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Conservative Fields and Path Independence · Tutorial 798 of 999

Workshop: Three-Dimensional Potential Fields

Practice testing and recovering potentials for three-dimensional fields, then use them to evaluate line integrals between endpoints.

Intermediate 9 min read

What You'll Learn

  • Compute the three components of the curl of a three-dimensional vector field.
  • Use zero curl as a test for whether a potential is possible.
  • Recover a scalar potential by integrating its partial derivatives.
  • Check a proposed potential by taking its gradient.
  • Evaluate a line integral from endpoint potential values.
  • Recognize when a nonzero curl prevents use of the endpoint shortcut.

From Three Components to One Potential

A three-dimensional vector field assigns a vector to each point in space. In a conservative field, those vectors are the gradient of a single scalar function. Finding that function can turn a line integral into a simple subtraction of endpoint values.

The practical challenge is to check the field carefully before using this shortcut. In three dimensions, curl has three components, and a missed component can lead to an incorrect conclusion. The examples below combine the curl test, potential recovery, and endpoint evaluation.

Definition: A vector field \(\mathbf{F}=(P,Q,R)\) on an open set \(D\subseteq\mathbb{R}^3\) is conservative if there is a scalar function \(\phi:D\to\mathbb{R}\) such that \(\nabla\phi=\mathbf{F}\). In components, this means \(\phi_x=P\), \(\phi_y=Q\), and \(\phi_z=R\).

For a field with continuous first partial derivatives, its curl is

$$ \nabla\times\mathbf{F} = \left(R_y-Q_z,\;P_z-R_x,\;Q_x-P_y\right). $$

If \(\mathbf{F}=\nabla\phi\) and \(\phi\) has continuous second partial derivatives, the mixed partial derivatives agree. Consequently, the curl of \(\mathbf{F}\) is zero. Thus zero curl is necessary for a potential under these smoothness assumptions. Whether zero curl is also sufficient depends on the domain; the examples with zero curl below are defined on all of \(\mathbb{R}^3\), and we construct their potentials directly.

Recovering a Potential and Checking the Integral

Worked Example: A Polynomial Field

Consider \(\mathbf{F}=(2xy,\;x^2+2yz,\;y^2+3)\) on \(\mathbb{R}^3\). First label the components \(P=2xy\), \(Q=x^2+2yz\), and \(R=y^2+3\). The curl components are

$$ R_y-Q_z=2y-2y=0,\qquad P_z-R_x=0-0=0,\qquad Q_x-P_y=2x-2x=0. $$

To recover a potential, integrate \(\phi_x=P=2xy\) with respect to \(x\), treating \(y,z\) as constants:

$$ \phi(x,y,z)=x^2y+g(y,z). $$

Differentiate this expression with respect to \(y\) and match it to \(Q\):

$$ \phi_y=x^2+g_y=x^2+2yz, \qquad g_y=2yz, \qquad g(y,z)=y^2z+h(z). $$

Now match the \(z\)-derivative to \(R\):

$$ \phi_z=y^2+h'(z)=y^2+3, \qquad h'(z)=3, \qquad h(z)=3z+C. $$

A potential is therefore \(\phi=x^2y+y^2z+3z\), with the arbitrary constant \(C\) omitted. Check it by differentiating all three ways:

$$ \nabla\phi = \left(2xy,\;x^2+2yz,\;y^2+3\right) = \mathbf{F}. $$

For a curve from \(\mathbf{a}=(1,0,0)\) to \(\mathbf{b}=(2,1,0)\), the endpoint values are

$$ \phi(\mathbf{b})=2^2(1)+(1)^2(0)+3(0)=4, \qquad \phi(\mathbf{a})=1^2(0)+(0)^2(0)+3(0)=0. $$

The line integral is \(4-0=4\). As a direct check, use the straight path \(\mathbf{r}(t)=(1+t,t,0)\), \(0\le t\le1\). Then \(\mathbf{r}'(t)=(1,1,0)\), and

$$ \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t) = 2(1+t)t+(1+t)^2 = 1+4t+3t^2. $$

Thus the line integral along this path is

$$ \int_0^1(1+4t+3t^2)\,dt = \left[t+2t^2+t^3\right]_0^1 = 1+2+1 = 4. $$

This agrees with the endpoint calculation.

Why the Endpoint Formula Works

Suppose \(\mathbf{F}=\nabla\phi\), and let \(\mathbf{r}(t)\), \(a\le t\le b\), be a piecewise smooth curve from \(\mathbf{r}(a)\) to \(\mathbf{r}(b)\). On each smooth piece, the chain rule gives

$$ \frac{d}{dt}\phi(\mathbf{r}(t)) = \nabla\phi(\mathbf{r}(t))\cdot\mathbf{r}'(t) = \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t). $$

Integrating over each piece and adding the results makes the intermediate endpoint values cancel. The result is the fundamental theorem for line integrals:

Theorem: If \(\mathbf{F}=\nabla\phi\) along a piecewise smooth curve \(C\) from \(A\) to \(B\), then \(\displaystyle \int_C\mathbf{F}\cdot d\mathbf{r}=\phi(B)-\phi(A)\).

The calculation uses the curve only to connect the endpoints. The integral’s value does not depend on which such curve is chosen within the region where the potential is defined.

A Field with Exponential and Trigonometric Terms

Worked Example: Recovering a Mixed-Function Potential

Let \(\mathbf{F}=(e^x\sin y+z^2,\;e^x\cos y,\;2xz)\) on \(\mathbb{R}^3\). Write its components as \(P,Q,R\). Compute each curl component:

$$ R_y-Q_z=0-0=0,\qquad P_z-R_x=2z-2z=0,\qquad Q_x-P_y=e^x\cos y-e^x\cos y=0. $$

Integrate \(\phi_x=P\) with respect to \(x\):

$$ \phi=e^x\sin y+xz^2+g(y,z). $$

Differentiate with respect to \(y\) and compare with \(Q\):

$$ \phi_y=e^x\cos y+g_y=e^x\cos y, \qquad g_y=0. $$

So \(g\) depends only on \(z\). Comparing the \(z\)-derivative with \(R\) gives

$$ \phi_z=2xz+g'(z)=2xz, \qquad g'(z)=0. $$

We can take \(\phi=e^x\sin y+xz^2\). Its gradient checks the recovery:

$$ \nabla\phi = \left(e^x\sin y+z^2,\;e^x\cos y,\;2xz\right) = \mathbf{F}. $$

Evaluate the line integral from \(A=(0,0,0)\) to \(B=(\ln 2,\pi/2,1)\):

$$ \phi(B)=e^{\ln 2}\sin\!\left(\frac{\pi}{2}\right)+(\ln 2)(1)^2 =2+\ln 2, \qquad \phi(A)=e^0\sin 0+0=0. $$

Therefore the integral is \(2+\ln 2\). To check this with a specific path, take \(\mathbf{r}(t)=(t\ln 2,t\pi/2,t)\), \(0\le t\le1\). Along it,

$$ \phi(\mathbf{r}(t)) = 2^t\sin\!\left(\frac{\pi t}{2}\right)+t^3\ln 2. $$

By the chain rule, the integrand \(\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\) is the derivative of this expression. Its integral is therefore its value at \(t=1\) minus its value at \(t=0\), namely \(2+\ln 2-0\), as required.

When a Curl Component Is Nonzero

A nonzero curl component rules out a potential, provided the field has the continuous derivatives needed for the mixed-partial argument. In that case, do not use endpoint values to evaluate a line integral. The integral can depend on the chosen path, as the next example demonstrates.

Worked Example: Detecting Path Dependence

Consider \(\mathbf{F}=(y,0,0)\). Here \(P=y\), \(Q=0\), and \(R=0\), so

$$ \nabla\times\mathbf{F} = (0-0,\;0-0,\;0-1) = (0,0,-1). $$

Because the curl is nonzero, this field cannot be the gradient of a twice continuously differentiable potential on a region containing the paths below. Compare two paths from \(A=(0,0,0)\) to \(B=(1,1,0)\).

For the straight path \(\mathbf{r}_1(t)=(t,t,0)\), \(0\le t\le1\), we have \(\mathbf{r}_1'(t)=(1,1,0)\) and \(\mathbf{F}(\mathbf{r}_1(t))=(t,0,0)\). Hence

$$ \int_{C_1}\mathbf{F}\cdot d\mathbf{r} = \int_0^1(t,0,0)\cdot(1,1,0)\,dt = \int_0^1t\,dt = \frac12. $$

For a second path, first travel from \((0,0,0)\) to \((1,0,0)\), then from \((1,0,0)\) to \((1,1,0)\). On the first segment, \(\mathbf{r}(t)=(t,0,0)\), so the field is \((0,0,0)\) and the integral is \(0\). On the second segment, \(\mathbf{r}(t)=(1,t,0)\), so \(\mathbf{r}'(t)=(0,1,0)\) and \(\mathbf{F}(\mathbf{r}(t))=(t,0,0)\). Its contribution is

$$ \int_0^1(t,0,0)\cdot(0,1,0)\,dt=0. $$

The total integral along the second path is \(0\), not \(1/2\). The different values confirm that the endpoint-only shortcut is unavailable for this field.

A Reliable Three-Dimensional Workflow

When working with a smooth field \(\mathbf{F}=(P,Q,R)\), keep the test and the recovery separate. A zero curl is a useful check, but a proposed potential must still have all three required partial derivatives. Conversely, if a curl component is nonzero, stop before applying the endpoint formula.

1
Compute all three curl components.
Evaluate \(R_y-Q_z\), \(P_z-R_x\), and \(Q_x-P_y\); do not check only one pair of mixed partials.
2
Decide whether potential recovery is appropriate.
A nonzero curl rules out a potential under the stated smoothness conditions. A zero curl permits further investigation, but domain conditions still matter.
3
Recover and verify the scalar function.
Integrate one component, allow the integration term to depend on the other variables, and use the remaining components to determine it. Then check the full gradient.
4
Subtract endpoint values.
For a curve in the region of a verified potential, calculate \(\phi(B)-\phi(A)\) in that order.

The common error is to infer from one matching pair of partial derivatives that a potential has been found. For example, in the field \((y,0,0)\), the check \(P_z=R_x\) gives \(0=0\), but the remaining curl component is \(Q_x-P_y=0-1=-1\). Checking just that first equality would miss the obstruction and could lead to an unjustified endpoint calculation. Always check every component, and verify any candidate by differentiating it.

Why This Method Is Useful

Directly computing a line integral requires describing and integrating along a curve. When a potential exists, the fundamental theorem for line integrals replaces that curve calculation with two evaluations of a scalar function. The curl test helps identify when this simplification may be available, while careful potential recovery establishes that the shortcut is justified.

Check Your Understanding

For each question, explain the checks or calculations that support your answer.

  1. For \(\mathbf{F}=(2xz,\;0,\;x^2)\), compute all three components of the curl.
  2. For \(\mathbf{F}=(y,\;x+z,\;y)\), find a potential and verify it by taking its gradient.
  3. If a verified potential satisfies \(\phi(A)=3\) and \(\phi(B)=-2\), what is the line integral from \(A\) to \(B\)?
  4. Why is checking only \(R_y=Q_z\) insufficient to establish that a three-dimensional field has zero curl?
  5. For \(\mathbf{F}=(0,x,0)\), calculate the curl and state whether the endpoint-potential method applies.

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