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Conservative Vector Fields

Describe simple and closed curves; define connected and simply connected regions.

Curves and Regions

Before continuing our study of conservative vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here.

We first define two special kinds of curves: closed curves and simple curves. As we have learned, a closed curve is one that begins and ends at the same point. A simple curve is one that does not cross itself. A curve that is both closed and simple is a simple closed curve ().

Example

Try it.

Is the curve with parameterization \(\text{r}(t)=〈\text{cos}\ t,\frac{\text{sin}(2t)}{2}〉,0\le t\le 2\pi\) a simple closed curve?

Solution

Note that \(\text{r}(0)=〈1,0〉=\text{r}(2\pi );\) therefore, the curve is closed. The curve is not simple, however. To see this, note that \(\text{r}(\frac{\pi }{2})=〈0,0〉=\text{r}(\frac{3\pi }{2}),\) and therefore the curve crosses itself at the origin ().

Many of the theorems in this chapter relate an integral over a region to an integral over the boundary of the region, where the region’s boundary is a simple closed curve or a union of simple closed curves. To develop these theorems, we need two geometric definitions for regions: that of a connected region and that of a simply connected region. A connected region is one in which there is a path in the region that connects any two points that lie within that region. A simply connected region is a connected region that does not have any holes in it. These two notions, along with the notion of a simple closed curve, allow us to state several generalizations of the Fundamental Theorem of Calculus later in the chapter. These two definitions are valid for regions in any number of dimensions, but we are only concerned with regions in two or three dimensions.

All simply connected regions are connected, but not all connected regions are simply connected ().

Condensed — the full section is in OpenStax Calculus Volume 3.

Fundamental Theorem for Line Integrals

Now that we understand some basic curves and regions, let’s generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function \(f\) has an antiderivative F, then the integral of \(f\) from a to b depends only on the values of F at a and at b—that is,

\[{\int }_{a}^{b}f(x)dx=\text{F}(b)-\text{F}(a).\]

If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. We show how this works using a motivational example.

Example

Try it.

Let \(\text{F}(x,y)=〈2x,4y〉.\) Calculate \({\int }_{C}\text{F}\cdot d\text{r},\) where C is the line segment from (0,0) to (2,2)().

Solution

We use to calculate \({\int }_{C}\text{F}\cdot d\text{r}.\) Curve C can be parameterized by \(\text{r}(t)=〈2t,2t〉,0\le t\le 1.\) Then, \(\text{F}(\text{r}(t))=〈4t,8t〉\) and \({\text{r}}^{'}(t)=〈2,2〉,\) which implies that

\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{0}^{1}〈4t,8t〉\cdot 〈2,2〉dt \\ & ={\int }_{0}^{1}(8t+16t)dt={\int }_{0}^{1}24t\ dt \\ & ={[12{t}^{2}]}_{0}^{1}=12.\end{array}\]

Notice that \(F=\text{∇}f,\) where \(f(x,y)={x}^{2}+2{y}^{2}.\) If we think of the gradient as a derivative, then \(f\) is an “antiderivative” of F. In the case of single-variable integrals, the integral of derivative \({g}^{'}(x)\) is \(g(b)-g(a),\) where a is the start point of the interval of integration and b is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral F to be \(f({P}_{1})-f({P}_{0}),\) where \({P}_{1}\) is the endpoint of the curve of integration and \({P}_{0}\) is the start point. Notice that this is the case for this example:

\[{\int }_{C}\text{F}\cdot d\text{r}={\int }_{C}\text{∇}f.d\text{r}=12\]

and

\[f(2,2)-f(0,0)=4+8-0=12.\]

In other words, the integral of a “derivative” can be calculated by evaluating an “antiderivative” at the endpoints of the curve and subtracting, just as for single-variable integrals.

The following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field. The same theorem holds for vector line integrals, which we call the Fundamental Theorem for Line Integrals.

Condensed — the full section is in OpenStax Calculus Volume 3.

Conservative Vector Fields and Potential Functions

As we have learned, the Fundamental Theorem for Line Integrals says that if F is conservative, then calculating \({\int }_{C}\text{F}\cdot dr\) has two steps: first, find a potential function \(f\) for F and, second, calculate \(f({P}_{1})-f({P}_{0}),\) where \({P}_{1}\) is the endpoint of C and \({P}_{0}\) is the starting point. To use this theorem for a conservative field F, we must be able to find a potential function \(f\) for F. Therefore, we must answer the following question: Given a conservative vector field F, how do we find a function \(f\) such that \(\text{∇}f=\text{F}?\) Before giving a general method for finding a potential function, let’s motivate the method with an example.

Example

Try it.

Find a potential function for \(\text{F}(x,y)=〈2x{y}^{3},3{x}^{2}{y}^{2}+\text{cos}(y)〉,\) thereby showing that F is conservative.

Solution

Suppose that \(f(x,y)\) is a potential function for F. Then, \(\text{∇}f=\text{F},\) and therefore

\[{f}_{x}=2x{y}^{3}\ \text{and}\ {f}_{y}=3{x}^{2}{y}^{2}+\text{cos}\ y.\]

Integrating the equation \({f}_{x}=2x{y}^{3}\) with respect to x yields the equation

\[f(x,y)={x}^{2}{y}^{3}+h(y).\]

Notice that since we are integrating a two-variable function with respect to x, we must add a constant of integration that is a constant with respect to x, but may still be a function of y. The equation \(f(x,y)={x}^{2}{y}^{3}+h(y)\) can be confirmed by taking the partial derivative with respect to x:

\[\frac{∂f}{∂x}=\frac{∂}{∂x}({x}^{2}{y}^{3})+\frac{∂}{∂x}(h(y))=2x{y}^{3}+0=2x{y}^{3}.\]

Since \(f\) is a potential function for F,

\[{f}_{y}=3{x}^{2}{y}^{2}+\text{cos}(y),\]

and therefore

\[3{x}^{2}{y}^{2}+{h}^{'}(y)=3{x}^{2}{y}^{2}+\text{cos}(y).\]

This implies that \(h'(y)=\text{cos}\ y,\) so \(h(y)=\text{sin}\ y+C.\) Therefore, any function of the form \(f(x,y)={x}^{2}{y}^{3}+\text{sin}(y)+C\) is a potential function. Taking, in particular, \(C=0\) gives the potential function \(f(x,y)={x}^{2}{y}^{3}+\text{sin}(y).\)

To verify that \(f\) is a potential function, note that \(\text{∇}f=〈2x{y}^{3},3{x}^{2}{y}^{2}+\text{cos}\ y〉=\text{F}.\)

The logic of the previous example extends to finding the potential function for any conservative vector field in \({ℝ}^{2}.\) Thus, we have the following problem-solving strategy for finding potential functions:

We can adapt this strategy to find potential functions for vector fields in \({ℝ}^{3},\) as shown in the next example.

\[\text{F}(x,y)=\text{-}G〈\frac{x}{{({x}^{2}+{y}^{2})}^{3\text{/}2}},\frac{y}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}〉,\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Testing a Vector Field

Until now, we have worked with vector fields that we know are conservative, but if we are not told that a vector field is conservative, we need to be able to test whether it is conservative. Recall that, if F is conservative, then F has the cross-partial property (see ). That is, if \(\text{F}=〈P,Q,R〉\) is conservative, then \({P}_{y}={Q}_{x},{P}_{z}={R}_{x},\) and \({Q}_{z}={R}_{y}.\) So, if F has the cross-partial property, then is F conservative? If the domain of F is open and simply connected, then the answer is yes.

Although a proof of this theorem is beyond the scope of the text, we can discover its power with some examples. Later, we see why it is necessary for the region to be simply connected.

Combining this theorem with the cross-partial property, we can determine whether a given vector field is conservative:

The version of this theorem in \({ℝ}^{2}\) is also true. If \(\text{F}=〈P,Q〉\) is a vector field on an open, simply connected domain in \({ℝ}^{2},\) then F is conservative if and only if \({P}_{y}={Q}_{x}.\)

Example

Try it.

Determine whether vector field \(\text{F}(x,y,z)=〈x{y}^{2}z,{x}^{2}yz,{z}^{2}〉\) is conservative.

Solution

Note that the domain of F is all of \({ℝ}^{2}\) and \({ℝ}^{3}\) is simply connected. Therefore, we can use to determine whether F is conservative. Let

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

Since \({Q}_{z}={x}^{2}y\) and \({R}_{y}=0,\) the vector field is not conservative.

Example

Try it.

Determine vector field \(\text{F}(x,y)=〈x\ \text{ln}(y),\frac{{x}^{2}}{2y}〉\) is conservative.

Solution

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

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

Then \({P}_{y}=\frac{x}{y}={Q}_{x}\) and thus F is conservative.

This vector field satisfies the cross-partial property, since

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

The theorems in this section require curves that are closed, simple, or both, and regions that are connected or simply connected.
The line integral of a conservative vector field can be calculated using the Fundamental Theorem for Line Integrals. This theorem is a generalization of the Fundamental Theorem of Calculus in higher dimensions. Using this theorem usually makes the calculation of the line integral easier.
Conservative fields are independent of path. The line integral of a conservative field depends only on the value of the potential function at the endpoints of the domain curve.
Given vector field F, we can test whether F is conservative by using the cross-partial property. If F has the cross-partial property and the domain is simply connected, then F is conservative (and thus has a potential function). If F is conservative, we can find a potential function by using the Problem-Solving Strategy.
The circulation of a conservative vector field on a simply connected domain over a closed curve is zero.

Key Equations

Fundamental Theorem for Line Integrals	\({\int }_{C}\text{∇}f\cdot d\text{r}=f(\text{r}(b))-f(\text{r}(a))\)
Circulation of a conservative field over curve C that encloses a simply connected region	\({\int }_{C}\text{∇}f\cdot d\text{r}=0\)

Conservative Vector Fields

For the following exercises, determine whether the vector field is conservative and, if it is, find the potential function.

For the following exercises, evaluate the line integrals using the Fundamental Theorem of Line Integrals.

For the following exercises, determine whether the vector field is conservative and, if so, find a potential function.

For the following exercises, evaluate the integral using the Fundamental Theorem of Line Integrals.

For the following exercises, let \(\text{F}(x,y)=2x{y}^{2}\text{i}+(2y{x}^{2}+2y)\text{j}\) and \(G(x,y)=(y+x)\text{i}+(y-x)\text{j},\) and let C₁ be the curve consisting of the circle of radius 2, centered at the origin and oriented counterclockwise, and C₂ be the curve consisting of a line segment from (0, 0) to (1, 1) followed by a line segment from (1, 1) to (3, 1).

For the following exercises, show that the following vector fields are conservative by using a computer. Calculate \({\int }_{C}\text{F}\cdot d\text{r}\) for the given curve.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Is the curve with parameterization \(\text{r}(t)=〈\text{cos}\ t,\frac{\text{sin}(2t)}{2}〉,0\le t\le 2\pi\) a simple closed curve?

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Note that \(\text{r}(0)=〈1,0〉=\text{r}(2\pi );\) therefore, the curve is closed. The curve is not simple, however. To see this, note that \(\text{r}(\frac{\pi }{2})=〈0,0〉=\text{r}(\frac{3\pi }{2}),\) and therefore the curve crosses itself at the origin ().
Is the curve given by parameterization \(\text{r}(t)=〈2\ \text{cos}\ t,3\ \text{sin}\ t〉,0\le t\le 6\pi ,\) a simple closed curve?

အဖြေကို ဖော်ပြပါ

Yes
Is the region in the below image connected? Is the region simply connected?

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The region in the figure is connected. The region in the figure is not simply connected.
Let \(\text{F}(x,y)=〈2x,4y〉.\) Calculate \({\int }_{C}\text{F}\cdot d\text{r},\) where C is the line segment from (0,0) to (2,2)().

အဖြေကို ဖော်ပြပါ

We use to calculate \({\int }_{C}\text{F}\cdot d\text{r}.\) Curve C can be parameterized by \(\text{r}(t)=〈2t,2t〉,0\le t\le 1.\) Then, \(\text{F}(\text{r}(t))=〈4t,8t〉\) and \({\text{r}}^{'}(t)=〈2,2〉,\) which implies that
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{0}^{1}〈4t,8t〉\cdot 〈2,2〉dt \\ & ={\int }_{0}^{1}(8t+16t)dt={\int }_{0}^{1}24t\ dt \\ & ={[12{t}^{2}]}_{0}^{1}=12.\end{array}\]
Notice that \(F=\text{∇}f,\) where \(f(x,y)={x}^{2}+2{y}^{2}.\) If we think of the gradient as a derivative, then \(f\) is an “antiderivative” of F. In the case of single-variable integrals, the integral of derivative \({g}^{'}(x)\) is \(g(b)-g(a),\) where a is the start point of the interval of integration and b is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral F to be \(f({P}_{1})-f({P}_{0}),\) where \({P}_{1}\) is the endpoint of the curve of integration and \({P}_{0}\) is the start point. Notice that this is the case for this example:
\[{\int }_{C}\text{F}\cdot d\text{r}={\int }_{C}\text{∇}f.d\text{r}=12\]
and
\[f(2,2)-f(0,0)=4+8-0=12.\]
In other words, the integral of a “derivative” can be calculated by evaluating an “antiderivative” at the endpoints of the curve and subtracting, just as for single-variable integrals.
Calculate integral \({\int }_{C}\text{F}\cdot d\text{r},\) where \(\text{F}(x,y,z)=〈2x\ \text{ln}\ y,\frac{{x}^{2}}{y}+{z}^{2},2yz〉\) and C is a curve with parameterization \(\text{r}(t)=〈{t}^{2},t,t〉,1\le t\le e\)
1. without using the Fundamental Theorem of Line Integrals and
2. using the Fundamental Theorem of Line Integrals.
အဖြေကို ဖော်ပြပါ
1. First, let’s calculate the integral without the Fundamental Theorem for Line Integrals and instead use :
  \[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{1}^{e}\text{F}(\text{r}(t)).{\text{r}}^{'}(t)dt \\ & ={\int }_{1}^{e}⟨2{t}^{2}\text{ln}\ t,\frac{{t}^{4}}{t}+{t}^{2},2{t}^{2}⟩.⟨2t,1,1⟩dt \\ & ={\int }_{1}^{e}(4{t}^{3}\text{ln}\ t+{t}^{3}+3{t}^{2})dt \\ & ={\int }_{1}^{e}4{t}^{3}\text{ln}\ tdt+{\int }_{1}^{e}({t}^{3}+3{t}^{2})dt \\ & ={\int }_{1}^{e}4{t}^{3}\text{ln}\ tdt+{[\frac{{t}^{4}}{4}+{t}^{3}]}_{1}^{e} \\ & =2{\int }_{1}^{e}{t}^{3}\text{ln}\ tdt+\frac{{e}^{4}}{4}+{e}^{3}-\frac{5}{4}.\end{array}\]
  Integral \({\int }_{1}^{e}{t}^{3}\text{ln}\ tdt\) requires integration by parts. Let \(u=\text{ln}\ t\) and \(dv={t}^{3}.\) Then \(u=\text{ln}\ t,dv={t}^{3}\)
  and
  \[du=\frac{1}{t}dt,v=\frac{{t}^{4}}{4}.\]
  Therefore,
  \[\begin{array}{ll}{\int }_{1}^{e}{t}^{3}\text{ln}\ tdt & ={[\frac{{t}^{4}}{4}\text{ln}\ t]}_{1}^{e}-\frac{1}{4}{\int }_{1}^{e}{t}^{3}dt \\ & =\frac{{e}^{4}}{4}-\frac{1}{4}(\frac{{e}^{4}}{4}-\frac{1}{4}).\end{array}\]
  Thus,
  \[\begin{array}{ll}{\int }_{C}\text{F}\cdot dr & =4{\int }_{1}^{e}{t}^{3}\text{ln}\ tdt+\frac{{e}^{4}}{4}+{e}^{3}-\frac{5}{4} \\ & =4(\frac{{e}^{4}}{4}-\frac{1}{4}(\frac{{e}^{4}}{4}-\frac{1}{4}))+\frac{{e}^{4}}{4}+{e}^{3}-\frac{5}{4} \\ & ={e}^{4}-\frac{{e}^{4}}{4}+\frac{1}{4}+\frac{{e}^{4}}{4}+{e}^{3}-\frac{5}{4} \\ & ={e}^{4}+{e}^{3}-1.\end{array}\]
2. Given that \(f(x,y,z)={x}^{2}\text{ln}\ y+y{z}^{2}\) is a potential function for F, let’s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that
  \[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{C}\text{∇}f.d\text{r} \\ & =f(\text{r}(e))-f(\text{r}(1)) \\ & =f({e}^{2},e,e)-f(1,1,1) \\ & ={e}^{4}+{e}^{3}-1.\end{array}\]
  This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.
Given that \(f(x,y)={(x-1)}^{2}y+{(y+1)}^{2}x\) is a potential function for \(\text{F}=〈2xy-2y+{(y+1)}^{2},{(x-1)}^{2}+2yx+2x〉,\) calculate integral \({\int }_{C}\text{F}\cdot d\text{r},\) where C is the lower half of the unit circle oriented counterclockwise.

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2
Use path independence to show that vector field \(\text{F}(x,y)=〈{x}^{2}y,y+5〉\) is not conservative.

အဖြေကို ဖော်ပြပါ

We can indicate that F is not conservative by showing that F is not path independent. We do so by giving two different paths, \({C}_{1}\) and \({C}_{2},\) that both start at \((0,0)\) and end at \((1,1),\) and yet \({\int }_{{C}_{1}}\text{F}\cdot dr\ne {\int }_{{C}_{2}}\text{F}\cdot d\text{r}.\)
Let \({C}_{1}\) be the curve with parameterization \({\text{r}}_{1}(t)=〈t,t〉,0\le t\le 1\) and let \({C}_{2}\) be the curve with parameterization \({\text{r}}_{2}(t)=〈t,{t}^{2}〉,0\le t\le 1\) (). Then
\[\begin{array}{ll}{\int }_{{C}_{1}}\text{F}\cdot d\text{r} & ={\int }_{0}^{1}\text{F}({\text{r}}_{1}(t))\cdot {\text{r}}_{1}{}^{'}(t)dt \\ & ={\int }_{0}^{1}〈{t}^{3},t+5〉\cdot 〈1,1〉dt={\int }_{0}^{1}({t}^{3}+t+5)dt \\ & ={[\frac{{t}^{4}}{4}+\frac{{t}^{2}}{2}+5t]}_{0}^{1}=\frac{23}{4}\end{array}\]
and
\[\begin{array}{ll}{\int }_{{C}_{2}}\text{F}\cdot d\text{r} & ={\int }_{0}^{1}\text{F}({\text{r}}_{2}(t))\cdot {\text{r}}_{2}'(t)dt \\ & ={\int }_{0}^{1}〈{t}^{4},{t}^{2}+5〉\cdot 〈1,2t〉dt={\int }_{0}^{1}({t}^{4}+2{t}^{3}+10t)dt \\ & ={[\frac{{t}^{5}}{5}+\frac{{t}^{4}}{2}+5{t}^{2}]}_{0}^{1}=\frac{57}{10}.\end{array}\]
Since \({\int }_{{C}_{1}}\text{F}\cdot d\text{r}\ne {\int }_{{C}_{2}}\text{F}\cdot d\text{r},\) the value of a line integral of F depends on the path between two given points. Therefore, F is not independent of path, and F is not conservative.
Show that \(\text{F}(x,y)=〈xy,{x}^{2}{y}^{2}〉\) is not path independent by considering the line segment from \((0,0)\) to \((2,2)\) and the piece of the graph of \(y=\frac{{x}^{2}}{2}\) that goes from \((0,0)\) to \((2,2).\)

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If \({C}_{1}\) and \({C}_{2}\) represent the two curves, then \({\int }_{{C}_{1}}\text{F}\cdot d\text{r}\ne {\int }_{{C}_{2}}\text{F}\cdot d\text{r}.\)
Find a potential function for \(\text{F}(x,y)=〈2x{y}^{3},3{x}^{2}{y}^{2}+\text{cos}(y)〉,\) thereby showing that F is conservative.

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Suppose that \(f(x,y)\) is a potential function for F. Then, \(\text{∇}f=\text{F},\) and therefore
\[{f}_{x}=2x{y}^{3}\ \text{and}\ {f}_{y}=3{x}^{2}{y}^{2}+\text{cos}\ y.\]
Integrating the equation \({f}_{x}=2x{y}^{3}\) with respect to x yields the equation
\[f(x,y)={x}^{2}{y}^{3}+h(y).\]
Notice that since we are integrating a two-variable function with respect to x, we must add a constant of integration that is a constant with respect to x, but may still be a function of y. The equation \(f(x,y)={x}^{2}{y}^{3}+h(y)\) can be confirmed by taking the partial derivative with respect to x:
\[\frac{∂f}{∂x}=\frac{∂}{∂x}({x}^{2}{y}^{3})+\frac{∂}{∂x}(h(y))=2x{y}^{3}+0=2x{y}^{3}.\]
Since \(f\) is a potential function for F,
\[{f}_{y}=3{x}^{2}{y}^{2}+\text{cos}(y),\]
and therefore
\[3{x}^{2}{y}^{2}+{h}^{'}(y)=3{x}^{2}{y}^{2}+\text{cos}(y).\]
This implies that \(h'(y)=\text{cos}\ y,\) so \(h(y)=\text{sin}\ y+C.\) Therefore, any function of the form \(f(x,y)={x}^{2}{y}^{3}+\text{sin}(y)+C\) is a potential function. Taking, in particular, \(C=0\) gives the potential function \(f(x,y)={x}^{2}{y}^{3}+\text{sin}(y).\)
To verify that \(f\) is a potential function, note that \(\text{∇}f=〈2x{y}^{3},3{x}^{2}{y}^{2}+\text{cos}\ y〉=\text{F}.\)
Find a potential function for \(\text{F}(x,y)=〈{e}^{x}{y}^{3}+y,3{e}^{x}{y}^{2}+x〉.\)

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\(f(x,y)={e}^{x}{y}^{3}+xy\)
Find a potential function for \(\text{F}(x,y,z)=〈2xy,{x}^{2}+2y{z}^{3},3{y}^{2}{z}^{2}+2z〉,\) thereby showing that \(\text{F}\) is conservative.

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Suppose that \(f\) is a potential function. Then, \(\text{∇}f=\text{F}\) and therefore \({f}_{x}=2xy.\) Integrating this equation with respect to x yields the equation \(f(x,y,z)={x}^{2}y+g(y,z)\) for some function g. Notice that, in this case, the constant of integration with respect to x is a function of y and z.
Since \(f\) is a potential function,
\[{x}^{2}+2y{z}^{3}={f}_{y}={x}^{2}+{g}_{y}.\]
Therefore,
\[{g}_{y}=2y{z}^{3}.\]
Integrating this function with respect to y yields
\[g(y,z)={y}^{2}{z}^{3}+h(z)\]
for some function \(h(z)\) of z alone. (Notice that, because we know that g is a function of only y and z, we do not need to write \(g(y,z)={y}^{2}{z}^{3}+h(x,z).)\) Therefore,
\[f(x,y,z)={x}^{2}y+g(y,z)={x}^{2}y+{y}^{2}{z}^{3}+h(z).\]
To find \(f\), we now must only find h. Since \(f\) is a potential function,
\[3{y}^{2}{z}^{2}+2z={g}_{z}=3{y}^{2}{z}^{2}+{h}^{'}(z).\]
This implies that \({h}^{'}(z)=2z,\) so \(h(z)={z}^{2}+C.\) Letting \(C=0\) gives the potential function
\[f(x,y,z)={x}^{2}y+{y}^{2}{z}^{3}+{z}^{2}.\]
To verify that \(f\) is a potential function, note that \(\text{∇}f=〈2xy,{x}^{2}+2y{z}^{3},3{y}^{2}{z}^{2}+2z〉=\text{F}.\)
Find a potential function for \(\text{F}(x,y,z)=〈12{x}^{2},\text{cos}\ y\ \text{cos}\ z,1-\text{sin}\ y\ \text{sin}\ z〉.\)

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\(f(x,y,z)=4{x}^{3}+\text{sin}\ y\ \text{cos}\ z+z\)
Find a potential function \(f\) for \(\text{F}(x,y)=\text{-}G〈\frac{x}{{({x}^{2}+{y}^{2})}^{3\text{/}2}},\frac{y}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}〉.\)

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Suppose that \(f\) is a potential function. Then, \(\text{∇}f=\text{F}\) and therefore
\[{f}_{x}=\frac{\text{-}Gx}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}.\]
To integrate this function with respect to x, we can use u-substitution. If \(u={x}^{2}+{y}^{2},\) then \(\frac{du}{2}=xdx,\) so
\[\begin{array}{ll}\int \frac{\text{-}Gx}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}dx & =\int \frac{\text{-}G}{2{u}^{3\text{/}2}}du \\ & =\frac{G}{\sqrt{u}}+h(y) \\ & =\frac{G}{\sqrt{{x}^{2}+{y}^{2}}}+h(y)\end{array}\]
for some function \(h(y).\) Therefore,
\[f(x,y)=\frac{G}{\sqrt{{x}^{2}+{y}^{2}}}+h(y).\]
Since \(f\) is a potential function for F,
\[{f}_{y}=\frac{\text{-}Gy}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}.\]
Since \(f(x,y)=\frac{G}{\sqrt{{x}^{2}+{y}^{2}}}+h(y),\) \({f}_{y}\) also equals \(\frac{\text{-}Gy}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}+{h}^{'}(y).\)
Therefore,
\[\frac{\text{-}Gy}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}+h'(y)=\frac{\text{-}Gy}{{({x}^{2}+{y}^{2})}^{3\text{/}2}},\]
which implies that \({h}^{'}(y)=0.\) Thus, we can take \(h(y)\) to be any constant; in particular, we can let \(h(y)=0.\) The function
\[f(x,y)=\frac{G}{\sqrt{{x}^{2}+{y}^{2}}}\]
is a potential function for the gravitational field F. To confirm that \(f\) is a potential function, note that
\[\begin{array}{ll}\text{∇}f & =〈-\frac{1}{2}\ \frac{G}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}(2x),-\frac{1}{2}\ \frac{G}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}(2y)〉 \\ & =〈\frac{\text{-}Gx}{{({x}^{2}+{y}^{2})}^{3\text{/}2}},\frac{\text{-}Gy}{{({x}^{2}+{y}^{2})}^{3\text{/}2}}〉 \\ & =\text{F}.\end{array}\]
Find a potential function \(f\) for the three-dimensional gravitational force \(\text{F}(x,y,z)=〈\frac{\text{-}Gx}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}},\frac{\text{-}Gy}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}},\frac{\text{-}Gz}{{({x}^{2}+{y}^{2}+{z}^{2})}^{3\text{/}2}}〉.\)

အဖြေကို ဖော်ပြပါ

\(f(x,y,z)=\frac{G}{\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}}\)
Determine whether vector field \(\text{F}(x,y,z)=〈x{y}^{2}z,{x}^{2}yz,{z}^{2}〉\) is conservative.

အဖြေကို ဖော်ပြပါ

Note that the domain of F is all of \({ℝ}^{2}\) and \({ℝ}^{3}\) is simply connected. Therefore, we can use to determine whether F is conservative. Let
\[P(x,y,z)=x{y}^{2}z,Q(x,y,z)={x}^{2}yz,\ \text{and}\ R(x,y,z)={z}^{2}.\]
Since \({Q}_{z}={x}^{2}y\) and \({R}_{y}=0,\) the vector field is not conservative.
Determine vector field \(\text{F}(x,y)=〈x\ \text{ln}(y),\frac{{x}^{2}}{2y}〉\) is conservative.

အဖြေကို ဖော်ပြပါ

Note that the domain of F is the part of \({ℝ}^{2}\) in which \(y>0.\) Thus, the domain of F is part of a plane above the x-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use to determine whether F is conservative. Let
\[P(x,y)=x\ \text{ln}(y)\ \text{and}\ Q(x,y)=\frac{{x}^{2}}{2y}.\]
Then \({P}_{y}=\frac{x}{y}={Q}_{x}\) and thus F is conservative.
Determine whether \(\text{F}(x,y)=〈\text{sin}\ x\ \text{cos}\ y,\text{cos}\ x\ \text{sin}\ y〉\) is conservative.

အဖြေကို ဖော်ပြပါ

It is conservative.
Calculate line integral \({\int }_{C}\text{F}\cdot d\text{r},\) where \(\text{F}(x,y,z)=〈2x{e}^{y}z+{e}^{x}z,{x}^{2}{e}^{y}z,{x}^{2}{e}^{y}+{e}^{x}〉\) and C is any smooth curve that goes from the origin to \((1,1,1).\)

အဖြေကို ဖော်ပြပါ

Before trying to compute the integral, we need to determine whether F is conservative and whether the domain of F is simply connected. The domain of F is all of \({ℝ}^{3},\) which is connected and has no holes. Therefore, the domain of F is simply connected. Let
\[P(x,y,z)=2x{e}^{y}z+{e}^{x}z,Q(x,y,z)={x}^{2}{e}^{y}z,\ \text{and}\ R(x,y,z)={x}^{2}{e}^{y}+{e}^{x}\]
so that \(\text{F}=〈P,Q,R〉.\) Since the domain of F is simply connected, we can check the cross partials to determine whether F is conservative. Note that
\[\begin{array}{lll}{P}_{y} & = & 2x{e}^{y}z={Q}_{x} \\ {P}_{z} & = & 2x{e}^{y}+{e}^{x}={R}_{x} \\ {Q}_{z} & = & {x}^{2}{e}^{y}={R}_{y}.\end{array}\]
Therefore, F is conservative.
To evaluate \({\int }_{C}\text{F}\cdot d\text{r}\) using the Fundamental Theorem for Line Integrals, we need to find a potential function \(f\) for F. Let \(f\) be a potential function for F. Then, \(\text{∇}f=\text{F},\) and therefore \({f}_{x}=2x{e}^{y}z+{e}^{x}z.\) Integrating this equation with respect to x gives \(f(x,y,z)={x}^{2}{e}^{y}z+{e}^{x}z+h(y,z)\) for some function h. Differentiating this equation with respect to y gives \({x}^{2}{e}^{y}z+{h}_{y}=Q={x}^{2}{e}^{y}z,\) which implies that \({h}_{y}=0.\) Therefore, h is a function of z only, and \(f(x,y,z)={x}^{2}{e}^{y}z+{e}^{x}z+h(z).\) To find h, note that \({f}_{z}={x}^{2}{e}^{y}+{e}^{x}+h'(z)=R={x}^{2}{e}^{y}+{e}^{x}.\) Therefore, \(h'(z)=0\) and we can take \(h(z)=0.\) A potential function for F is \(f(x,y,z)={x}^{2}{e}^{y}z+{e}^{x}z.\)
Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{C}\text{∇}f\cdot d\text{r} \\ & =f(1,1,1)-f(0,0,0) \\ & =2e.\end{array}\]
Calculate integral \({\int }_{C}\text{F}\cdot d\text{r},\) where \(\text{F}(x,y)=〈\text{sin}\ x\ \text{sin}\ y,5-\text{cos}\ x\ \text{cos}\ y〉\) and C is a semicircle with starting point \((0,\pi )\) and endpoint \((0,\text{-}\pi ).\)

အဖြေကို ဖော်ပြပါ

\(-10\pi\)
Let \(\text{F}(x,y)=〈2x{y}^{2},2{x}^{2}y〉\) 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 F does positive work on the particle.

အဖြေကို ဖော်ပြပါ

We show that F does positive work on the particle by showing that F is conservative and then by using the Fundamental Theorem for Line Integrals.
To show that F is conservative, suppose \(f(x,y)\) were a potential function for F. Then, \(\text{∇}f=\text{F}=〈2x{y}^{2},2{x}^{2}y〉\) and therefore \({f}_{x}=2x{y}^{2}\) and \({f}_{y}=2{x}^{2}y.\) Equation \({f}_{x}=2x{y}^{2}\) implies that \(f(x,y)={x}^{2}{y}^{2}+h(y).\) Deriving both sides with respect to y yields \({f}_{y}=2{x}^{2}y+{h}^{'}(y).\) Therefore, \({h}^{'}(y)=0\) and we can take \(h(y)=0.\)
If \(f(x,y)={x}^{2}{y}^{2},\) then note that \(\text{∇}f=〈2x{y}^{2},2{x}^{2}y〉=\text{F},\) and therefore \(f\) is a potential function for F.
Let \((a,b)\) be the point at which the particle stops is motion, and let C denote the curve that models the particle’s motion. The work done by F on the particle is \({\int }_{C}\text{F}\cdot d\text{r}.\) By the Fundamental Theorem for Line Integrals,
\[\begin{array}{ll}{\int }_{C}\text{F}\cdot d\text{r} & ={\int }_{C}\text{∇}f\cdot d\text{r} \\ & =f(a,b)-f(0,0) \\ & ={a}^{2}{b}^{2}.\end{array}\]
Since \(a\ne 0\) and \(b\ne 0,\) by assumption, \({a}^{2}{b}^{2}>0.\) Therefore, \({\int }_{C}\text{F}\cdot d\text{r}>0,\) and F does positive work on the particle.
Let \(\text{F}(x,y)=〈4{x}^{3}{y}^{4},4{x}^{4}{y}^{3}〉,\) and suppose that a particle moves from point \((4,4)\) to \((1,1)\) along any smooth curve. Is the work done by F on the particle positive, negative, or zero?

အဖြေကို ဖော်ပြပါ

Negative
True or False? If vector field F is conservative on the open and connected region D, then line integrals of F are path independent on D, regardless of the shape of D.

အဖြေကို ဖော်ပြပါ

True
True or False? Function \(\text{r}(t)=\text{a}+t(\text{b}-\text{a}),\) where \(0\le t\le 1,\) parameterizes the straight-line segment from \(\text{a}\ \text{to}\ \text{b}.\)
True or False? Vector field \(\text{F}(x,y,z)=(y\ \text{sin}\ z)\text{i}+(x\ \text{sin}\ z)\text{j}+(xy\ \text{cos}\ z)\text{k}\) is conservative.

အဖြေကို ဖော်ပြပါ

True
True or False? Vector field \(\text{F}(x,y,z)=y\text{i}+(x+z)\text{j}-y\text{k}\) is conservative.
Verify the Fundamental Theorem of Line Integrals for \({\int }_{C}\text{F}\cdot d\text{r}\) by computing the integral using a parameterization and, separately, by finding a potential function for the case when \(\text{F}(x,y)=(2x+2y)\text{i}+(2x+2y)\text{j}\) and C is a portion of circle \({x}^{2}+{y}^{2}=25\) oriented counterclockwise from (5, 0) to (3, 4).

အဖြေကို ဖော်ပြပါ

\({\int }_{C}^{}\text{F}\cdot d\text{r}=24\)
[T] Find \({\int }_{C}^{}\text{F}\cdot d\text{r},\) where \(\text{F}(x,y)=(y{e}^{xy}+\text{cos}\ x)\text{i}+(x{e}^{xy}+\frac{1}{{y}^{2}+1})\text{j}\) and C is a portion of curve \(y=\text{sin}\ x\) from \(x=0\) to \(x=\frac{\pi }{2}.\)
[T] Evaluate line integral \({\int }_{C}^{}\text{F}\cdot d\text{r},\) where \(\text{F}(x,y)=({e}^{x}\text{sin}\ y-y)\text{i}+({e}^{x}\text{cos}\ y-x-2)\text{j},\) and C is the path given by \(r(t)=[{t}^{3}\text{sin}\ \frac{\pi t}{2}]\text{i}-[\frac{\pi }{2}\text{cos}(\frac{\pi t}{2}+\frac{\pi }{2})]\text{j}\) for \(0\le t\le 1.\)

အဖြေကို ဖော်ပြပါ

\({\int }_{C}^{}\text{F}\cdot d\text{r}=e-\frac{3\pi }{2}\)
\(\text{F}(x,y)=2x{y}^{3}\text{i}+3{y}^{2}{x}^{2}\text{j}\)
\(\text{F}(x,y)=(\text{-}y+{e}^{x}\text{sin}\ y)\text{i}+[(x+2){e}^{x}\text{cos}\ y]\text{j}\)

အဖြေကို ဖော်ပြပါ

Not conservative
\(\text{F}(x,y)=({e}^{2x}\text{sin}\ y)\text{i}+[{e}^{2x}\text{cos}\ y]\text{j}\)
\(\text{F}(x,y)=(6x+5y)\text{i}+(5x+4y)\text{j}\)

အဖြေကို ဖော်ပြပါ

Conservative, \(f(x,y)=3{x}^{2}+5xy+2{y}^{2}\)
\(\text{F}(x,y)=[2x\ \text{cos}(y)-y\ \text{cos}(x)]\text{i}+[\text{-}{x}^{2}\text{sin}(y)-\text{sin}(x)]\text{j}\)
\(\text{F}(x,y)=[y{e}^{x}+\text{sin}(y)]\text{i}+[{e}^{x}+x\ \text{cos}(y)]\text{j}\)

အဖြေကို ဖော်ပြပါ

Conservative, \(f(x,y)=y{e}^{x}+x\ \text{sin}(y)\)
\({\int }_{C}(y\text{i}+x\text{j})\cdot d\text{r},\) where C is any path from (0, 0) to (2, 4)
\({\int }_{C}(2ydx+2xdy),\) where C is the line segment from (0, 0) to (4, 4)

အဖြေကို ဖော်ပြပါ

\({\int }_{C}(2ydx+2xdy)=32\)
[T] \({\int }_{C}[\text{arctan}\ \frac{y}{x}-\frac{xy}{{x}^{2}+{y}^{2}}]dx+[\frac{{x}^{2}}{{x}^{2}+{y}^{2}}+{e}^{\text{-}y}(1-y)]dy,\) where C is any smooth curve from (1, 1) to \((-1,2)\)
Find the conservative vector field for the potential function
\[f(x,y)=5{x}^{2}+3xy+{10y}^{2}.\]

အဖြေကို ဖော်ပြပါ

\(\text{F}(x,y)=(10x+3y)\text{i}+(3x+20y)\text{j}\)
\(\text{F}(x,y)=(12xy)\text{i}+6({x}^{2}+{y}^{2})\text{j}\)
\(\text{F}(x,y)=({e}^{x}\text{cos}\ y)\text{i}+6({e}^{x}\text{sin}\ y)\text{j}\)

အဖြေကို ဖော်ပြပါ

F is not conservative.

Symbols used here

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Conservative Vector Fields

Describe simple and closed curves; define connected and simply connected regions.
Explain how to find a potential function for a conservative vector field.
Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.
Explain how to test a vector field to determine whether it is conservative.
without using the Fundamental Theorem of Line Integrals and
using the Fundamental Theorem of Line Integrals.
First, let’s calculate the integral without the Fundamental Theorem for Line Integrals and instead use
Given that

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

သင့်ရဲ့ကိုယ်ပိုင်စမ်းသပ်

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ပိုပြီး Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems