082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

y
1
0
G
D
R
27.6 Double and triple integrals 887
Solution (a) The path of integration is shown in Figure 27.15. To apply Green’s theorem the
path of integration should be followed in such a way that the region of integration
isalways toits left.We musttherefore travel anticlockwise aroundC.
1
F
E
Figure27.15
ThepathC ischosen so
that the regionRis
always to itsleft.
x
OnDE,y=0,dy=0and0x1.
OnEF,x=1,dx=0and0y1.
On FG,y = 1, dy = 0 andxdecreases from1to0.
On GD,x = 0, dx = 0 andydecreases from1to0.
The integral around the curveC can thenbe written as
∮
C
=
∫ E
D
+
∫ F
E
+
∫ G
Therefore
∮
xydx+x 2 dy=0+
C
F
∫ D
+
G
∫ 1
0
= [y] 1 0 + [ x
2
1dy+
2
] 0
1
∫ 0
1
xdx+0
= 1 − 1 2
= 1 2
(b) Applying Green’s theorem withP(x,y) = xy andQ(x,y) = x 2 we can convert the
line integral into a double integral. Note that ∂Q ∂P
= 2x and = x. Clearly the
∂x ∂y
region of integration isthe squareR. We find
∮
∫ ∫
xydx+x 2 dy= (2x −x)dxdy
R
C
=
=
=
∫ 1 ∫ 1
0
∫ 1
0
∫ 1
0
0
[ x
2
2
1
2 dy
[ ] 1 1
=
2 y 0
xdxdy
] 1
dy
0
= 1 2
We see that the same result as that in part (a) is obtained and so Green’s theorem
has been verified.