OBJECTIVES

When you finish this section, you should be able to:

1. Identify a conservative vector field and its potential function (p. 990)
2. Use the Fundamental Theorem of Line Integrals (p. 991)
3. Reconstruct a function from its gradient: Finding the potential function for a conservative vector field (p. 995)
4. Determine whether a vector field is conservative (p. 996)

spanDEFINITIONspan Conservative Vector Field; Potential Function

\(\mathbf{F}\) is a conservative vector field if \(\mathbf{F}\) is the gradient of some function \(f\) that has continuous first-order partial derivatives. That is, \(\mathbf{F}\) is a conservative vector field if \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] {{\bf{F}} = \nabla f}} \]

The function \(f\) is called a potential function for \(\mathbf{F}\).

NEED TO REVIEW?

The gradient is discussed in Section 13.1, pp. 866-871.

Identifying a Conservative Vector Field and Its Potential Function

\(\mathbf{F}(x,y)=2xy\mathbf{i}+x^{2}\mathbf{j}\) is a conservative vector field, since \(\mathbf{F}\) is the gradient of the function \(f(x,y)=x^{2}y\). That is, \[ \nabla\! \ f(x,y)=f_{x}(x,y)\mathbf{i}+f_{y}(x,y)\mathbf{j}=2xy\mathbf{i}+x^{2}\mathbf{j}=\mathbf{F}(x,y) \]

The function \(f(x,y)=x^{2}y\) is the potential function for \(\mathbf{F}\).

THEOREM Fundamental Theorem of Line Integrals

Let \(\mathbf{F}=\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}\) be a vector field, where the functions \(P(x,y)\) and \(Q(x,y)\) are continuous on some open region \(R\) containing the points \((x_{0},y_{0})\) and \((x_{1},y_{1})\). Let \(C\) be a piecewise-smooth curve beginning at the point \((x_{0},y_{0})\) and ending at the point \((x_{1},y_{1})\) that lies entirely in \(R\). If \(\mathbf{F}\) is a conservative vector field on \(R,\) that is, if \(\mathbf{F}\) is the gradient of some function \(f\) so that \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \mathbf{F}(x,y)=\nabla\! \ f(x,y)}} \]

throughout \(R\), then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}\nabla\! \ f\,{\cdot}\, d\mathbf{r}=\int_{(x_{0},y_{0})}^{(x_{1},y_{1})}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\big[ f(x,y)\big]_{(x_{0},y_{0})}^{(x_{1},y_{1})}=f(x_{1},y_{1})-f(x_{0},y_{0}) }} \] \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \int_{C}(P\,dx+Q\,dy)=\int_{(x_{0},y_{0})}^{(x_{1},y_{1})}(P \,dx+Q\,dy)=\big[ f(x,y)\big] _{(x_{0},y_{0})}^{(x_{1},y_{1})}=f(x_{1},y_{1})-f(x_{0},y_{0}) }} \]

Proof

This proof is for a smooth curve \(C\). (The proof for a piecewise-smooth curve is obtained by considering one piece of the curve at a time.) Let \(x=x(t)\) and \(y=y(t)\), \(a\leq t\leq b\), be parametric equations of a smooth curve \(C\). The initial point and endpoint of \(C\) can then be written as \[ (x_{0},y_{0})=(x(a),y(a))\qquad \hbox{and}\qquad (x_{1},y_{1})=(x(b),y(b)) \]

Since \(\mathbf{F}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}\) is a conservative vector field, then \(\mathbf{F}=\nabla\! \ f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}\) for some function \(f\). Since \(\mathbf{F}=P(x,y) \mathbf{i}+Q(x,y) \mathbf{j}\), it follows that \(P(x,y) =\frac{\partial f}{\partial x}\) and \(Q(x,y) =\frac{\partial f}{\partial y}\). Then

\[ \begin{eqnarray*} \int_{C}(P\,dx+Q\,dy)& =&\int_{C}\left( \frac{\partial f}{\partial x}\,dx+ \frac{\partial f}{\partial y}\,dy\right)=\int_{a}^{b}\left( \frac{\partial f }{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}\right) \ \!dt\\ & =&\int_{a}^{b}\frac{d}{dt}[f(x(t),y(t))]\,dt\qquad {\color{#0066A7}{\hbox{Use the Chain Rule I: \(\frac{d}{dt} f (x ( t) ,y ( t))=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}.\)}} }\\ & =&\big[ f(x(t),y(t))\big] _{t=a}^{t=b}\qquad\qquad\enspace {\color{#0066A7}{\hbox{Use the Fundamental Theorem of Calculus.}}}\\ & =&f(x(b),y(b))-f(x(a),y(a))=f(x_{1},y_{1})-f(x_{0},y_{0}) \end{eqnarray*} \]

Using the Fundamental Theorem of Line Integrals

\(\mathbf{F}=\mathbf{F}(x,y)=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j}\) is a conservative vector field, since \(\mathbf{F}\) is the gradient of \(f(x,y)=x^{2}y+12x^{2}+16y\). Use this fact to find \[ \int_{C}\left[ (2xy+24x)\,dx+(x^{2}+16)\,dy\right] \]

where \(C\) is any piecewise-smooth curve joining the points \((1,1)\) and \((2,4)\).

Solution We use two methods to find \(\int_{C}[(2xy+24x)\,dx+(x^{2}+16)\,dy]\).

Method I uses the potential function \(f(x,y)=x^{2}y+12x^{2}+16y\) whose gradient is \[ \nabla\! \ f=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j}=\mathbf{F}(x,y) \]

and the Fundamental Theorem of Line Integrals. \[\begin{eqnarray*} \hspace{-250pt}\int_{C}[(2xy+24x)\,dx+(x^{2}+16)\,dy]=\big[ f(x,y)\big]_{(1,1)}^{(2,4)}=f(2,4)-f(1,1)\\[4pt] \hspace{0pt}= [ ( 2^{2}) ( 4) +12(2^{2}) +16( 4) ] -[ ( 1^{2}) (1) +12( 1^{2}) +16( 1) ] =99 \end{eqnarray*}\]

Figure 23

Method II uses the fact that the given line integral is independent of the path, so it can be integrated along any piecewise-smooth curve joining \((1,1)\) and \((2,4)\). We choose the path shown in Figure 23 since it makes the integration easy. So, along \(C_{1},\) \(dy=0\),

and along \(C_{2},\) \(dx=0.\) Then \[\begin{eqnarray*} \hspace{-1.0pc}\int_{C}[(2xy+24x)\,dx+(x^{2}+16)\,dy]&=&\int_{C_{1}}[(2xy+24x)\,dx+(x^{2}+16)\,dy]\\[4pt] &&+\int_{C_{2}}[(2xy+24x)\,dx+(x^{2}+16)\,dy] \notag \\[4pt] &=& \int_{1}^{2}(2x+24x)\,dx+\int_{1}^{4}(4+16)\,dy=39+60=99 \end{eqnarray*}\]

NOW WORK

Problem 19.

Finding \(\int_{C}\mathbf{F}\,{\cdot}\, d \mathbf{r}=\int_{C}(P\,dx+Q\,dy)\) for a Conservative Vector Field

Method I: Use the Fundamental Theorem of Line Integrals directly. That is, if \(\mathbf{F}=\nabla\! \ f,\) \[ \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}(P\,dx+Q\,dy)=f(x_{1},y_{1})-f(x_{0},y_{0}) \]

Method II: Use the fact that \(\int_{C}(P\,dx+Q\,dy)\) is independent of the path and select some suitable path joining the endpoints of \(C\) that makes finding the line integral easy.

COROLLARY Value of a Line Integral over a Closed Curve

Suppose \(\mathbf{F}\) is a conservative vector field on some open region \(R,\) and \(C\) is a closed, piecewise-smooth curve that lies entirely in \(R\). Then \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}(P\,dx+Q\,dy)=0 }} \]

Finding a Line Integral over a Closed Curve

\(\mathbf{F}=\mathbf{F}(x,y)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}}\) is the gradient of \(f(x,y) =\tan ^{-1}\frac{y}{x},\) \(x\neq 0,\) since \[\begin{equation*} \nabla\! \ f=\frac{\partial }{\partial x}\tan ^{-1}\frac{y}{x}\mathbf{i}+\frac{\partial }{\partial y}\tan ^{-1}\frac{y}{x}\mathbf{j}=\frac{ \frac{-y}{x^{2}}}{1+\frac{y^{2}}{x^2}}\mathbf{i}+\frac{\frac{1}{x}}{1+\frac{y^{2}}{x^2}}\mathbf{j}=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}} \end{equation*}\]

So, \(\mathbf{F}\) is a conservative vector field on any region \(R\) that contains no points on the \(y\)-axis (\(x=0\)). Then by the corollary \[\begin{equation*} \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=0 \end{equation*}\]

along any closed, piecewise-smooth curve \(C\) that does not cross or touch the \(y\)-axis.

NOW WORK

Problem 27.

NOTE

The converse of the conditional statement “If \(p,\) then \(q\)” is “If \(q\), then \(p\).”

THEOREM The Converse of the Fundamental Theorem of Line Integrals

Let \(\mathbf{F}=\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}\) be a vector field, where the functions \(P(x,y)\) and \(Q(x,y)\) are continuous on an open, connected region \(R\) containing the point \((x_{0},y_{0})\). If the line integral \(\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}= \int_{C}(P\,dx+Q\,dy)\) for every piecewise-smooth curve \(C\) in \(R\), is independent of the path, then the line integral \(\int_{C}\mathbf{F}\,\cdot\, d\mathbf{r}\) defines a function \(f(x,y)\) such that \(\nabla f\)=\(\mathbf{F}\). In other words, \(\mathbf{F}\) is a conservative vector field on \(R\).

Proof

Suppose \(\int_{C}(P\,dx+Q\,dy)\) has the same value for every piecewise-smooth curve \(C\) in \(R\) that joins \((x_{0},y_{0})\) to \((x,y)\). That is, suppose \(\int_{C}(P\,dx+Q\,dy)\) is independent of the path. If \(C\) is any piecewise-smooth curve in \(R\) joining a fixed point \((x_{0},y_{0})\) to an arbitrary point \((x,y)\), the line integral \(\int_{C}(P\,dx+Q\,dy)\) defines a function \(f\) that depends only on \((x,y)\) and not on \(C\). And, we can define \[\begin{equation*} f(x,y)=\int_{(x_{0},y_{0})}^{(x,y)}(P\,dx+Q\,dy) \end{equation*}\]

Since any piecewise-smooth curve \(C\) that lies entirely in \(R\) is allowed, we choose two paths, one in which a horizontal line segment is used, and another in which a vertical line segment is used. As Figure 26 illustrates:

Figure 26

\(C_{1}\): Consists of a piecewise-smooth curve joining \((x_{0},y_{0})\) to \(( x_{1},y)\) plus a horizontal line segment joining \(( x_{1},y)\) to \((x,y)\).

\(C_{2}\): Consists of a piecewise-smooth curve joining \((x_{0},y_{0})\) to \(( x,y_{1})\) plus a vertical line segment joining \(( x,y_{1})\) to \((x,y)\).

For path \(C_{1}\), \[ f(x,y)=\int_{(x_{0},y_{0})}^{(x,y)}(P\,dx+Q\,dy)=\int_{(x_{0},y_{0})}^{(x_{1},y)}(P\,dx+Q\,dy)+\int_{(x_{1},y)}^{(x,y)}(P \,dx+Q\,dy) \]

The first integral on the right is a function of \(y\) alone, since \(x_{0}\) and \(x_{1}\) are constants. So, its partial derivative with respect to \(x\) equals \(0\). That is, \[ \frac{\partial }{\partial x}\int_{(x_{0},y_{0})}^{(x_{1},y)}(P\,dx+Q\,dy)=0 \]

In the second integral on the right, the value of \(y\) is the same in both the lower and upper limits of integration, so it follows that \(dy=0\). That is, \[ \frac{\partial }{\partial x}\int_{(x_{1},y)}^{(x,y)}(P\,dx+Q\,dy)=\frac{\partial }{\partial x}\int_{x_{1}}^{x}\left[ P\,dx+0\right] =P(x,y) \]

Combining these two results, we find \[\begin{eqnarray*} \frac{\partial }{\partial x} \ f(x,y) &=&\frac{\partial }{\partial x} \int_{(x_{0},y_{0})}^{(x_{1},y)}(P\,dx+Q\,dy)+\frac{\partial }{\partial x} \int_{(x_{1},y)}^{(x,y)}(P\,dx+Q\,dy) = P(x,y) \end{eqnarray*}\]

For path \(C_{2}\), \[ f(x,y)=\int_{(x_{0},y_{0})}^{(x,y_{1})}(P\,dx+Q\,dy)+\int_{(x,y_{1})}^{(x,y)}(P\,dx+Q\,dy) \]

The first integral on the right is a function of \(x\) alone, since \(y_{0}\) and \(y_{1}\) are constants. So, the partial derivative with respect to \(y\) equals \(0\). That is, \[\begin{equation*} \frac{\partial }{\partial y}\int_{(x_{0},y_{0})}^{(x,y_{1})}(P\,dx+Q\,dy)=0 \end{equation*}\]

In the second integral on the right, the value of \(x\) is the same in both the lower and upper limits of integration, so it follows that \(dx=0\). That is, \[ \frac{\partial }{\partial y}\int_{(x,y_{1})}^{(x,y)}(P\,dx+Q\,dy)=\frac{ \partial }{\partial y}\int_{y_{1}}^{y}\left[ 0+Q(x,y)\,dy\right] =Q(x,y) \]

Combining these two results, we find \[\begin{eqnarray*} \frac{\partial }{\partial y} \ f(x,y) &=&\frac{\partial }{\partial y} \int_{(x_{0},y_{0})}^{(x,y_1)}(P\,dx+Q\,dy)+\frac{\partial }{\partial y} \int_{(x,y_{1})}^{(x,y)}(P\,dx+Q\,dy) = Q(x,y) \end{eqnarray*}\]

Finally, since \(\frac{\partial }{\partial x} \ f(x,y)=P(x,y)\) and \(\frac{\partial }{\partial y} \ f(x,y)=Q(x,y)\), we have \[\begin{equation*} \nabla\! \ f=\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}=\mathbf{F} \end{equation*}\]

That is, the vector field \(\mathbf{F}\) is conservative.

Finding a Potential Function for a Conservative Vector Field

If it is known that \[\begin{equation*} \mathbf{F}=\mathbf{F}(x,y)=(6xy+y^{3})\mathbf{i}+(3x^{2}+3xy^{2})\mathbf{j} \end{equation*}\]

is a conservative vector field, find a potential function \(f\) for \(\mathbf{F}\).

Solution We seek a function \(f(x,y)\) for which \[\begin{equation*} \nabla\! \ f=\mathbf{F}=(6xy+y^{3})\mathbf{i}+(3x^{2}+3xy^{2})\mathbf{j} \end{equation*}\]

Let \[ P(x,y)=6xy+y^{3}\qquad \hbox{and}\qquad Q(x,y)=3x^{2}+3xy^{2} \]

Since \[ \nabla\! \ f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{ \partial y}\mathbf{j}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j} \]

we have \[\begin{eqnarray*} \dfrac{\partial f}{\partial x}=6xy+y^{3}\qquad \hbox{and}\qquad \dfrac{\partial f}{\partial y}=3x^{2}+3xy^{2}\tag{1} \end{eqnarray*}\]

Now integrate the left equation partially with respect to \(x\). \[ f(x,y) =\int ( 6xy+y^{3})\, dx=3x^{2}y+xy^{3}+h(y) \]

where the “constant of integration,” \(h(y)\), is a function of \(y\).

Next differentiate \(f\) with respect to \(y\): \[\begin{equation*} \dfrac{\partial f}{\partial y}=3x^{2}+3xy^{2}+h' (y) \end{equation*}\]

From (1), we have \[\begin{equation*} \dfrac{\partial f}{\partial y}=3x^{2}+3xy^{2} \end{equation*}\]

So, \(h' (y)=0\) or \(h(y) =K,\) where \(K\) is a constant. A potential function \(f\) for \(\mathbf{F}\) is \[ f(x,y)=3x^{2}y+xy^{3}+K \]

NOW WORK

Problem 29.

spanDEFINITIONspan Closed Curve; Simple Curve

Let \(\mathbf{r}=\mathbf{r}(t)\), \(a\leq t\leq b\), trace out a piecewise-smooth curve \(C\).

• \(C\) is closed if \(\mathbf{r}(a)=\mathbf{r}(b).\)
• \(C\) is simple if it does not intersect itself for \(a<t<b\).

IN WORDS

A curve \(C\) is closed if its initial point and its terminal point coincide.

spanDEFINITIONspan Simply Connected

Let \(R\) denote a set of points in the plane. Then \(R\) is called simply connected if its boundary consists of a single, simple closed curve, and the interior of any simple closed curve \(C\) in \(R\) contains only points of \(R\).

THEOREM Conservative Vector Field Test

Let \(\mathbf{F}=\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}\) be a vector field, where the functions \(P\) and \(Q\) are continuous on some simply connected, open region \(R\). Suppose \(\dfrac{\partial P}{\partial y}\) and \(\dfrac{\partial Q}{\partial x}\) are also continuous on \(R\). Then \(\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}\) is a conservative vector field on \(R\) if and only if \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{ \partial x} }} \]

throughout \(R\).

Proof

Suppose that \(\mathbf{F}=\mathbf{F}(x,y)=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j}\) is a conservative vector field. Then \(\mathbf{F}\) is the gradient of some function \(f\). That is, \[ \nabla\! \ f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{ \partial y}\mathbf{j}=P(x,y)\mathbf{i}+Q(x,y)\mathbf{j} \]

Then \[ \dfrac{\partial f}{\partial x}=P(x,y)\qquad \hbox{and}\qquad \dfrac{\partial f}{\partial y}=Q(x,y) \]

so that \[ \dfrac{\partial ^{2}f}{\partial y\partial x}=\dfrac{\partial P}{\partial y} \qquad \hbox{and}\qquad \dfrac{\partial ^{2}f}{\partial x\partial y}=\dfrac{\partial Q}{ \partial x} \]

Since it is assumed that \(\dfrac{\partial P}{\partial y}\) and \(\dfrac{\partial Q}{\partial x}\) are continuous in \(R\), we can use the Equality of Mixed Partials Theorem (page 835), which states that if the first-order partial derivatives and the second-order mixed partial derivatives exist at each point in some disk centered at \((x_{0},y_{0})\) and if the mixed partial derivatives are continuous at \((x_{0},y_{0})\), then \(f_{xy}(x_{0},y_{0})=f_{yx}(x_{0},y_{0})\). As a result, \[ \dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x} \]

Determining Whether \(\mathbf{F}\) Is a Conservative Vector Field

(a) \(\mathbf{F}=2xy\mathbf{i}+(x^{2}+1)\mathbf{j}\) is a conservative vector field on the entire \(xy\)-plane since \[\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}(2xy)=2x\qquad \hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}(x^{2}+1)=2x \end{equation*}\]

are equal for any choice of \((x,y)\).

(b) The vector field \(\mathbf{F}=\dfrac{x}{y^{2}}\mathbf{i}-\dfrac{x^{2}}{y^{3}}\mathbf{j}\) is conservative on any connected region not containing points on the \(x\)-axis \((y=0)\) since \[\begin{equation*} \dfrac{\partial P}{\partial y}=\dfrac{\partial }{\partial y}\dfrac{x}{y^{2}} =-\dfrac{2x}{y^{3}} \qquad \hbox{and}\qquad \dfrac{\partial Q}{\partial x}=\dfrac{\partial }{\partial x}\left( -\dfrac{x^{2}}{y^{3}}\right) =-\dfrac{2x}{y^{3}} \end{equation*}\]

are equal, provided \(y\neq 0\). Because \(\mathbf{F}=\dfrac{x}{y^{2}}\mathbf{i}-\dfrac{x^{2}}{y^{3}}\mathbf{j}\) is conservative for \(y\neq 0,\) the line integral \(\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}\left[ \dfrac{x}{y^{2}}dx-\dfrac{x^{2}}{y^{3}}dy\right]\) is independent of the path in any simply connected region not containing points on the \(x\)-axis.

NOW WORK

Problem 35.

Finding a Potential Function for a Conservative Vector Field

1. Show that the line integral \(\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}[(2xy+24x)\,dx+(x^{2}+16)\,dy]\) is independent of the path.
2. Find a function \(f\) for which \(\nabla \! \ f=(2xy+24x) \mathbf{i}+(x^{2}+16)\mathbf{j}\).
3. Find \(\int_{C}[(2xy+24x)\,dx+(x^{2}+16)\,dy]\), where \(C\) is any piecewise-smooth curve joining \((0,1)\) to \((1,2)\).

Solution (a) Let \(P(x,y)=2xy+24x\) and \(Q(x,y)=x^{2}+16\). Then \(\dfrac{\partial P}{\partial y}=2x\) and \(\dfrac{\partial Q}{\partial x}=2x\). Since \(P,\) \( Q,\) \( \dfrac{\partial P}{\partial y},\) and \(\dfrac{\partial Q }{\partial x}\) are continuous everywhere in the \(xy\)-plane and since \(\dfrac{ \partial P}{\partial y}=\dfrac{\partial Q}{\partial x}\), then \(\mathbf{F}=\mathbf{F}(x,y)=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j}\) is a conservative vector field. By the Fundamental Theorem of Line Integrals, \(\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}\) is independent of the path.

(b)Since \(\mathbf{F}=\mathbf{F}(x,y)=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j}\) is a conservative vector field, there is a potential function \(f\) for \(\mathbf{F}\) for which \[ \nabla\! \ f=\mathbf{F}=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j} \]

Since \(\nabla\! \ f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{\partial y}\mathbf{j}\), then \[\begin{equation*} \dfrac{\partial f}{\partial x}=2xy+24x\qquad \hbox{and}\qquad \dfrac{\partial f}{\partial y}=x^{2}+16 \end{equation*}\]

We integrate the first of these partial derivatives partially with respect to \(x\). \[\begin{equation*} f(x,y)=\int (2xy+24x)\,dx=x^{2}y+12x^{2}+h(y) \end{equation*}\]

where the “constant of integration,” \(h(y)\), is a function of \(y\). Now we differentiate \(f\) with respect to \(y\), obtaining \[\begin{equation*} \dfrac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}[x^{2}y+12x^{2}+h(y)] =x^{2}+h' (y) \end{equation*}\]

But \(\dfrac{\partial f}{\partial y}=x^{2}+16\), so \(h' (y)=16\). Now we integrate \(h'\) with respect to \(y\); then \[ h(y)=16y+K \]

where \(K\) is a constant. So, \[ f(x,y)=x^{2}y+12x^{2}+16y+K \]

is a potential function for \(\mathbf{F}.\)

(c) Since \(\mathbf{F}\) is independent of the path, we can use the Fundamental Theorem of Line Integrals. Since \(f(x,y)=x^{2}y+12x^{2}+16y + K\) and \(\mathbf{F}=\nabla\! \ f,\) \[\begin{equation*} \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}\nabla\! \ f\,{\cdot}\, d\mathbf{r}=\big[ f(x,y)\big] _{(0,1)}^{(1,2)}=f(1,2)-f(0,1)=46-16=30 \end{equation*}\]

for any piecewise-smooth curve \(C\) connecting \(( 0,1)\) to \(( 1,2)\).

NOW WORK

Problem 37.

Finding a Potential Function for a Conservative Vector Field

1. Show that the line integral \[\begin{equation*} \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}[(y\cos x+2xe^{y})\,dx+(\sin x+x^{2}e^{y}+4)\,dy] \end{equation*}\]

   is independent of the path.

2. Find a potential function \(f\) for which \[\begin{equation*} \nabla\! \ f=(y\cos x+2xe^{y})\mathbf{i}+(\sin x+x^{2}e^{y}+4)\mathbf{j} \end{equation*}\]

Solution (a) Let \(P(x,y)=y\cos x+2xe^{y}\) and \(Q(x,y)=\sin x+x^{2}e^{y}+4\). Then \[\begin{equation*} \dfrac{\partial P}{\partial y}=\cos x+2xe^{y}=\dfrac{\partial Q}{\partial x} \end{equation*}\]

Since \(P,\) \(\ Q,\) \(\ \dfrac{\partial P}{\partial y}\), and \(\dfrac{\partial Q }{\partial x}\) are continuous everywhere in the \(xy\)-plane, and since \(\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}\), the vector field \(\mathbf{F}=\mathbf{F}(x,y)=(y\cos x+2xe^{y})\mathbf{i}+(\sin x+x^{2}e^{y}+4)\mathbf{j}\) is conservative and \(\int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}\) is independent of the path.

(b) Since \(\mathbf{F}\) is a conservative vector field, there is a potential function \(f\) for \(\mathbf{F}\) for which \[\begin{equation*} \nabla\! \ f=\mathbf{F}=(y\cos x+2xe^{y})\mathbf{i}+(\sin x+x^{2}e^{y}+4)\mathbf{j} \end{equation*}\]

Since \(\nabla \! \ f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{\partial y}\mathbf{j}\), \[\begin{equation*} \dfrac{\partial f}{\partial x}=y\cos x+2xe^{y}\qquad \hbox{and}\qquad \dfrac{\partial f}{\partial y}=\sin x+x^{2}e^{y}+4 \end{equation*}\]

We integrate the function on the right partially with respect to \(y.\) Then \[\begin{equation*} f(x,y)=\int ( \sin x+x^{2}e^{y}+4)\, dy=y\sin x+x^{2}e^{y}+4y+k(x) \end{equation*}\]

in which the “constant of integration,” \(k(x)\), is a function of \(x\). Now we differentiate \(f\) with respect to \(x\) to get \[\begin{equation*} \dfrac{\partial f}{\partial x}=y\cos x+2xe^{y}+k' (x) \end{equation*}\]

But \(\dfrac{\partial f}{\partial x}=y\cos x+2xe^{y}.\) So, \[\begin{eqnarray*} y\cos x+2xe^{y}&=& y\cos x+2xe^{y}+k' (x) \\[4pt] k' (x)& =&0 \end{eqnarray*}\]

Then \[ k(x)=K\qquad \hbox{where \(K\) is a constant} \]

Therefore, \[ f(x,y)=y\sin x+x^{2}e^{y}+4y+K \]

is a potential function for \(\mathbf{F}\).

NOW WORK

Problem 49.

Determining Whether a Line Integral Is Independent of the Path

Determine if \(\int_{C}(x^{2}y\,dx+xy^{2}\,dy)\) is independent of the path anywhere in the plane.

Solution Let \(P=x^{2}y\) and \(Q=xy^{2}\). Then \(\dfrac{\partial P}{\partial y}=x^{2}\) and \(\dfrac{\partial Q}{\partial x}=y^{2}\). Since these two functions are not equal (except on the graph of the equation \(x^{2}=y^{2}\), which is not an open set), the line integral is not independent of the path anywhere in the \(xy\)-plane.