Determine Whether The Following Vector Field Is Conservative On

Introduction

A vector field (\mathbf{F}(\mathbf{r})) is called conservative when it can be expressed as the gradient of a scalar potential function ( \phi ); that is,

[ \mathbf{F} = \nabla \phi . ]

Conservative fields possess two fundamental properties that make them especially useful in physics and engineering: the line integral of (\mathbf{F}) between two points depends only on the endpoints (path‑independence), and the circulation of (\mathbf{F}) around any closed curve is zero. Now, determining whether a given vector field is conservative therefore answers the question “does a potential function exist? ” and, if so, how to find it.

The following sections outline a systematic procedure for testing conservativeness, discuss the underlying theory, and work through a concrete example step‑by‑step. By the end of this article you will be able to decide quickly whether a vector field is conservative and, when it is, construct its potential function.

1. Theoretical Criteria

1.1 Curl Test (Three‑Dimensional Case)

For a continuously differentiable vector field (\mathbf{F} = \langle P, Q, R\rangle) defined on an open, simply‑connected region (D \subset \mathbb{R}^3), the curl must vanish:

[ \nabla \times \mathbf{F} = \mathbf{0} \quad\Longleftrightarrow\quad \begin{cases} \displaystyle \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}=0,\[4pt] \displaystyle \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}=0,\[4pt] \displaystyle \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0. \end{cases} ]

If the curl is zero everywhere in (D) and (D) is simply connected (no holes), then (\mathbf{F}) is guaranteed to be conservative.

1.2 Gradient Test (Two‑Dimensional Case)

In the plane, a vector field (\mathbf{F} = \langle P(x,y),,Q(x,y)\rangle) is conservative on a simply‑connected region if

[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} ]

holds throughout the region. This condition is simply the two‑dimensional version of the curl test because (\nabla \times \mathbf{F}) reduces to a scalar (\displaystyle \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}).

1.3 Domain Considerations

Even when the curl (or mixed partials) vanishes, the field may fail to be conservative if the domain is not simply connected. Classic counter‑examples involve fields like

[ \mathbf{F} = \left\langle \frac{-y}{x^{2}+y^{2}},; \frac{x}{x^{2}+y^{2}} \right\rangle, ]

which has zero curl everywhere except at the origin, yet no single‑valued potential exists on (\mathbb{R}^{2}\setminus{(0,0)}) because the domain contains a hole.

1.4 Summary of Decision Flow

Check differentiability – ensure all component functions have continuous first partial derivatives on the region of interest.
Identify the domain – verify that the region is open and simply connected.
Compute the curl (3‑D) or mixed partials (2‑D).
If the curl is identically zero, proceed to find a potential function (\phi).
If the curl is non‑zero, the field is not conservative.

2. Step‑by‑Step Procedure

Below is a practical checklist that can be applied to any vector field (\mathbf{F}).

Step	Action	What to look for
1	Verify (C^{1}) continuity	All partial derivatives of (P, Q, R) exist and are continuous. In real terms,
3	Compute (\nabla \times \mathbf{F}) (or (\partial P/\partial y, \partial Q/\partial x) in 2‑D).
5	Differentiate (\phi) with respect to the remaining variables to determine those “functions”.
2	Determine the domain	Is the region open?
6	Verify (\nabla \phi = \mathbf{F}). On the flip side, does it contain holes? Now,	Match the obtained partial derivatives with the original components (Q, R). Practically speaking,
4	Integrate component‑wise to obtain (\phi).	If the result is the zero vector (or zero scalar), continue; otherwise stop – the field is non‑conservative.

3. Worked Example

Consider the vector field

[ \mathbf{F}(x,y,z)=\bigl\langle 2xy,e^{z},; x^{2}e^{z}+ \cos y,; x^{2}y,e^{z}\bigr\rangle . ]

We will determine whether (\mathbf{F}) is conservative on (\mathbb{R}^{3}).

3.1 Verify Differentiability

All component functions are products and sums of elementary functions ((e^{z}, \cos y, xy)), each of which is infinitely differentiable. Hence (\mathbf{F}) is (C^{1}) on the whole space.

3.2 Identify the Domain

The domain is (\mathbb{R}^{3}), an open, simply‑connected region (no holes). The topological condition is satisfied.

3.3 Compute the Curl

[ \nabla \times \mathbf{F}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\[2pt] \displaystyle\frac{\partial}{\partial x}&\displaystyle\frac{\partial}{\partial y}&\displaystyle\frac{\partial}{\partial z}\[4pt] 2xy,e^{z}&x^{2}e^{z}+\cos y&x^{2}y,e^{z} \end{vmatrix}. ]

Evaluating each component:

(i)-component

[ \frac{\partial}{\partial y}\bigl(x^{2}y,e^{z}\bigr)-\frac{\partial}{\partial z}\bigl(x^{2}e^{z}+\cos y\bigr) = x^{2}e^{z}-x^{2}e^{z}=0. ]

(j)-component (note the sign change)

[ \frac{\partial}{\partial z}\bigl(2xy,e^{z}\bigr)-\frac{\partial}{\partial x}\bigl(x^{2}y,e^{z}\bigr) = 2xy,e^{z}-2xy,e^{z}=0. ]

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(k)-component

[ \frac{\partial}{\partial x}\bigl(x^{2}e^{z}+\cos y\bigr)-\frac{\partial}{\partial y}\bigl(2xy,e^{z}\bigr) = 2xe^{z}-2xe^{z}=0. ]

Thus (\nabla \times \mathbf{F} = \mathbf{0}) everywhere. The curl test is satisfied, so (\mathbf{F}) is conservative.

3.4 Find a Potential Function (\phi)

We need (\nabla\phi = \mathbf{F}), i.e.

[ \phi_{x}=2xy,e^{z},\qquad \phi_{y}=x^{2}e^{z}+\cos y,\qquad \phi_{z}=x^{2}y,e^{z}. ]

Integrate with respect to (x):

[ \phi(x,y,z)=\int 2xy,e^{z},dx = x^{2}y,e^{z}+g(y,z), ]

where (g) is a “constant” with respect to (x) (may depend on (y) and (z)).

Differentiate this expression with respect to (y) and compare to (\phi_{y}):

[ \phi_{y}=x^{2}e^{z}+g_{y}(y,z). ]

But the given (\phi_{y}) must equal (x^{2}e^{z}+\cos y). Hence

[ g_{y}(y,z)=\cos y \quad\Longrightarrow\quad g(y,z)=\sin y + h(z), ]

where (h(z)) is a function of (z) alone.

Now (\phi) becomes

[ \phi(x,y,z)=x^{2}y,e^{z}+\sin y + h(z). ]

Finally, differentiate with respect to (z) and match (\phi_{z}):

[ \phi_{z}=x^{2}y,e^{z}+h'(z). ]

The original (\phi_{z}) is (x^{2}y,e^{z}); therefore (h'(z)=0) and (h(z)=C) (constant).

Thus a potential function is

[ \boxed{;\phi(x,y,z)=x^{2}y,e^{z}+\sin y + C;} ]

and indeed (\nabla\phi = \mathbf{F}).

3.5 Verification

A quick check:

(\phi_{x}=2xy,e^{z}) ✓
(\phi_{y}=x^{2}e^{z}+\cos y) ✓
(\phi_{z}=x^{2}y,e^{z}) ✓

All three components match, confirming conservativeness.

4. Common Pitfalls and How to Avoid Them

Ignoring domain topology – Even with zero curl, a field defined on a region with a hole may not be conservative. Always sketch or describe the domain.
Assuming continuity without proof – A missing partial derivative at a single point can break the theorem. Verify continuity of each partial derivative on the entire region.
Dropping “functions of the other variables” – When integrating component‑wise, treat the “constant of integration” as a function of the remaining variables; forgetting this leads to inconsistent potentials.
Mismatched dimensions – In 2‑D problems, the curl reduces to a scalar; using the 3‑D determinant formula can cause confusion. Stick to the appropriate version.

5. Frequently Asked Questions

Q1: If the curl is zero only on a subset of the domain, can the field still be conservative?
Only if the subset is itself simply connected and contains the entire region of interest. If the curl vanishes on a region that excludes a singular point, the field may be non‑conservative around loops that encircle that point.

Q2: How does the concept of a path‑independent line integral relate to conservativeness?
A vector field is conservative iff the line integral (\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}) depends solely on the endpoints of curve (C). So naturally, the integral over any closed curve is zero.

Q3: Can a vector field be conservative in some regions but not others?
Yes. As an example, the field (\mathbf{F} = \langle -y/(x^{2}+y^{2}),; x/(x^{2}+y^{2})\rangle) is not conservative on (\mathbb{R}^{2}\setminus{0}) because the domain is not simply connected, yet it becomes conservative if we restrict it to a region that does not surround the origin (e.g., a half‑plane that excludes the origin).

Q4: What is the physical meaning of a conservative field?
In physics, a conservative force (e.g., gravity, electrostatic force) stores energy in a potential function. Work done by such a force along a closed path is zero, and the potential energy difference between two points equals the negative of the work done.

Q5: Does the existence of a potential guarantee uniqueness?
The potential function is unique up to an additive constant: if (\phi) is a potential, then (\phi + C) (with (C) constant) is also a valid potential because its gradient is unchanged.

6. Conclusion

Determining whether a vector field is conservative hinges on three pillars: smoothness, domain topology, and the vanishing curl (or equality of mixed partials). By systematically applying the curl test, confirming that the region is simply connected, and then constructing a scalar potential through integration, one can conclusively classify the field. The example presented illustrates every step—from computing the curl to integrating component‑wise and verifying the result—providing a template that can be adapted to any vector field encountered in mathematics, physics, or engineering.

Remember that the absence of a curl is a necessary condition, but not always sufficient; the domain’s shape can still obstruct the existence of a global potential. Keeping these nuances in mind will empower you to tackle more complex fields, recognize hidden pitfalls, and appreciate the elegant link between vector calculus and the physical world.