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### Section 5-5 : Fundamental Theorem for Line Integrals

In Calculus I we had the Fundamental Theorem of Calculus that told us how to evaluate definite integrals. This told us,

$\int_{{\,a}}^{{\,b}}{{F'\left( x \right)dx}} = F\left( b \right) - F\left( a \right)$

It turns out that there is a version of this for line integrals over certain kinds of vector fields. Here it is.

#### Theorem

Suppose that $$C$$ is a smooth curve given by $$\vec r\left( t \right)$$, $$a \le t \le b$$. Also suppose that $$f$$ is a function whose gradient vector, $$\nabla f$$, is continuous on $$C$$. Then,

$\int\limits_{C}{{\nabla f\centerdot \,d\,\vec r}} = f\left( {\vec r\left( b \right)} \right) - f\left( {\vec r\left( a \right)} \right)$

Note that $$\vec r\left( a \right)$$ represents the initial point on $$C$$ while $$\vec r\left( b \right)$$ represents the final point on $$C$$. Also, we did not specify the number of variables for the function since it is really immaterial to the theorem. The theorem will hold regardless of the number of variables in the function.

#### Proof

This is a fairly straight forward proof.

For the purposes of the proof we’ll assume that we’re working in three dimensions, but it can be done in any dimension.

Let’s start by just computing the line integral.

\begin{align*}\int\limits_{C}{{\nabla f\centerdot d\,\vec r}} & = \int_{{\,a}}^{{\,b}}{{\nabla f\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right)\,dt}}\\ & = \int_{{\,a}}^{{\,b}}{{\left( {\frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial f}}{{\partial z}}\frac{{dz}}{{dt}}} \right)\,dt}}\end{align*}

Now, at this point we can use the Chain Rule to simplify the integrand as follows,

\begin{align*}\int\limits_{C}{{\nabla f\centerdot d\,\vec r}} & = \int_{{\,a}}^{{\,b}}{{\left( {\frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial f}}{{\partial z}}\frac{{dz}}{{dt}}} \right)\,dt}}\\ & = \int_{{\,a}}^{{\,b}}{{\frac{d}{{dt}}\left[ {f\left( {\vec r\left( t \right)} \right)} \right]\,dt}}\end{align*}

To finish this off we just need to use the Fundamental Theorem of Calculus for single integrals.

$\int\limits_{C}{{\nabla f\centerdot d\,\vec r}} = f\left( {\vec r\left( b \right)} \right) - f\left( {\vec r\left( a \right)} \right)$

Let’s take a quick look at an example of using this theorem.

Example 1 Evaluate $$\displaystyle \int\limits_{C}{{\nabla f\centerdot \,d\,\vec r}}$$ where $$f\left( {x,y,z} \right) = \cos \left( {\pi x} \right) + \sin \left( {\pi y} \right) - xyz$$ and $$C$$ is any path that starts at $$\left( {1,\frac{1}{2},2} \right)$$ and ends at $$\left( {2,1, - 1} \right)$$.
Show Solution

First let’s notice that we didn’t specify the path for getting from the first point to the second point. The reason for this is simple. The theorem above tells us that all we need are the initial and final points on the curve in order to evaluate this kind of line integral.

So, let $$\vec r\left( t \right)$$, $$a \le t \le b$$ be any path that starts at $$\left( {1,\frac{1}{2},2} \right)$$ and ends at $$\left( {2,1, - 1} \right)$$. Then,

$\vec r\left( a \right) = \left\langle {1,\frac{1}{2},2} \right\rangle \hspace{0.5in}\vec r\left( b \right) = \left\langle {2,1, - 1} \right\rangle$

The integral is then,

\begin{align*}\int\limits_{C}{{\nabla f\centerdot \,d\,\vec r}} & = f\left( {2,1, - 1} \right) - f\left( {1,\frac{1}{2},2} \right)\\ & = \cos \left( {2\pi } \right) + \sin \pi - 2\left( 1 \right)\left( { - 1} \right) - \left( {\cos \pi + \sin \left( {\frac{\pi }{2}} \right) - 1\left( {\frac{1}{2}} \right)\left( 2 \right)} \right)\\ & = 4\end{align*}

Notice that we also didn’t need the gradient vector to actually do this line integral. However, for the practice of finding gradient vectors here it is,

$\nabla f = \left\langle { - \pi \sin \left( {\pi x} \right) - yz,\pi \cos \left( {\pi y} \right) - xz, - xy} \right\rangle$

The most important idea to get from this example is not how to do the integral as that’s pretty simple, all we do is plug the final point and initial point into the function and subtract the two results. The important idea from this example (and hence about the Fundamental Theorem of Calculus) is that, for these kinds of line integrals, we didn’t really need to know the path to get the answer. In other words, we could use any path we want and we’ll always get the same results.

In the first section on line integrals (even though we weren’t looking at vector fields) we saw that often when we change the path we will change the value of the line integral. We now have a type of line integral for which we know that changing the path will NOT change the value of the line integral.

Let’s formalize this idea up a little. Here are some definitions. The first one we’ve already seen before, but it’s been a while and it’s important in this section so we’ll give it again. The remaining definitions are new.

#### Definitions

First suppose that $$\vec F$$ is a continuous vector field in some domain $$D$$.

1. $$\vec F$$ is a conservative vector field if there is a function $$f$$ such that $$\vec F = \nabla f$$. The function $$f$$ is called a potential function for the vector field. We first saw this definition in the first section of this chapter.

2. $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}}$$ is independent of path if $$\displaystyle \int\limits_{{{C_1}}}{{\vec F\centerdot \,d\,\vec r}} = \int\limits_{{{C_2}}}{{\vec F\centerdot \,d\,\vec r}}$$ for any two paths $${C_1}$$ and $${C_2}$$ in $$D$$ with the same initial and final points.

3. A path $$C$$ is called closed if its initial and final points are the same point. For example, a circle is a closed path.

4. A path $$C$$ is simple if it doesn’t cross itself. A circle is a simple curve while a figure 8 type curve is not simple.

5. A region $$D$$ is open if it doesn’t contain any of its boundary points.

6. A region $$D$$ is connected if we can connect any two points in the region with a path that lies completely in $$D$$.

7. A region $$D$$ is simply-connected if it is connected and it contains no holes. We won’t need this one until the next section, but it fits in with all the other definitions given here so this was a natural place to put the definition.

With these definitions we can now give some nice facts.

#### Facts

1. $$\displaystyle \int\limits_{C}{{\nabla f\centerdot \,d\,\vec r}}$$ is independent of path.

This is easy enough to prove since all we need to do is look at the theorem above. The theorem tells us that in order to evaluate this integral all we need are the initial and final points of the curve. This in turn tells us that the line integral must be independent of path.

2. If $$\vec F$$ is a conservative vector field then $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}}$$ is independent of path.

This fact is also easy enough to prove. If $$\vec F$$ is conservative then it has a potential function, $$f$$, and so the line integral becomes $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}} = \int\limits_{C}{{\nabla f\centerdot \,d\,\vec r}}$$. Then using the first fact we know that this line integral must be independent of path.

3. If $$\vec F$$ is a continuous vector field on an open connected region $$D$$ and if$$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}}$$ is independent of path (for any path in $$D$$) then $$\vec F$$ is a conservative vector field on $$D$$.

4. If $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}}$$ is independent of path then $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}} = 0$$ for every closed path $$C$$.

5. If $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}} = 0$$ for every closed path $$C$$ then $$\displaystyle \int\limits_{C}{{\vec F\centerdot \,d\,\vec r}}$$ is independent of path.

These are some nice facts to remember as we work with line integrals over vector fields. Also notice that 2 & 3 and 4 & 5 are converses of each other.

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