∂φ
∂y =F^2 (x, y, z^0 ) +∫zz 0∂F 2 (x, y, z )
∂z dz
=F 2 (x, y, z 0 ) + F 2 (x, y, z )z
z 0
=F 2 (x, y, z 0 ) + F 2 (x, y, z )−F 2 (x, y, z 0 )
=F 2 (x, y, z).Thus, from the hypothesis that ∇× F = 0 ,one finds that F = grad φ=∇φ. Conse-
quently, one can say that an irrotational vector field is derivable from a potential
function φ.
Example 9-1. Show that
F = (y^2 +z)ˆe 1 + (2 xy +z^2 )ˆe 2 + (2 yz +x)ˆe 3is an irrotational vector field and find the corresponding potential function from
which F is derivable.
Solution: It is readily verified that curl F =
∣∣
∣∣
∣∣ˆe 1 ˆe 2 ˆe 3
∂
∂x∂
∂y∂
∂z
(y^2 +z) (2 xy +z^2 ) (2 yz +x)∣∣
∣∣∣∣=^0 and
hence F is irrotational. Two methods of finding the corresponding potential function
are as follows.
Method 1 By line integral integration, where the path of integration consists of
the straight-line segments illustrated in figure 9-3, one can show
φ(x, y, z)−φ(x 0 , y 0 , z 0 ) =∫xx 0(y^20 +z 0 )dx +∫yy 0(2 xy +z 02 )dy +∫zz 0(2 yz +x)dz= (y 02 x+z 0 x)x
x 0+ (xy^2 +z 02 y)y
y 0+ (yz^2 +xz )z
z 0
= (xy^2 +yz^2 +xz )−(x 0 y 02 +y 0 z 02 +x 0 z 0 )where in the second integral xis held constant and in the third integral both xand
yare held constant. The resulting integral implies
φ(x, y, z) = xy^2 +yz^2 +xz.Method 2 The components of the relation F= grad φproduce the scalar equations
∂φ
∂x =y(^2) +z, ∂φ
∂y = 2xy +z
(^2) , ∂φ
∂z = 2yz +x.
Integrating the first equation with respect to x, the second equation with respect to
yand the third equation with respect to zproduces
φ=y^2 x+zx +f 1 (y, z), φ =y^2 x+z^2 y+f 2 (x, z ), φ =xz +z^2 y+f 3 (x, y )