Mathematical Physics Vol 1

4.6 Examples

165

By solving the equation (4.195) we obtain

f ( y , z )= c 1 + g ( z ) ,

(4.196)

where c 1 is the integration constant. From (4.196) and (4.193) we obtain φ = x 2 y + xz 3 + g ( z )+ c 1 ,

(4.197)

By differentiating the equation (4.197) by z we obtain ∂ f ( y , z ) ∂ z = 3 xz 2 + ∂ g ( z ) ∂ z . From equations (4.192) and (4.198) we obtain d g ( z ) d z = 0 .

(4.198)

(4.199)

By solving the equation (4.199) we obtain

g ( z )= c 2 ,

(4.200)

where c 2 is the integration constant. By substituting g ( z ) from (4.200) into equation (4.197) we obtain the final solution φ = x 2 y + xz 3 + c ,

where c = c 1 + c 2 is an arbitrary constant. Thework A of the force F is

Q Z P

F · d r

A =

Q Z P ( 2 xy + z 3 ) d x + x 2 d y + 3 xz 2 d z

=

Q Z P

( 3 , 1 , 4 )

d ( x 2 y + xz 3 )= x 2 y + xz 3

= 202

=

( 1 , − 2 , 1 )

The work can be expressed as a difference of potentials

Q Z P

Q Z P d ϕ = φ ( Q ) − φ ( P )= 201 − ( − 1 )= 202 .

∇ ϕ · d r =

A =

(4.201)