the area enclosed by the x-axis and the upper half of circle x 2 + y 2 = a 2 .
(M.D.U. Dec. 2009, U.P., I Sem., Dec. 2004)
2 2 2 2
Solution. [(2 x y ) dx ( x y ) dy]
C
By Green’s Theorem, we’ve ( dx
dy)
=
C
S
C
dx dy
x
y
2