Testing a Vector Field

Until now, we have worked with vector fields that we know are conservative. But what if we’re not told whether a vector field is conservative? We need to be able to test it ourselves.

Recall that if [latex]\mathbf{F}[/latex] is conservative, then [latex]\mathbf{F}[/latex] has the cross-partial property. That is, if [latex]\mathbf{F}=\langle P,Q,R\rangle[/latex] is conservative, then [latex]P_y=Q_x[/latex], [latex]P_z=R_x[/latex], and [latex]Q_z=R_y[/latex]. This raises a natural question: If [latex]\mathbf{F}[/latex] has the cross-partial property, is [latex]\mathbf{F}[/latex] conservative? If the domain of [latex]\mathbf{F}[/latex] is open and simply connected, then the answer is yes.

the cross-partial test for conservative fields

If [latex]\mathbf{F}=\langle P,Q,R\rangle[/latex] is a vector field on an open, simply connected region [latex]D[/latex] and [latex]P_y=Q_x[/latex], [latex]P_z=R_x[/latex], and [latex]Q_z=R_y[/latex] throughout [latex]D[/latex], then [latex]\mathbf{F}[/latex] is conservative.

Although a proof of this theorem is beyond the scope of this course, we can see its power through examples. Later, we’ll see why it’s necessary for the region to be simply connected.

Combining this theorem with the cross-partial property gives us a powerful tool for determining whether a vector field is conservative.

the cross-partial property of conservative fields

Let [latex]\mathbf{F}=\langle P,Q,R\rangle[/latex] be a vector field on an open, simply connected region [latex]D[/latex]. Then [latex]P_y=Q_x[/latex], [latex]P_z=R_x[/latex], and [latex]Q_z=R_y[/latex] throughout [latex]D[/latex] if and only if [latex]\mathbf{F}[/latex] is conservative.

The version of this theorem in [latex]\mathbb{R}^2[/latex] is also true. If [latex]\mathbf{F}=\langle P,Q\rangle[/latex] is a vector field on an open, simply connected domain in [latex]\mathbb{R}^2[/latex], then [latex]\mathbf{F}[/latex] is conservative if and only if [latex]P_y=Q_x[/latex].

Determine whether vector field [latex]{\bf{F}}(x,y,z)=\langle{x}y^2z,x^2yz,z^2\rangle[/latex] is conservative.

Show Solution

Note that the domain of [latex]{\bf{F}}[/latex] is all of [latex]\mathbb{R}^2[/latex] and [latex]\mathbb{R}^3[/latex] is simply connected. Therefore, we can use the cross-partial property of conservative fields theorem to determine whether [latex]{\bf{F}}[/latex] is conservative. Let

[latex]P(x,y,z)=xy^2z,\text{ }Q(x,y,z)=x^2yz,\text{ and }R(x,y,z)=z^2[/latex].

Since [latex]Q_z=x^{2}y[/latex] and [latex]R_y=0[/latex], the vector field is not conservative.

Determine vector field [latex]{\bf{F}}(x,y)=\left\langle{x}\ln{(y)},\dfrac{x^2}{2y}\right\rangle[/latex] is conservative.

Show Solution

Note that the domain of [latex]{\bf{F}}[/latex] is the part of [latex]\mathbb{R}^2[/latex] in which [latex]y>0[/latex]. Thus, the domain of [latex]{\bf{F}}[/latex] is part of a plane above the [latex]x[/latex]-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use the cross-partial property of conservative fields theorem to determine whether [latex]{\bf{F}}[/latex] is conservative. Let

[latex]P(x,y)=x\ln{(y)}\text{ and }Q(x,y)=\dfrac{x^2}{2y}[/latex].

Then [latex]P_y=\dfrac{x}y=Q_x[/latex] and thus [latex]{\bf{F}}[/latex] is conservative.

When the Domain is Not Simply Connected

When using the Cross-Partial Property of Conservative Fields theorem, it’s important to remember that a theorem is a tool—and like any tool, it can only be applied under the right conditions. In this case, the theorem only applies if the domain of the vector field is simply connected.

To see what can go wrong when misapplying the theorem, consider the vector field:

[latex]{\bf{F}}(x,y)=\dfrac{y}{x^2+y^2}{\bf{i}}+\dfrac{-x}{x^2+y^2}{\bf{j}}[/latex].

This vector field satisfies the cross-partial property, since

[latex]\dfrac{\partial}{\partial y}\left(\frac{y}{x^2+y^2}\right)=\dfrac{(x^2+y^2)-y(2y)}{(x^2+y^2)^2}=\dfrac{x^2-y^2}{(x^2+y^2)^2}[/latex]

and

[latex]\dfrac{\partial}{\partial x}\left(\dfrac{-x}{x^2+y^2}\right)=\dfrac{-(x^2+y^2)+x(2x)}{(x^2+y^2)^2}=\dfrac{x^2-y^2}{(x^2+y^2)^2}[/latex].

Since [latex]\mathbf{F}[/latex] satisfies the cross-partial property, we might be tempted to conclude that [latex]\mathbf{F}[/latex] is conservative. However, [latex]\mathbf{F}[/latex] is not conservative. To see this, let

[latex]\mathbf{r}(t)=\langle\cos t,\sin t\rangle\text{, }0\leq t\leq\pi[/latex]

be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this [latex]C_1[/latex]) and let

[latex]\mathbf{s}(t)=\langle\cos t,-\sin t\rangle\text{, }0\leq t\leq\pi[/latex]

be a parameterization of the lower half of a unit circle oriented clockwise (denote this [latex]C_2[/latex]). Notice that [latex]C_1[/latex] and [latex]C_2[/latex] have the same starting point and endpoint. Since [latex]\sin^2t+\cos^2t=1[/latex],

[latex]\mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}^\prime(t)=\langle\sin(t),-\cos(t)\rangle\cdot\langle-\sin(t),\cos(t)\rangle=-1[/latex]

and

[latex]\begin{aligned} \mathbf{F}(\mathbf{s}(t))\cdot\mathbf{s}^\prime(t)&=\langle-\sin t,-\cos t\rangle\cdot\langle-\sin t,-\cos t\rangle \\ &=\sin^2t+\cos^2t \\ &=1 \end{aligned}[/latex]

Therefore,

[latex]\displaystyle\int_{C_1}\mathbf{F}\cdot d\mathbf{r}=\displaystyle\int_0^\pi-1 dt = -\pi\text{ and }\displaystyle\int_{C_2}\mathbf{F}\cdot d\mathbf{r}=\displaystyle\int_0^\pi1 dt=\pi[/latex].

Thus, [latex]C_1[/latex] and [latex]C_2[/latex] have the same starting point and endpoint, but [latex]\displaystyle\int_{C_1}\mathbf{F}\cdot d\mathbf{r}\ne\displaystyle\int_{C_2}\mathbf{F}\cdot d\mathbf{r}[/latex]. Therefore, [latex]\mathbf{F}[/latex] is not independent of path and [latex]\mathbf{F}[/latex] is not conservative.

To summarize: [latex]\mathbf{F}[/latex] satisfies the cross-partial property and yet [latex]\mathbf{F}[/latex] is not conservative. What went wrong? Does this contradict the cross-partial property of conservative fields theorem ? The issue is that the domain of [latex]\mathbf{F}[/latex] is all of [latex]\mathbb{R}^2[/latex] except for the origin. In other words, the domain of [latex]\mathbf{F}[/latex] has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, the cross-partial property of conservative fields theorem does not apply to [latex]\mathbf{F}[/latex].

We can now summarize a useful strategy for calculating vector line integrals. If you’re asked to calculate an integral of the form [latex]\displaystyle\int_{C}\mathbf{F}\cdot d\mathbf{r}[/latex], your first question should be: Is [latex]\mathbf{F}[/latex] conservative?

If the answer is yes, find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals can’t help you, and you’ll need to use other methods to compute the integral.

Calculate line integral [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}r[/latex], where [latex]{\bf{F}}(x,y,z)=\langle2xe^yz+e^xz,x^2e^yz,x^2e^y+e^x\rangle[/latex] and [latex]C[/latex] is any smooth curve that goes from the origin to [latex](1, 1, 1)[/latex].

Show Solution

Before trying to compute the integral, we need to determine whether [latex]{\bf{F}}[/latex] is conservative and whether the domain of [latex]{\bf{F}}[/latex] is simply connected. The domain of [latex]{\bf{F}}[/latex] is all of [latex]\mathbb{R}^3[/latex], which is connected and has no holes. Therefore, the domain of [latex]{\bf{F}}[/latex] is simply connected. Let

[latex]\large{P(x,y,z)=2xe^yz+e^xz,\text{ }Q(x,y,z)=x^2e^yz,\text{and }R(x,y,z)=x^2e^y+e^x}[/latex]

so that [latex]{\bf{F}}=\langle{P},Q,R\rangle[/latex]. Since the domain of [latex]{\bf{F}}[/latex] is simply connected, we can check the cross partials to determine whether [latex]{\bf{F}}[/latex] is conservative. Note that

[latex]\begin{aligned} P_y&=2xe^yz=Q_x \\ P_z&=2xe^y+e^x=R_x \\ Q_z&=x^2e^y=R_y \end{aligned}[/latex].

Therefore, [latex]{\bf{F}}[/latex] is conservative.

To evaluate [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}r[/latex] using the Fundamental Theorem for Line Integrals, we need to find a potential function [latex]f[/latex] for [latex]{\bf{F}}[/latex]. Let [latex]f[/latex] be a potential function for [latex]{\bf{F}}[/latex]. Then, [latex]\nabla{f}={\bf{F}}[/latex], and therefore [latex]f_x=2xe^yz+e^xz[/latex]. Integrating this equation with respect to [latex]x[/latex] gives [latex]f(x,y,z)=x^2e^yz+e^xz+h(y,z)[/latex] for some function [latex]h[/latex]. Differentiating this equation with respect to [latex]y[/latex] gives [latex]x^2e^yz+h_y=Q=x^2e^yz[/latex], which implies that [latex]h_y=0[/latex]. Therefore, [latex]h[/latex] is a function of [latex]z[/latex] only, and [latex]f(x,y,z)=x^2e^yz+e^x+h(z)[/latex]. To find [latex]h[/latex], note that [latex]f_z=x^2e^y+e^x+h^\prime(z)[/latex]. Therefore, [latex]h^\prime(z)=0[/latex] and we can take [latex]h(z)=0[/latex]. A potential function for [latex]{\bf{F}}[/latex] is [latex]f(x,y,z)=x^2e^yz+e^xz[/latex].

Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,

[latex]\begin{aligned} \displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_C\nabla{f}\cdot{d}{\bf{r}} \\ &=f(1,1,1)-f(0,0,0) \\ &=2e \end{aligned}[/latex].

Analysis

Notice that if we hadn’t recognized that [latex]{\bf{F}}[/latex] is conservative, we would have had to parameterize [latex]C[/latex] and use [latex]\displaystyle\int_{C}{\bf{F}}\cdot{\bf{T}}ds=\displaystyle\int_{a}^{b}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^{\prime}(t)dt[/latex]. Since curve [latex]C[/latex] is unknown, using the Fundamental Theorem for Line Integrals is much simpler.

Let [latex]{\bf{F}}(x,y)=\langle2xy^2,2x^2y\rangle[/latex] be a force field. Suppose that a particle begins its motion at the origin and ends its movement at any point in a plane that is not on the x-axis or the y-axis. Furthermore, the particle’s motion can be modeled with a smooth parameterization. Show that [latex]{\bf{F}}[/latex] does positive work on the particle.

Show Solution

We show that [latex]{\bf{F}}[/latex] does positive work on the particle by showing that [latex]{\bf{F}}[/latex] is conservative and then by using the Fundamental Theorem for Line Integrals.

To show that [latex]{\bf{F}}[/latex] is conservative, suppose [latex]f(x, y)[/latex] were a potential function for [latex]{\bf{F}}[/latex]. Then, [latex]\nabla{f}={\bf{F}}=\langle2xy^2,2x^2y\rangle[/latex] and therefore [latex]f_x=2xy^{2}[/latex] and [latex]f_y=2x^{2}y[/latex]. Equation [latex]f_x=2xy^{2}[/latex] implies that [latex]f(x, y)=x^{2}y^{2}+h(y)[/latex]. Deriving both sides with respect to [latex]y[/latex] yields [latex]f_y=2x^{2}y+h'(y)[/latex]. Therefore, [latex]h'(y)[/latex] and we can take [latex]h(y)=0[/latex].

If [latex]f(x, y)=x^{2}y^{2}[/latex], then note that [latex]\nabla{f}=\langle2xy^2,2x^2y\rangle={\bf{F}}[/latex], and therefore [latex]f[/latex] is a potential function for [latex]{\bf{F}}[/latex].

Let [latex](a, b)[/latex] be the point at which the particle stops is motion, and let [latex]C[/latex] denote the curve that models the particle’s motion. The work done by [latex]{\bf{F}}[/latex] on the particle is [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}[/latex]. By the Fundamental Theorem for Line Integrals,

[latex]\begin{aligned} \displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_c\nabla{f}\cdot{d}{\bf{r}} \\ &=f(a,b)-f(0,0) \\ &=a^2b^2 \end{aligned}[/latex].

Since [latex]a\ne0[/latex] and [latex]b\ne0[/latex], by assumption, [latex]a^2b^2>0[/latex]. Therefore, [latex]\displaystyle\int_{C}{\bf{F}}\cdot{d}{\bf{r}}>0[/latex], and [latex]{\bf{F}}[/latex] does positive work on the particle.

Analysis

Notice that this problem would be much more difficult without using the Fundamental Theorem for Line Integrals. To apply the tools we have learned, we would need to give a curve parameterization and use [latex]\displaystyle\int_{C}{\bf{F}}\cdot{\bf{T}}ds=\displaystyle\int_{a}^{b}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^{\prime}(t)dt[/latex]. Since the path of motion [latex]C[/latex] can be as exotic as we wish (as long as it is smooth), it can be very difficult to parameterize the motion of the particle.