Kenneth Kuttler

23.4. A GENERAL GREEN’S THEOREM 445

Example 23.4.6 Let F (x,y,z) =(x,y2x,z

). Find a scalar potential for F if it exists.

If φ exists, then φ x = x and so φ = x2

2 +ψ (y,z). Then φ y = ψy (y,z) = xy2 but thisis impossible because the left side depends only on y and z while the right side dependsalso on x. Therefore, this vector field is not conservative and there does not exist a scalarpotential.

Definition 23.4.7 A set of points in three dimensional space V is simply connected if everypiecewise smooth closed curve C is the edge of a surface S which is contained entirelywithin 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 23.4.8 For a three dimensional simply connected open set V and F a C1 vectorfield 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 23.4.9 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

23.4. A GENERAL GREEN’S THEOREM 445Example 23.4.6 Let F (x,y,z) = (x, y°x,2). Find a scalar potential for F if it exists.If @ exists, then @, =x and so @ = x + y(y,z). Then 9, = Wy (y,z) = xy* but thisis impossible because the left side depends only on y and z while the right side dependsalso on x. Therefore, this vector field is not conservative and there does not exist a scalarpotential.Definition 23.4.7 A set of points in three dimensional space V is simply connected if everypiecewise smooth closed curve C is the edge of a surface S which is contained entirelywithin 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 23.4.8 For a three dimensional simply connected open set V and F aC! vectorfield 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= [vx F-ndA= | P-aR.S CcThus F is conservative.Example 23.4.9 Determine whether the vector field(4x7 +2 (cos (x +27))x, 1,2 (cos (x +2)) 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.a j kOx dy Oz4x3 +2 (cos(x?+27))x 1 2(cos(x?+z7))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’)) Zzand 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? + z*)) x = g! (x) + 2xcos (x* +2”)and so it suffices to take g (x) =x*. Hence @ (x,y,z) = y+x4 +sin (x? +27).