Conservative Vector Fields

In this section, we continue the study of conservative vector fields. We examine the Fundamental Theorem for Line Integrals, which is a useful generalization of the Fundamental Theorem of Calculus to line integrals of conservative vector fields. We also discover show how to test whether a given vector field is conservative, and determine how to build a potential function for a vector field known to be conservative.

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 ([link]).

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 ([link]).

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

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.

Evaluating a Line Integral and the Antiderivatives of the Endpoints

Let $F (x, y) = 〈 2 x, 4 y 〉 .$

Calculate $\int_{C} F • d r,$

where C is the line segment from (0,0) to (2,2)([link]).

We use [link] to calculate $\int_{C} F • d r .$

Curve C can be parameterized by $r (t) = 〈 2 t, 2 t 〉, 0 \leq t \leq 1 .$

Then, $F (r (t)) = 〈 4 t, 8 t 〉$

and $r^{'} (t) = 〈 2, 2 〉,$

which implies that

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

A vector field in two dimensions. The arrows are longer the further away from the origin they are. They stretch out from the origin, forming a rectangular pattern. A line segment is drawn from P_0 at (0,0) to P_1 at (2,2).

Notice that $F = ∇ f,$

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} F • d r = \int_{C} ∇ f • d 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.

Proof

We know that if F is a conservative vector field, there are potential functions

f

Therefore

\int_{C} F \cdot d r = \int_{C} ∇ f \cdot d r = f (r (b)) - f (r (a)) .

In other words, just as with the Fundamental Theorem of Calculus, computing the line integral

\int_{C} F \cdot d r,

where F is conservative, is a two-step process: (1) find a potential function (“antiderivative”)

f

Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals. A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.

Applying the Fundamental Theorem

Calculate integral $\int_{C} F • d r,$

where $F (x, y, z) = 〈 2 x ln y, \frac{x^{2}}{y} + z^{2}, 2 y z 〉$

and C is a curve with parameterization $r (t) = 〈 t^{2}, t, t 〉, 1 \leq t \leq e$

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 [link]:

$\begin{matrix} \int_{C} F • d r & = \int_{1}^{e} F (r (t)) • r^{'} (t) d t \\ = \int_{1}^{e} 〈 2 t^{2} ln t, \frac{t^{4}}{t} + t^{2}, 2 t^{2} 〉 • 〈 2 t, 1, 1 〉 d t \\ = \int_{1}^{e} (4 t^{3} ln t + t^{3} + 3 t^{2}) d t \\ = \int_{1}^{e} 4 t^{3} ln t d t + \int_{1}^{e} (t^{3} + 3 t^{2}) d t \\ = \int_{1}^{e} 4 t^{3} ln t d t + {[\frac{t^{4}}{4} + t^{3}]}_{1}^{e} \\ = r \int_{1}^{e} t^{3} ln t d t + \frac{e^{4}}{4} + e^{3} - \frac{5}{4} . \end{matrix}$

Integral
$\int_{1}^{e} t^{3} ln t d t$
requires integration by parts. Let
$u = ln t$
and
$d v = t^{3} .$
Then
$u = ln t, d v = t^{3}$

and

$d u = \frac{1}{t} d t, v = \frac{t^{4}}{4} .$

Therefore,

$\begin{matrix} \int_{1}^{e} t^{3} ln t d t & = {[\frac{t^{4}}{4} ln t]}_{1}^{e} - \frac{1}{4} \int_{1}^{e} t^{3} d t \\ = \frac{e^{4}}{4} - \frac{1}{r} (\frac{e^{4}}{4} - \frac{1}{4}) . \end{matrix}$

Thus,

$\begin{matrix} \int_{C} F • d r & = 4 \int_{1}^{e} t^{3} ln t d t + \frac{e^{4}}{4} + e^{3} - \frac{5}{4} \\ = 4 (\frac{e^{4}}{4} - \frac{1}{4} (\frac{e^{4}}{4} - \frac{1}{4})) + \frac{e^{4}}{4} + e^{3} - \frac{5}{4} \\ = e^{4} - \frac{e^{4}}{4} + \frac{1}{4} + \frac{e^{4}}{4} + e^{3} - \frac{5}{4} \\ = e^{4} + e^{3} - 1. \end{matrix}$
Given that $f (x, y, z) = x^{2} ln y + y z^{2}$
is a potential function for F, let’s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that

$\begin{matrix} \int_{C} F • d r & = \int_{C} ∇ f • d r \\ = f (r (e)) - f (r (1)) \\ = f (e^{2}, e, e) - f (1, 1, 1) \\ = e^{4} + e^{3} - 1. \end{matrix}$

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.

[link] illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.

The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if F is conservative and C is a closed curve, then the circulation of F along C is zero—that is,

\int_{C} F \cdot d r = 0 .

be a potential function for F. Since C is a closed curve, the terminal point r(b) of C is the same as the initial point r(a) of C—that is,

r (a) = r (b) .

Recall that the reason a conservative vector field F is called “conservative” is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field F in integral

\oint_{C} F \cdot d r

follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.

The second important consequence of the Fundamental Theorem for Line Integrals is that line integrals of conservative vector fields are independent of path—meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.

Proof

be two paths in D with the same initial and terminal points ([link]). Call the initial point

P_{1}

To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.

We have shown that if F is conservative, then F is independent of path. It turns out that if the domain of F is open and connected, then the converse is also true. That is, if F is independent of path and the domain of F is open and connected, then F is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.

Proof

makes sense only because F is independent of path. If F was not independent of path, then it might be possible to find another path

C^{'}

A similar argument using a vertical line segment rather than a horizontal line segment shows that

f_{y} = Q (x, y) .

We have spent a lot of time discussing and proving [link] and [link], but we can summarize them simply: a vector field F on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.

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} F \cdot d r

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

Before giving a general method for finding a potential function, let’s motivate the method with an example.

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},

Finding a Potential Function in

ℝ^{3}

Find a potential function for $F (x, y) = 〈 2 x y, x^{2} + 2 y z^{3}, 3 y^{2} z^{2} + 2 z 〉,$

thereby showing that F is conservative.

Suppose that $f$

is a potential function. Then, $∇ f = F$

and therefore $f_{x} = 2 x y .$

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} + 2 y z^{3} = f_{y} = x^{2} + g_{y} .

Therefore,

g_{y} = 2 y 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} + 2 z = g_{z} = 3 y^{2} z^{2} + h^{'} (z) .

This implies that $h^{'} (z) = 2 z,$

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 $∇ f = 〈 2 x y, x^{2} + 2 y z^{3}, 3 y^{2} z^{2} + 2 z 〉 = F .$

We can apply the process of finding a potential function to a gravitational force. Recall that, if an object has unit mass and is located at the origin, then the gravitational force in

ℝ^{2}

where G is the universal gravitational constant. In the next example, we build a potential function for F, thus confirming what we already know: that gravity is conservative.

Finding a Potential Function

Find a potential function $f$

for $F (x, y) = − G 〈 \frac{x}{{(x^{2} + y^{2})}^{3 / 2}}, \frac{y}{{(x^{2} + y^{2})}^{3 / 2}} 〉 .$

Suppose that $f$

is a potential function. Then, $∇ f = F$

and therefore

f_{x} = \frac{− G x}{{(x^{2} + y^{2})}^{3 / 2}} .

To integrate this function with respect to x, we can use u-substitution. If $u = x^{2} + y^{2},$

then $\frac{d u}{2} = x d x,$

\begin{matrix} \int \frac{− G x}{{(x^{2} + y^{2})}^{3 / 2}} d x & = \int \frac{− G}{2 u^{3 / 2}} d u \\ = \frac{G}{\sqrt{u}} + h (y) \\ = \frac{G}{\sqrt{x^{2} + y^{2}}} + h (y) \end{matrix}

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{− G y}{{(x^{2} + y^{2})}^{3 / 2}} .

Since $f (x, y) = \frac{G}{\sqrt{x^{2} + y^{2}}} + h (y),$

f_{y}

also equals $\frac{− G y}{{(x^{2} + y^{2})}^{3 / 2}} + h^{'} (y) .$

Therefore,

\frac{− G y}{{(x^{2} + y^{2})}^{3 / 2}} + h' (y) = \frac{− G y}{{(x^{2} + y^{2})}^{3 / 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{matrix} ∇ f & = 〈 - \frac{1}{2} \frac{G}{{(x^{2} + y^{2})}^{3 / 2}} (2 x), - \frac{1}{2} \frac{G}{{(x^{2} + y^{2})}^{3 / 2}} (2 y) 〉 \\ = 〈 \frac{− G x}{{(x^{2} + y^{2})}^{3 / 2}}, \frac{− G y}{{(x^{2} + y^{2})}^{3 / 2}} 〉 \\ = F . \end{matrix}

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 [link]). That is, if

F = 〈 P, Q, R 〉

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:

When using [link], it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of [link], the theorem can be applied only if the domain of the vector field is simply connected.

To see what can go wrong when misapplying the theorem, consider the vector field from [link]:

Since F satisfies the cross-partial property, we might be tempted to conclude that F is conservative. However, F is not conservative. To see this, let

be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this

C_{1})

be a parameterization of the lower half of a unit circle oriented clockwise (denote this

C_{2}) .

have the same starting point and endpoint. Since

{sin}^{2} t + {cos}^{2} t = 1,

have the same starting point and endpoint, but

\int_{C_{1}} F \cdot d r \neq \int_{C_{2}} F \cdot d r .

To summarize: F satisfies the cross-partial property and yet F is not conservative. What went wrong? Does this contradict [link]? The issue is that the domain of F is all of

ℝ^{2}

except for the origin. In other words, the domain of F has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, [link] does not apply to F.

We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form

\int_{C} F \cdot d r,

then our first question should be: Is F conservative? If the answer is yes, then we should 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 us and we have to use other methods, such as using [link].

Using the Fundamental Theorem for Line Integrals

Calculate line integral $\int_{C} F \cdot d r,$

where $F (x, y, z) = 〈 2 x e^{y} z + e^{x} z, x^{2} e^{y} z, x^{2} e^{y} + e^{x} 〉$

and C is any smooth curve that goes from the origin to $(1, 1, 1) .$

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) = 2 x e^{y} z + e^{x} z, Q (x, y, z) = x^{2} e^{y} z, and R (x, y, z) = x^{2} e^{y} + e^{x}

so that $F = 〈 P, Q, R 〉 .$

Since the domain of F is simply connected, we can check the cross partials to determine whether F is conservative. Note that

\begin{matrix} P_{y} & = & 2 x e^{y} z = Q_{x} \\ P_{z} & = & 2 x e^{y} + e^{x} = R_{x} \\ Q_{z} & = & x^{2} e^{y} = R_{y} . \end{matrix}

Therefore, F is conservative.

To evaluate $\int_{C} F \cdot d r$

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, $∇ f = F,$

and therefore $f_{x} = 2 x e^{y} z + e^{x} z .$

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{matrix} \int_{C} F \cdot d r & = \int_{C} ∇ f \cdot d r \\ = f (1, 1, 1) - f (0, 0, 0) \\ = 2 e . \end{matrix}

Analysis

Notice that if we hadn’t recognized that F is conservative, we would have had to parameterize C and use [link]. Since curve C is unknown, using the Fundamental Theorem for Line Integrals is much simpler.

Work Done on a Particle

Let $F (x, y) = 〈 2 x y^{2}, 2 x^{2} y 〉$

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.

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, $∇ f = F = 〈 2 x y^{2}, 2 x^{2} y 〉$

and therefore $f_{x} = 2 x y^{2}$

and $f_{y} = 2 x^{2} y .$

Equation $f_{x} = 2 x y^{2}$

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 $∇ f = 〈 2 x y^{2}, 2 x^{2} y 〉 = F,$

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 $\int_{C} F \cdot d r .$

By the Fundamental Theorem for Line Integrals,

\begin{matrix} \int_{C} F \cdot d r & = \int_{C} ∇ f \cdot d r \\ = f (a, b) - f (0, 0) \\ = a^{2} b^{2} . \end{matrix}

Since $a \neq 0$

and $b \neq 0,$

by assumption, $a^{2} b^{2} > 0 .$

Therefore, $\int_{C} F \cdot d r > 0,$

and F 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 [link]. Since the path of motion C 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.

Key Concepts

Key Equations

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.

True

True or False? Function $r (t) = a + t (b - a),$

where $0 \leq t \leq 1,$

parameterizes the straight-line segment from $a to b .$

True or False? Vector field $F (x, y, z) = (y sin z) i + (x sin z) j + (x y cos z) k$

is conservative.

True

True or False? Vector field $F (x, y, z) = y i + (x + z) j - y k$

is conservative.

Justify the Fundamental Theorem of Line Integrals for $\int_{C} F \cdot d r$

in the case when $F (x, y) = (2 x + 2 y) i + (2 x + 2 y) j$

and C is a portion of the positively oriented circle $x^{2} + y^{2} = 25$

from (5, 0) to (3, 4).

\int_{C}^{} F \cdot d r = 24

[T] Find $\int_{C}^{} F \cdot d r,,]$

where $F (x, y) = (y e^{x y} + cos x) i + (x e^{x y} + \frac{1}{y^{2} + 1}) j$

and C is a portion of curve $y = sin x$

from $x = 0$

to $x = \frac{π}{2} .$

[T] Evaluate line integral $\int_{C}^{} F \cdot d r,$

where $F (x, y) = (e^{x} sin y - y) i + (e^{x} cos y - x - 2) j,$

and C is the path given by $r (t) = [t^{3} sin \frac{π t}{2}] i - [\frac{π}{2} cos (\frac{π t}{2} + \frac{π}{2})] j$

for $0 \leq t \leq 1 .$

A vector field in three dimensions. The arrows are roughly the same length and all point up into the z-plane. A curve is drawn seemingly parallel to the (x,y)-plane. In the (x,y)-plane, it would look like a decreasing concave down curve in quadrant 1.

\int_{C}^{} F \cdot d r = e - \frac{3 π}{2}

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

F (x, y) = 2 x y^{3} i + 3 y^{2} x^{2} j

F (x, y) = (− y + e^{x} sin y) i + [(x + 2) e^{x} cos y] j

Not conservative

F (x, y) = (e^{2 x} sin y) i + [e^{2 x} cos y] j

F (x, y) = (6 x + 5 y) i + (5 x + 4 y) j

Conservative, $f (x, y) = 3 x^{2} + 5 x y + 2 y^{2}$

F (x, y) = [2 x cos (y) - y cos (x)] i + [− x^{2} sin (y) - sin (x)] j

F (x, y) = [y e^{x} + sin (y)] i + [e^{x} + x cos (y)] j

Conservative, $f (x, y) = y e^{x} + x sin (y)$

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

\oint_{C} (y i + x j) \cdot d r,

where C is any path from (0, 0) to (2, 4)

\oint_{C} (2 y d x + 2 x d y),

where C is the line segment from (0, 0) to (4, 4)

\oint_{C} (2 y d x + 2 x d y) = 32

[T] $\oint_{C} [arctan \frac{y}{x} - \frac{x y}{x^{2} + y^{2}}] d x + [\frac{x^{2}}{x^{2} + y^{2}} + e^{− y} (1 - y)] d y,$

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} + 3 x y + {10 y}^{2} .

F (x, y) = (10 x + 3 y) i + (3 x + 10 y) j

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

F (x, y) = (12 x y) i + 6 (x^{2} + y^{2}) j

F (x, y) = (e^{x} cos y) i + 6 (e^{x} sin y) j

F is not conservative.

F (x, y) = (2 x y e^{x^{2} y}) i + 6 (x^{2} e^{x^{2} y}) j

F (x, y, z) = (y e^{z}) i + (x e^{z}) j + (x y e^{z}) k

F is conservative and a potential function is $f (x, y, z) = x y e^{z} .$

F (x, y, z) = (sin y) i - (x cos y) j + k

F (x, y, z) = (\frac{1}{y}) i + (\frac{x}{y^{2}}) j + (2 z - 1) k

F is conservative and a potential function is $f (x, y, z) = z .$

F (x, y, z) = 3 z^{2} i - cos y j + 2 x z k

F (x, y, z) = (2 x y) i + (x^{2} + 2 y z) j + y^{2} k

F is conservative and a potential function is $f (x, y, z) = x^{2} y + y^{2} z .$

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

F (x, y) = (e^{x} cos y) i + 6 (e^{x} sin y) j

F (x, y) = (2 x y e^{x^{2} y}) i + 6 (x^{2} e^{x^{2} y}) j

F is conservative and a potential function is $f (x, y) = e^{x^{2} y}$

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

Evaluate $\int_{C} \nabla f \cdot d r,$

where $f (x, y, z) = cos (π x) + sin (π y) - x y z$

and C is any path that starts at $(1, \frac{1}{2}, 2)$

and ends at $(2, 1, −1) .$

[T] Evaluate $\int_{C} \nabla f \cdot d r,$

where $f (x, y) = x y + e^{x}$

and C is a straight line from $(0, 0)$

to $(2, 1) .$

\int_{C} F \cdot d r = e^{2} + 1

[T] Evaluate $\int_{C} \nabla f \cdot d r,$

where $f (x, y) = x^{2} y - x$

and C is any path in a plane from (1, 2) to (3, 2).

Evaluate $\int_{C} \nabla f \cdot d r,$

where $f (x, y, z) = x y z^{2} - y z$

and C has initial point (1, 2) and terminal point (3, 5).

\int_{C} F \cdot d r = 41

For the following exercises, let $F (x, y) = 2 x y^{2} i + (2 y x^{2} + 2 y) j$

and $G (x, y) = (y + x) i + (y - x) j,$

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).

A vector field has the same curves C_1 and C_2. However, the arrows are different. Here, the arrows spiral out from the origin in a clockwise manner. The further away they are from the origin, the longer they become. They are largely horizontal in quadrants 1 and 3 and largely vertical in quadrants 2 and 4.

Calculate the line integral of F over C₁.

Calculate the line integral of G over C₁.

\oint_{C_{1}} G \cdot d r = −8 π

Calculate the line integral of F over C₂.

Calculate the line integral of G over C₂.

\oint_{C_{2}} F \cdot d r = 7

[T] Let $F (x, y, z) = x^{2} i + z sin (y z) j + y sin (y z) k .$

Calculate $\oint_{C} F \cdot d r,$

where C is a path from $A = (0, 0, 1)$

to $B = (3, 1, 2) .$

[T] Find line integral $\oint_{C} F \cdot d r$

of vector field $F (x, y, z) = 3 x^{2} z i + z^{2} j + (x^{3} + 2 y z) k$

along curve C parameterized by $r (t) = (\frac{ln t}{ln 2}) i + t^{3 / 2} j + t cos (π t), 1 \leq t \leq 4 .$

\int_{C} F \cdot d r = 150

For the following exercises, show that the following vector fields are conservative by using a computer. Calculate $\int_{C} F \cdot d r$

for the given curve.

F = (x y^{2} + 3 x^{2} y) i + (x + y) x^{2} j;

C is the curve consisting of line segments from $(1, 1)$

to $(0, 2)$

to $(3, 0) .$

F = \frac{2 x}{y^{2} + 1} i - \frac{2 y (x^{2} + 1)}{{(y^{2} + 1)}^{2}} j;

C is parameterized by $x = t^{3} - 1, y = t^{6} - t, 0 \leq t \leq 1 .$

\int_{C} F \cdot d r = −1

[T] $F = [cos (x y^{2}) - x y^{2} sin (x y^{2})] i - 2 x^{2} y sin (x y^{2}) j;$

C is curve $(e^{t}, e^{t + 1}), −1 \leq t \leq 0 .$

The mass of Earth is approximately $6 \times 10^{27} g$

and that of the Sun is 330,000 times as much. The gravitational constant is $6.7 \times 10^{−8} {cm}^{3} / s^{2} \cdot g .$

The distance of Earth from the Sun is about $1.5 \times 10^{12} cm .$

Compute, approximately, the work necessary to increase the distance of Earth from the Sun by $1 cm .$

4 \times 10^{31} erg

[T] Let $F = (x, y, z) = (e^{x} sin y) i + (e^{x} cos y) j + z^{2} k .$

Evaluate the integral $\int_{C} F \cdot d s,$

where $c (t) = (\sqrt{t}, t^{3}, e^{\sqrt{t}}), 0 \leq t \leq 1 .$

[T] Let $c : [1, 2] \to ℝ^{2}$

be given by $x = e^{t - 1}, y = sin (\frac{π}{t}) .$

Use a computer to compute the integral $\int_{C} F \cdot d s = \int_{C} 2 x cos y d x - x^{2} sin y d y,$

where $F = (2 x cos y) i - (x^{2} sin y) j .$

\int_{C} F \cdot d s = 0.4687

[T] Use a computer algebra system to find the mass of a wire that lies along curve $r (t) = (t^{2} - 1) j + 2 t k, 0 \leq t \leq 1,$

if the density is $\frac{3}{2} t .$

Find the circulation and flux of field $F = − y i + x j$

around and across the closed semicircular path that consists of semicircular arch $r_{1} (t) = (a cos t) i + (a sin t) j, 0 \leq t \leq π,$

followed by line segment $r_{2} (t) = t i, − a \leq t \leq a .$

A vector field in two dimensions. The arrows are shorter the closer they are to the origin. They surround the origin in a counterclockwise radial pattern. The upper half of a circle with radius 2 and center at the origin is drawn. (-2,0) and (2,0) are labeled as –a and a, respectively, and the curve is labeled r_1.

circulation = π a^{2} and flux = 0

Compute $\int_{C} cos x cos y d x - sin x sin y d y,$

where $c (t) = (t, t^{2}), 0 \leq t \leq 1 .$

Complete the proof of [link] by showing that $f_{y} = Q (x, y) .$

Conservative Vector Fields

Curves and Regions

Fundamental Theorem for Line Integrals

Proof

Proof

Proof