30.5. CONSERVATIVE VECTOR FIELDS 607

30.5.1 Some TerminologyIf F = (P,Q,R) is a vector field. Then the statement that F is conservative is the same assaying the differential form Pdx+Qdy+Rdz is exact. Some people like to say things interms of vector fields and some say it in terms of differential forms. In Example 30.5.8, thedifferential form

(4x3 +2

(cos(x2 + z2

))x)

dx+dy+(2(cos(x2 + z2

))z)

dz is exact.

Definition 30.5.6 A set of points in three dimensional space V is simply connectedif every piecewise smooth closed curve C is the edge of a surface S which is containedentirely within V in such a way that Stokes theorem holds for the surface S and its edge, C.

C

S

This is like a sock. The surface is the sock and the curve C goes around the opening ofthe sock.

As an application of Stoke’s theorem, here is a useful theorem which gives a way tocheck whether a vector field is conservative.

Theorem 30.5.7 For a three dimensional simply connected open set V and F a C1

vector field defined in V , F is conservative if ∇×F = 0 in V .

Proof: If ∇×F = 0 then taking an arbitrary closed curve C, and letting S be a surfacebounded by C which is contained in V , Stoke’s theorem implies

0 =∫

S∇×F ·ndA =

∫CF ·dR.

Thus F is conservative.

Example 30.5.8 Determine whether the vector field(4x3 +2

(cos(x2 + z2))x,1,2

(cos(x2 + z2))z

)is conservative.

Since this vector field is defined on all of R3, it only remains to take its curl and see ifit is the zero vector.∣∣∣∣∣∣

i j k∂x ∂y ∂z

4x3 +2(cos(x2 + z2

))x 1 2

(cos(x2 + z2

))z

∣∣∣∣∣∣ .This is obviously equal to zero. Therefore, the given vector field is conservative. Can youfind a potential function for it? Let φ be the potential function. Then φ z = 2

(cos(x2 + z2

))z

and so φ (x,y,z) = sin(x2 + z2

)+g(x,y). Now taking the derivative of φ with respect to y,

you see gy = 1 so g(x,y) = y+ h(x). Hence φ (x,y,z) = y+ g(x)+ sin(x2 + z2

). Taking

the derivative with respect to x, you get 4x3 +2(cos(x2 + z2

))x = g′ (x)+2xcos

(x2 + z2

)and so it suffices to take g(x) = x4. Hence φ (x,y,z) = y+ x4 + sin

(x2 + z2

).

30.5. CONSERVATIVE VECTOR FIELDS 60730.5.1 Some TerminologyIf F = (P,Q,R) is a vector field. Then the statement that F' is conservative is the same assaying the differential form Pdx + Qdy + Rdz is exact. Some people like to say things interms of vector fields and some say it in terms of differential forms. In Example 30.5.8, thedifferential form (4x7 +2 (cos (x? +-z”)) x) dx + dy + (2 (cos (x +z”)) z) dz is exact.Definition 30.5.6 A sez of points in three dimensional space V is simply connectedif every piecewise smooth closed curve C is the edge of a surface S which is containedentirely within V in such a way that Stokes theorem holds for the surface S and its edge, C.This is like a sock. The surface is the sock and the curve C goes around the opening ofthe sock.As an application of Stoke’s theorem, here is a useful theorem which gives a way tocheck whether a vector field is conservative.Theorem 30.5.7 For a three dimensional simply connected open set V and F aC!vector field defined in V, F is conservative if V x F = OinV.Proof: If V x F' = 0 then taking an arbitrary closed curve C, and letting S be a surfacebounded by C which is contained in V, Stoke’s theorem implies0= | vxF-ndA= | Far.S CcThus F' is conservative. §jExample 30.5.8 Determine whether the vector field(4x° +2 (cos (x? +z7)) x, 1,2 (cos (x? +z”) z)is conservative.Since this vector field is defined on all of R°, it only remains to take its curl and see ifit is the zero vector.4 j kOn dy 0.4x3 +2 (cos(x?+27))x 1 2(cos(x?+27))zThis is obviously equal to zero. Therefore, the given vector field is conservative. Can youfind a potential function for it? Let @ be the potential function. Then @, =2 (cos (x? + z’)) V4and so @ (x,y,z) = sin (x? + z?) + g(x,y). Now taking the derivative of @ with respect to y,you see gy = 1 so g(x,y) =y+h(x). Hence @ (x,y,z) =y+g(x) +sin(x? +2). Takingthe derivative with respect to x, you get 4x° + 2 (cos (x? + 2?) ) x = g' (x) + 2xcos (x? + 2?)and so it suffices to take g(x) =x*. Hence @ (x,y,z) = y+x4 +sin (x? +27),