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

898 Chapter 27 Line integrals and multiple integrals
On II,y = 2, dy = 0 and hence v·dr = (x +2)dx. Noting thatxvaries from 0
to2,the contribution tothe lineintegral is
∫ 2
[ ] x
2 2
x+2dx=
0 2 +2x = 6
0
On III,x = 2, dx = 0 and hence v·dr = (y +2)dy. Hereyvaries from 2 to 0.
This contribution tothe lineintegral is
∫ 0
[ ] y
2 0
y+2dy=
2 2 +2y = −6
2
On IV,y = 0, dy = 0 and hence v·dr = xdx. Herexvaries from 2 to 0. The
contribution tothe lineintegral is
∫ 0
[ x
2
xdx= = −2
2
2
] 0
2
Putting all these results together we find
∮
v·dr=6+6−6−2=4
C
i j k
(b) curlv =
∂ ∂ ∂
∂x ∂y ∂z
=−i−j−k
∣
x+yy+zx+z
∣
(c) Now we calculate the surface integral which must be performed over the five solid
surfaces separately. Refer to Figure 27.26. On surfaceA, the front face lying in the
planex = 2,dS = dydzi. Hence curlv·dS = −dydz. Then
∫ ∫ z=2 ∫ y=2
curlv·dS= −dydz
A
=
=
z=0
∫ z=2
z=0
∫ z=2
z=0
= [−2z] 2 0
= −4
y=0
[−y] 2 0 dz
−2dz
On B, the back face lying in the plane x = 0, dS = −dydzi. It follows that
curlv·dS = dydz. The required integral overBis
∫ z=2 ∫ y=2
z=0
y=0
dydz=4
OnC, the right-hand face, dS = dxdzj. Hence curlv·dS = −dxdz. The required
integral overC is
∫ z=2 ∫ x=2
z=0
x=0
−dxdz = −4