Fundamental Theorem Of Line Integrals (w/ Step-by-Step Examples!)

What determines the work performed by a vector field?

Jenn (B.S., M.Ed.) of Calcworkshop® teaching fundamental theorem of line integrals

Jenn, Founder Calcworkshop^®, 15+ Years Experience (Licensed & Certified Teacher)

Does the work only depend on the endpoints, or does changing the path while keeping the endpoints fixed make a difference?

In other words, knowing that a line integral gives us insight as to the work performed by a vector field on an object moving along a parametric curve $C$ , would the work change if we altered the course but kept the endpoints the same?

To answer such a question, we need the fundamental theorem of line integrals and a look into a special type of vector field called conservative vector fields.

Let’s investigate.

Fundamental Theorem Of Line Integrals

Think about a smooth curve $C$ given by the vector equation $\vec{r} (t)$ from $t \in [a, b]$ and a differentiable function $f$ of two or three variables whose gradient vector $\nabla f$ is continuous on $C$ .

$Then \int_{C} \nabla f \cdot d \vec{r} = \int_{a}^{b} \nabla f (\vec{r} (t)) \cdot {\vec{r}}^{'} (t) d t = f (\vec{r} (b)) - f (\vec{r} (a))$

Notice how this is just an extension of the fundamental theorem of calculus (FTC) to line integrals.

In fact, this is explicitly saying that a line integral in a conservative vector field is independent of path.

Path Independence Of Line Integrals

Let me explain further.

Recall from our lesson on vector fields, $\vec{F}$ is called a conservative vector field, or a gradient vector field, if $\vec{F} = \nabla f$ where $f$ is called the potential function of $\vec{F}$ .

But here’s what cool. The fundamental theorem for line integrals states that $\vec{F} = \nabla f$ if and only if $\int_{C} \vec{F} \cdot d \vec{r}$ is independent of path

But what does path independence really mean?

Suppose $C_{1}$ and $C_{2}$ are two different piecewise smooth curves that have the same initial point $A$ and terminal point $B$ , as shown in the graphic below.

Path Independence Line Integrals

The starting point and stopping point are fixed, but the path or route we take to get from point A to point B is different.

But here’s what’s important…it doesn’t matter what path we choose because they both start and stop at the same place.

So, if a vector field is path-independent, then

$\int_{C_{1}} \vec{F} \cdot d \vec{r} = \int_{C_{2}} \vec{F} \cdot d \vec{r}$

More specifically…

$\int_{A}^{B} \vec{F} \cdot d \vec{r} = f (B) - f (A)$

Regardless of whether you take path $C_{1}$ or $C_{2}$ .

This means we can now easily answer our earlier question.

In a conservative force field, the amount of work required to move an object from point $a$ to point $b$ only depends on those two endpoints, not on the path taken between them.

Cool, right?

Please note that in a non-conservative vector field (i.e., a general vector field), $\int_{C_{1}} \vec{F} \cdot d \vec{r} \neq \int_{C_{2}} \vec{F} \cdot d \vec{r}$ but in a conservative vector field, for any two paths $C_{1}$ and $C_{2}$ that have the same starting and ending points, then $\int_{C_{1}} \vec{F} \cdot d \vec{r} = \int_{C_{2}} \vec{F} \cdot d \vec{r}$ .

Example

Let’s look at an example.

Suppose $\vec{F} = ⟨ 12 x^{2} + 3 y^{2} + 5 y, 6 x y - 3 y^{2} + 5 x ⟩$ , knowing that $\vec{F}$ is conservative and independent of path with potential function $f (x, y) = 4 x^{3} + 3 y^{2} x + 5 x y - y^{3}$ .

Calculate $\int_{C} \vec{F} \cdot d \vec{r}$ where $C$ is any path from $(0, 0)$ to $(2, 1)$ .

Because we know that $\vec{F}$ is conservative and independent of the path, we can use the fundamental theorem of line integrals.

$\int_{C} \vec{F} \cdot d \vec{r} = \int_{C} \nabla f (\vec{r} (t)) \cdot d \vec{r} = f (B) - f (A)$

$\int_{C} \vec{F} \cdot d \vec{r} = \int_{(0, 0)}^{(2, 1)} \vec{F} \cdot d \vec{r}$

$= {f (x, y) |}_{(0, 0)}^{(2, 1)}$

$= f (2, 1) - f (0, 0)$

$= {(4 x^{3} + 3 y^{2} x + 5 x y - y^{3}]}_{(0, 0)}^{(2, 1)}$

$= (4 (2)^{3} + 3 (1)^{2} (2) + 5 (2) (1) - (1)^{3}) - 0$

$= 47$

Isn’t this a fantastic concept?

Because as Paul’s Online Notes so accurately summarizes, we now see that for these kinds of line integrals, we didn’t need to know the curve to get the answer because if we can prove the independence of path, we can use any path we want, and we’ll always get the same results.

Is A Vector Field Conservative?

But how do we determine whether a vector field is conservative and thus independent of the path?

Well, as it turns out, the only vector fields that are conservative satisfy the following theorem:

Suppose $\vec{F}$ is a vector field that is continuous on a region $D$ , where $D$ is open and connected. $\int_{C} \vec{F} \cdot d \vec{r}$ is independent of path if and only if $\int_{C} \vec{F} \cdot d \vec{r} = 0$ for every closed path $C$ . And if $\int_{C} \vec{F} \cdot d \vec{r}$ is independent of path in $D$ , then $\vec{F}$ is a conservative vector field in $D$ and there exists a function $f$ such that $\vec{F} = \nabla f$ .

While this is all well and good and extremely important to note about conservative vector fields, it’s not a useful computational tool.

Thankfully, there’s an easy way for us to determine if $\vec{F}$ is conservative!

Assume $\vec{F} = ⟨ P, Q ⟩$ is a vector field on region $D$ and that $P$ and $Q$ have continuous first order partial derivatives. Then $\vec{F}$ is conservative if and only if $P_{y} = Q_{x}$ .

And for three-dimensional vector fields such as $\vec{F} = ⟨ P, Q, R ⟩$ where $P$ , $Q$ , and $R$ have continuous first partials then $\vec{F}$ is conservative if and only if $c u r l \vec{F} = \nabla \times \vec{F} = \vec{0}$

$curl \vec{F} = \nabla \times \vec{F} = | \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{array} |$

We will predominately work with two-dimensional vector fields and leave three-dimensional vector fields, and the concept of the “curl” for a later lesson.

Example

Okay, so let’s put our new theorem to work with an example.

Is $\vec{F} = ⟨ 12 x^{2} + 3 y^{2} + 5 y, 6 x y - 3 y^{2} + 5 x ⟩$ conservative?

First, we identify $P$ and $Q$ .

$\vec{F} = ⟨ \underset{P}{\underset{⏟}{12 x^{2} + 3 y^{2} + 5 y}}, \underset{Q}{\underset{⏟}{6 x y - 3 y^{2} + 5 x}} ⟩$

Next, we take our partial derivatives and check to see that they are equal.

$P_{y} = 6 y + 5 and Q_{x} = 6 y + 5$

Because $P_{y} = Q_{x}$ we know that $\vec{F}$ is conservative.

Easy!

Finding Potential Functions

And now that we know how to determine whether or not a vector field is independent of path, we now need to learn how to find a potential function.

There are several ways this can be computed, but the easiest method is the one outlined below:

Show that $\vec{F}$ is conservative on $R^{2}$ by checking that $P_{y} = Q_{x}$ or conservative on $R^{3}$ by confirming that $c u r l \vec{F} = \nabla \times \vec{F} = \vec{0}$ .
Find $f$ by integrating $P$ with respect to $x$ and $Q$ with respect to $y$ , separately.
All terms create the potential function $f$ , however, any duplicate terms are only considered once.

While these steps may seem strange at first, I promise you that they are pretty easy to perform.

Example

Let’s work through an example to see how.

Given $\vec{F} = ⟨ 12 x^{2} + 3 y^{2} + 5 y, 6 x y - 3 y^{2} + 5 x ⟩$ , find $f$ such that $\nabla f = \vec{F}$ .

First, we must confirm that $\vec{F}$ is a conservative vector field.

$\begin{aligned} \vec{F} = ⟨ \underset{P}{\underset{⏟}{12 x^{2} + 3 y^{2} + 5 y}}, \underset{Q}{\underset{⏟}{6 x y - 3 y^{2} + 5 x}} ⟩ \\ P_{y} = 6 y + 5 = Q_{x} \end{aligned}$

Next, we will integrate $P$ with respect to $x$ and $Q$ with respect to $y$ .

$\begin{array}{cc} \int (12 x^{2} + 3 y^{2} + 5 y) d x & \int (6 x y - 3 y^{2} + 5 x) d y \\ = 4 x^{3} + 3 x y^{2} + 5 x y & = 3 x y^{2} - y^{3} + 5 x y \end{array}$

Now, all we have to do is put our terms together, only counting duplicate terms once.

$\begin{array}{cc} \int (12 x^{2} + 3 y^{2} + 5 y) d x & \int (6 x y - 3 y^{2} + 5 x) d y \\ = 4 x^{3} + \underset{―}{3 x y^{2}} + 5 x y & = \underset{―}{3 x y^{2}} - y^{3} + \underset{―}{\underset{―}{5 x y}} \end{array}$

$f (x, y) = 4 x^{3} - y^{3} + 3 x y^{2} + 5 x y + c$ , where $c$ is a constant.

That’s it!

And our solution can be easily verified as true by checking that $\nabla f = \vec{F}$ !

Together we will learn how to prove a vector field is conservative; thus, showing path independence, find a potential function $f (x, y)$ , and then evaluate using the Fundamental Theorem of line integrals.

Video Tutorial w/ Full Lesson & Detailed Examples (Video)

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