Evaluate the line integral where is the parameterized path

Note that this is different from the double integrals that we were working with in the previous chapter where the points came out of some two-dimensional region. We will assume that the curve is smooth defined shortly and is given by the parametric equations,. We will often want to write the parameterization of the curve as a vector function. In this case the curve is given by,. If you recall from Calculus II when we looked at the arc length of a curve given by parametric equations we found it to be,.

So, to compute a line integral we will convert everything over to the parametric equations. The line integral is then,. If we use the vector form of the parameterization we can simplify the notation up somewhat by noticing that,. Using this notation, the line integral becomes,. Next we need to talk about line integrals over piecewise smooth curves.

Below is an illustration of a piecewise smooth curve. Evaluation of line integrals over piecewise smooth curves is a relatively simple thing to do. All we do is evaluate the line integral over each of the pieces and then add them up. The line integral for some function over the above piecewise curve would be,.

Notice that we put direction arrows on the curve in the above example. The direction of motion along a curve may change the value of the line integral as we will see in the next section.

Cubing it out is not that difficult, but it is more work than a simple substitution. So, the previous two examples seem to suggest that if we change the path between two points then the value of the line integral with respect to arc length will change. In a later section we will investigate this idea in more detail. Here is the parameterization of the curve. A breakdown of the steps:. This definition is not very useful by itself for finding exact line integrals. If data is provided, then we can use it as a guide for an approximate answer.

Fortunately, there is an easier way to find the line integral when the curve is given parametrically or as a vector valued function.

We are now ready to state the theorem that shows us how to compute a line integral. The main application of line integrals is finding the work done on an object in a force field. As usual, we add up all the small pieces of work and take the limit as the pieces get small to end up with an integral. Notice that work done by a force field on an object moving along a curve depends on the direction that the object goes.

In fact the opposite direction will produce the negative of the work done in the original direction. This is clear from the fact that everything is the same except the order which we write a and b. Find the value of integral where is the upper half of the unit circle. The integrand is Figure shows the graph of curve C , and the sheet formed by them. Notice that this sheet has the same area as a rectangle with width and length 2. To see that using the definition of line integral, we let be a parameterization of C.

Then, for any number in the domain of r. Find the value of where is the curve parameterized by. Find the shape formed by C and the graph of function. Note that in a scalar line integral, the integration is done with respect to arc length s , which can make a scalar line integral difficult to calculate.

To make the calculations easier, we can translate to an integral with a variable of integration that is t. Let for be a parameterization of Since we are assuming that is smooth, is continuous for all in In particular, and exist for all in According to the arc length formula, we have.

If width is small, then function is almost constant over the interval Therefore,. See Figure. Note that. In other words, as the widths of intervals shrink to zero, the sum converges to the integral Therefore, we have the following theorem.

Let be a continuous function with a domain that includes the smooth curve with parameterization Then. Although we have labeled Figure as an equation, it is more accurately considered an approximation because we can show that the left-hand side of Figure approaches the right-hand side as In other words, letting the widths of the pieces shrink to zero makes the right-hand sum arbitrarily close to the left-hand sum.

Let be a continuous function with a domain that includes the smooth curve C with parameterization Then. Note that a consequence of this theorem is the equation In other words, the change in arc length can be viewed as a change in the t domain, scaled by the magnitude of vector.

Find the value of integral where is part of the helix parameterized by. To compute a scalar line integral, we start by converting the variable of integration from arc length s to t. Then, we can use Figure to compute the integral with respect to t. Note that and. Notice that Figure translated the original difficult line integral into a manageable single-variable integral. Evaluate where C is the curve with parameterization. Use the two-variable version of Figure.

Find the value of integral where is part of the helix parameterized by Notice that this function and curve are the same as in the previous example; the only difference is that the curve has been reparameterized so that time runs twice as fast. As with the previous example, we use Figure to compute the integral with respect to t.

Notice that this agrees with the answer in the previous example. Changing the parameterization did not change the value of the line integral. Scalar line integrals are independent of parameterization, as long as the curve is traversed exactly once by the parameterization.

Evaluate line integral where is the line with parameterization Reparameterize C with parameterization recalculate line integral and notice that the change of parameterization had no effect on the value of the integral. Both line integrals equal. Use Figure. Now that we can evaluate line integrals, we can use them to calculate arc length.

If then. Therefore, is the arc length of. A wire has a shape that can be modeled with the parameterization Find the length of the wire. The length of the wire is given by where C is the curve with parameterization r.

Find the length of a wire with parameterization. The second type of line integrals are vector line integrals, in which we integrate along a curve through a vector field. For example, let. To answer this question, first note that a particle could travel in two directions along a curve: a forward direction and a backward direction. The work done by the vector field depends on the direction in which the particle is moving. Therefore, we must specify a direction along curve C ; such a specified direction is called an orientation of a curve.

The specified direction is the positive direction along C ; the opposite direction is the negative direction along C. When C has been given an orientation, C is called an oriented curve Figure. The work done on the particle depends on the direction along the curve in which the particle is moving. A closed curve is one for which there exists a parameterization such that and the curve is traversed exactly once.

In other words, the parameterization is one-to-one on the domain. Let be a parameterization of C for such that the curve is traversed exactly once by the particle and the particle moves in the positive direction along C. Divide the parameter interval into n subintervals of equal width. Denote the endpoints of by Points P i divide C into n pieces. If is small, then as the particle moves from to along C , it moves approximately in the direction of the unit tangent vector at the endpoint of Let denote the endpoint of Then, the work done by the force vector field in moving the particle from to P i is so the total work done along C is.

Letting the arc length of the pieces of C get arbitrarily small by taking a limit as gives us the work done by the field in moving the particle along C. Therefore, the work done by F in moving the particle in the positive direction along C is defined as.

The vector line integral of vector field F along oriented smooth curve C is. With scalar line integrals, neither the orientation nor the parameterization of the curve matters. As long as the curve is traversed exactly once by the parameterization, the value of the line integral is unchanged. With vector line integrals, the orientation of the curve does matter.

If we think of the line integral as computing work, then this makes sense: if you hike up a mountain, then the gravitational force of Earth does negative work on you. In other words, reversing the path changes the work value from negative to positive in this case.

As with scalar line integrals, it is easier to compute a vector line integral if we express it in terms of the parameterization function r and the variable t. To translate the integral in terms of t , note that unit tangent vector T along C is given by assuming Since as we saw when discussing scalar line integrals, we have. Because of Figure , we often use the notation for the line integral.

If then d r denotes vector differential. Find the value of integral where is the semicircle parameterized by and. We can use Figure to convert the variable of integration from s to t. We then have. Reversing Orientation Find the value of integral where is the semicircle parameterized by and. Notice that this is the same problem as Figure , except the orientation of the curve has been traversed.

In this example, the parameterization starts at and ends at By Figure ,. Notice that this is the negative of the answer in Figure. Then, the previous two examples illustrate the following fact:. That is, reversing the orientation of a curve changes the sign of a line integral. Let be a vector field and let C be the curve with parameterization for Which is greater: or. Imagine moving along the path and computing the dot product as you go.

Another standard notation for integral is In this notation, P , Q , and R are functions, and we think of d r as vector To justify this convention, recall that Therefore,. If then which implies that Therefore. Find the value of integral where C is the curve parameterized by.

As with our previous examples, to compute this line integral we should perform a change of variables to write everything in terms of t. In this case, Figure allows us to make this change:.

Write the integral in terms of t using Figure. We have learned how to integrate smooth oriented curves. Now, suppose that C is an oriented curve that is not smooth, but can be written as the union of finitely many smooth curves. In this case, we say that C is a piecewise smooth curve. To be precise, curve C is piecewise smooth if C can be written as a union of n smooth curves such that the endpoint of is the starting point of Figure. When curves satisfy the condition that the endpoint of is the starting point of we write their union as.

The next theorem summarizes several key properties of vector line integrals. Let F and G be continuous vector fields with domains that include the oriented smooth curve C. Notice the similarities between these items and the properties of single-variable integrals.

Properties i. Property iii. If we think of the integral as computing the work done on a particle traveling along C , then this makes sense. If the particle moves backward rather than forward, then the value of the work done has the opposite sign.

This is analogous to the equation Finally, if are intervals, then. Find the value of integral where C is the rectangle oriented counterclockwise in a plane with vertices and where Figure. Note that curve C is the union of its four sides, and each side is smooth.

Therefore C is piecewise smooth. Let represent the side from to let represent the side from to let represent the side from to and let represent the side from to Figure. We want to compute each of the four integrals on the right-hand side using Figure. Before doing this, we need a parameterization of each side of the rectangle. Here are four parameterizations note that they traverse C counterclockwise :. Notice that the value of this integral is positive, which should not be surprising.

As we move along curve C 1 from left to right, our movement flows in the general direction of the vector field itself.