Calculus of One and Several Variables

28.4.1 Conservative Vector Fields

Conservative vector fields are important because of the following theorem, sometimes called the fundamental theorem for line integrals.

Proof: To save space and fussing over things which are unimportant, denote by p

a polygonal curve from x₀ to x. Thus the orientation is such that it

goes from x₀ to x. The curve p

denotes the same set of points but in the opposite order. Suppose first F is conservative. Fix x₀ ∈ U and let

∫ ϕ(x) ≡ p(x0,x)F ⋅dR.

This is well defined because if q

is another polygonal curve joining x₀ to x, Then the curve obtained by following p from x₀ to x and then from x to x₀ along q is a closed piecewise smooth curve and so by assumption, the line integral along this closed curve equals 0. However, this integral is just

∫ ∫ ∫ ∫ F ⋅dR+ F⋅dR = F⋅dR − F ⋅dR p(x0,x) q(x,x0) p(x0,x) q(x0,x)

which shows

∫ ∫ F⋅dR = F⋅dR p(x0,x) q(x0,x)

and that ϕ is well defined. For small t,

Since U is open, for small t, the ball of radius centered at x is contained in U. Therefore, the line segment from x to x + te_i is also contained in U and so one can take p = x + s for s ∈. Therefore, the above difference quotient reduces to by the mean value theorem for integrals. Here s_t is some number between 0 and 1. By continuity of F, this converges to F_i as t → 0. Therefore, ∇ϕ = F as claimed.

Conversely, if ∇ϕ = F, then if R :

→ ℝ^p is any C¹ curve joining x to y,

and this verifies 28.5 in the case where the curve joining the two points is smooth. The general case follows immediately from this by using this result on each of the pieces of the piecewise smooth curve. For example if the curve goes from x to p and then from p to y, the above would imply the integral over the curve from x to p is ϕ − ϕ while from p to y the integral would yield ϕ − ϕ. Adding these gives ϕ − ϕ. The formula 28.5 implies the line integral over any closed curve equals zero because the starting and ending points of such a curve are the same. ■

The specifics of the curve are not given so the problem is nonsense unless the vector field is conservative. Therefore, it is reasonable to look for the function ϕ satisfying ∇ϕ = F. Such a function satisfies

ϕx = cosx− y (sinxz)z

and so, assuming ϕ exists,

ϕ(x,y,z) = sin x+ ycos(xz)+ ψ (y,z).

I have to add in the most general thing possible, ψ

to ensure possible solutions are not being thrown out. It wouldn’t be good at this point to only add in a constant since the answer could involve a function of either or both of the other variables. Now from what was just obtained,

ϕy = cos(xz)+ ψy = cosxz

and so it is possible to take ψ_y = 0. Consequently, ϕ, if it exists is of the form

ϕ(x,y,z) = sinx + ycos(xz)+ ψ (z).

Now differentiating this with respect to z gives

ϕ = − yx sin(xz)+ ψ = − yx sin(xz) z z

and this shows ψ does not depend on z either. Therefore, it suffices to take ψ = 0 and

ϕ(x,y,z) = sin(x)+ ycos(xz).

Therefore, the desired line integral equals

sin (π-)+ 3cos(π)− (sin (π) +cos(π)) = − 1. 2

The above process for finding ϕ will not lead you astray in the case where there does not exist a scalar potential. As an example, consider the following.

If ϕ exists, then ϕ_x = x and so ϕ =

+ ψ. Then ϕ_y = ψ_y = xy² but this is impossible because the left side depends only on y and z while the right side depends also on x. Therefore, this vector field is not conservative and there does not exist a scalar potential.

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

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

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

∫ ∫ 0 = ∇ × F ⋅ndA = F ⋅dR. S C

Thus F is conservative. ■

Since this vector field is defined on all of ℝ³, it only remains to take its curl and see if it 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 you find a potential function for it? Let ϕ be the potential function. Then ϕ_z = 2

z and so ϕ = sin + g. Now taking the derivative of ϕ with respect to y, you see g_y = 1 so g = y + h. Hence ϕ = y + g + sin. Taking the derivative with respect to x, you get 4x³ + 2x = g^′ + 2xcos and so it suffices to take g = x⁴. Hence ϕ = y + x⁴ + sin.