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)

DEFINITION 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.

$\mathbf{F}(x,y)=2xy\mathbf{i}+x^{2}\mathbf{j}$ 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*}$$

$\mathbf{F}=\mathbf{F}(x,y)=(2xy+24x)\mathbf{i}+(x^{2}+16)\mathbf{j}$ 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 $$\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt] { \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=\int_{C}(P\,dx+Q\,dy)=0 }}$$

$\mathbf{F}=\mathbf{F}(x,y)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}}$ 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.

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.

DEFINITION 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.

DEFINITION 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, $$\dfrac{\partial P}{\partial y}=\dfrac{\partial Q}{\partial x}$$

(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.

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.

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.

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.