Mathematical Methods for Physics and Engineering : A Comprehensive Guide

LINE, SURFACE AND VOLUME INTEGRALS

which shows that we requirea·drto be an exact differential: condition (iv). From

(10.27) we can writedφ=∇φ·dr, and so we have

(a−∇φ)·dr=0.

Sincedris arbitrary, we find thata=∇φ; this immediately implies∇×a= 0 ,

condition (iii) (see (10.37)).

Alternatively, if we suppose that there exists a single-valued function of position

φsuch thata=∇φthen∇×a= 0 follows as before. The line integral around a

closed loop then becomes
∮

C

a·dr=

∮

C

∇φ·dr=

∮

dφ.

Since we definedφto be single-valued, this integral is zero as required.

Now suppose∇×a= 0. From Stoke’s theorem, which is discussed in sec-

tion 11.9, we immediately obtain

∮
Ca·dr=0;thena=∇φanda·dr=dφfollow
as above.

Finally, let us supposea·dr=dφ. Then immediately we havea=∇φ,andthe

other results follow as above.

Evaluate the line integralI=

∫B

Aa·dr,wherea=(xy

(^2) +z)i+(x (^2) y+2)j+xk,Ais the
point(c, c, h)andBis the point(2c, c/ 2 ,h), along the different paths
(i)C 1 ,givenbyx=cu,y=c/u,z=h,
(ii)C 2 ,givenby 2 y=3c−x,z=h.
Show that the vector fieldais in fact conservative, and findφsuch thata=∇φ.
Expanding out the integrand, we have

I=

∫(2c, c/ 2 ,h)

(c, c, h)

[

(xy^2 +z)dx+(x^2 y+2)dy+xdz

]

, (11.7)

which we must evaluate along each of the pathsC 1 andC 2.
(i) AlongC 1 we havedx=cdu,dy=−(c/u^2 )du,dz= 0, and on substituting in (11.7)
and finding the limits onu,weobtain

I=

∫ 2

1

c

(

h−

2

u^2

)

du=c(h−1).

(ii) AlongC 2 we have 2dy=−dx,dz= 0 and, on substituting in (11.7) and using the
limits onx,weobtain

I=

∫ 2 c

c

( 1

2 x

(^3) − 9
4 cx
(^2) + 9
4 c
(^2) x+h− 1 )dx=c(h−1).
Hence the line integral has the same value along pathsC 1 andC 2. Taking the curl ofa,
we have
∇×a=(0−0)i+(1−1)j+(2xy− 2 xy)k= 0 ,
soais a conservative vector field, and the line integral between two points must be