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Lecture Comments (6)
 3 answersLast reply by: Joel FredinSun Apr 6, 2014 5:52 PMPost by Joel Fredin on March 29, 2014I don't really understand what you mean by saying "the parametrization doesn't really matter, it's the same integral". For a circle, can i choose another param. instead of x=rcos(t) and y=rsin(t)? Could you give me one or two examples please, i'm lost right now :(Joel 1 answerLast reply by: Professor HovasapianSat May 4, 2013 5:32 AMPost by Eunhee Kim on May 4, 2013@11:46, c(t) = (4cost, 2sint) is a parametrization specifying a counterclockwise direction. What would the parametrization be if I were to parametrize in a clockwise direction?

More on Line Integrals

Find ∫C F given F(x,y) = (2x, − 2y), x = t − 1, y = [t/2] and 0 ≤ t ≤ 2.
• Since x and y are parametrized, we let C(t) = ( t − 1,[t/2] ) and find the line integral ∫C F by computing ∫C F = ∫ab F(C(t)) ×C(t)dt
• Now, F(C(t)) = ( 2t − 2, − t ) and C′(t) = ( 1,[1/2] ) so that ∫02 ( 2t − 2, − t ) ×( 1,[1/2] )dt = ∫02 ( [3t/2] − 2 ) dt
Integrating yields ∫C F = ( [3/4]t2 − 2t ) |02 = − 1.
Find ∫C F given F(x,y) = ( − xy2,x2 + y2), x = t2, y = 2t and 0 ≤ t ≤ [1/4].
• Since x and y are parametrized, we let C(t) = ( t2,2t ) and find the line integral ∫C F by computing ∫C F = ∫ab F(C(t)) ×C(t)dt
• Now, F(C(t)) = ( − 4t4,t4 + 4t2 ) and C′(t) = ( 2t,2 ) so that ∫01 \mathord/ phantom 1 4 4 ( − 4t4,t4 + 4t2 ) ×( 2t,2 )dt = ∫01 \mathord/ phantom 1 4 4 ( − 8t5 + 2t4 + 8t2 ) dt
Integrating yields ∫C F = ( − [4/3]t6 + [2/5]t5 + [8/3]t3 ) |01 \mathord/ phantom 1 4 4 = [641/15,360].
Find the curve integral of the vector field F(x,y) = ( x2,[1/(y2)] ) along the line y = 2x − 1 from (0, − 1) to (1,1).
• To find the curve integral ∫C F along a line, we parametrize the line using two points, P and Q, on the line so that C(t) = P + t(Q − P) from 0 ≤ t ≤ 1.
• Let P = (0, − 1) and Q = (1,1). Then C(t) = (0, − 1) + t[(1,1) − (0, − 1)] = (0, − 1) + t(1,2) so C(t) = (t, − 1 + 2t).
• Now we find the line integral ∫C F by computing ∫C F = ∫ab F(C(t)) ×C(t)dt .
• F(C(t)) = ( t2,[1/(( − 1 + 2t)2)] ) and C′(t) = ( 1,2 ) so that ∫01 ( t2,[1/(( − 1 + 2t)2)] ) ×( 1,2 )dt = ∫01 ( t2 + [2/(( − 1 + 2t)2)] ) dt
Integrating yields ∫C F = ( [(t3)/3] ) |01 + ( − [1/( − 1 + 2t)] ) |01 = − [5/3].
Find the curve integral of the vector field F(x,y) = ( x − y,2y − x ) along the line 0 = [3/4]x − [1/2]y − 4 from (2, − 5) to ( − 2, − 11).
• To find the curve integral ∫C F along a line, we parametrize the line using two points, P and Q, on the line so that C(t) = P + t(Q − P) from 0 ≤ t ≤ 1.
• Let P = (2, − 5) and Q = ( − 2, − 1). Then C(t) = (2, − 5) + t[( − 2, − 1) − (2, − 5)] = (2, − 5) + t( − 4,4) so C(t) = (2 − 4t, − 5 + 4t).
• Now we find the line integral ∫C F by computing ∫C F = ∫ab F(C(t)) ×C(t)dt .
• F(C(t)) = ( 7 − 8t, − 12 + 12t ) and C′(t) = ( − 4,4 ) so that ∫01 ( 7 − 8t, − 12 + 12t ) ×( − 4,4 )dt = ∫01 ( − 76 + 80t ) dt
Integrating yields ∫C F = ( − 76t + 40t2 ) |01 = − 36.
Find the curve integral of the vector field F(x,y) = (√{xy} ,x2y2) along the curve y = x2 from (0,0) to (1,1).
• We parametrize our curve by letting x = t so that y = t2 and C(t) = ( t,t2 ).
• Now we find the line integral ∫C F by computing ∫C F = ∫ab F(C(t)) ×C(t)dt . Note that in this case 0 ≤ t ≤ 1.
• Now, F(C(t)) = ( t3 \mathord/ phantom 3 2 2,t6 ) and C′(t) = ( 1,2t ) so that ∫01 ( t3 \mathord/ phantom 3 2 2,t6 ) ×( 1,2t )dt = ∫01 ( t3 \mathord/ phantom 3 2 2 + 2t7 )dt
Integrating yields ∫C F = ( [2/5]t5 \mathord/ phantom 5 2 2 + [1/4]t8 ) |01 = [13/20].
Find the curve integral of the vector field F(x,y) = (xy2,x2y) along the curve x2 + y2 = 4 from (2,0) to ( − 2,0).
• Since our curve is a circle, we parametrize by letting x = rcos(t) and y = rsin(t) with 0 ≤ t ≤ 2p and r the radius of the circle.
• Then x = 2cos(t) and y = 2sin(t) so C(t) = ( 2cos(t),2sin(t) ). Note that in this case we just want the top semicircle so 0 ≤ t ≤ p.
• Now we find the line integral ∫C F by computing ∫C F = ∫ab F(C(t)) ×C(t)dt .
• Now, F(C(t)) = ( 8cos(t)sin2(t),8cos2(t)sin(t) ) and C′(t) = ( − 2sin(t),2cos(t) ) so that ∫0p ( 8cos(t)sin2(t),8cos2(t)sin(t) ) ×( − 2sin(t),2cos(t) )dt = ∫0p ( − 16cos(t)sin3(t) + 16cos3(t)sin(t) )dt
Integrating yields ∫C F = ( − 4sin4(t) ) |0p + ( − 4cos4(t) ) |0p = 0.
Find ∫C F given F(x,y) = ( [1/2]x, − [1/2]y ) and C(t) = (cos(t),sin(t)) from − [p/2] ≤ t ≤ [p/2].
• To find the line integral ∫C F we compute ∫C F = ∫ab F(C(t)) ×C(t)dt .
• F(C(t)) = ( [1/2]cos(t), − [1/2]sin(t) ) and C′(t) = ( − sin(t),cos(t)). Then ∫C F = ∫ − p \mathord/ phantom − p 2 2p \mathord/ phantom p 2 2 ( [1/2]cos(t), − [1/2]sin(t) ) ×( − sin(t),cos(t))dt = ∫ − p \mathord/ phantom − p 2 2p \mathord/ phantom p 2 2 ( − [1/2]cos(t)sin(t) − [1/2]sin(t)cos(t) )dt
Integrating yields ∫C F = ( − [1/2]sin2(t) ) | − p \mathord/ phantom − p 2 2p \mathord/ phantom p 2 2 = 0.
Find ∫C F given F(x,y) = ( ln(x + y),x3 − y2 ) and C(t) = (t,1 − t) from 0 ≤ t ≤ 1.
• To find the line integral ∫C F we compute ∫C F = ∫ab F(C(t)) ×C(t)dt .
• F(C(t)) = ( 0,t3 − t2 + 2t − 1 ) and C′(t) = (1, − 1). Then ∫C F = ∫01 ( 0,t3 − t2 + 2t − 1 ) ×(1, − 1)dt = ∫01 ( − t3 + t2 − 2t + 1 )dt
Integrating yields ∫C F = ( − [1/4]t4 + [1/3]t3 − t2 + t ) |01 = [1/12].
Find ∫C F given F(x,y) = ( ex,ey ) and C(t) = (t2, − t2) from − 4 ≤ t ≤ 0.
• To find the line integral ∫C F we compute ∫C F = ∫ab F(C(t)) ×C(t)dt .
• F(C(t)) = ( et2,e − t2 ) and C′(t) = (2t, − 2t). Then ∫C F = ∫ − 40 ( et2,e − t2 ) ×(2t, − 2t)dt = ∫ − 40 ( 2tet2 − 2te − t2 )dt
Integrating yields ∫C F = ( et2 + e − t2 ) | − 40 = 2 − e16 − e − 16.
Find ∫C F given F(x,y) = ( 2x2y, − 3y ) and C(t) = ( [t/2], − [t/3] ) from − 2 ≤ t ≤ 2.
• To find the line integral ∫C F we compute ∫C F = ∫ab F(C(t)) ×C(t)dt .
• F(C(t)) = ( − [(t3)/6],t ) and C′(t) = ( [1/2], − [1/3] ). Then ∫C F = ∫ − 22 ( − [(t3)/6],t ) ×( [1/2], − [1/3] )dt = ∫ − 22 ( − [(t3)/12] − [t/3] )dt
Integrating yields ∫C F = ( − [(t4)/48] − [(t2)/6] ) | − 22 = 0.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

More on Line Integrals

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• More on Line Integrals 0:10
• Line Integrals Notation
• Curve Given in Non-parameterized Way: In General
• Curve Given in Non-parameterized Way: For the Circle of Radius r
• Curve Given in Non-parameterized Way: For a Straight Line Segment Between P & Q
• The Integral is Independent of the Parameterization Chosen
• Example 1: Find the Integral on the Ellipse Centered at the Origin
• Example 2: Find the Integral of the Vector Field
• Discussion of Result and Vector Field for Example 2
• Graphical Example

Transcription: More on Line Integrals

Hello, and welcome back to educator.com and multivariable calculus.0000

Today we are going to continue our discussion of line integrals, so, let us just jump right on in.0004

We are going to talk about notation, and as it turns out, you know, the line integral notation, there are several different notations for it.0010

So, I am going to introduce a pretty standard notation and it is going to be the one we end up using more and more especially when we end up talking about double integrals and Green's theorem later on, which is the fundamental theorem of calculus in two dimensions.0019

So, let us introduce this notation, so that we are aware of it. Actually, you know what, I think I am going to do this lesson in blue. There we go. Okay.0032

So, notation. Okay. Let f(x,y) = f and g. So, this is a function. It is a function of 2 variables, it is a vector field. This is the first coordinate function, this is the second coordinate function, but notice I have not included the f(x,y), g(x,y), but it is implicit there.0045

We are going to start to use a slightly shortened notation, and c(t), the curve over which we integrate, or the line, is going to be some x of t, so x is a function of t and y is a function of t. Something like that.0070

Therefore, then the integral of f along c is equal to... we write it as f dx + g dy. That is the notation right here. fdx + gdy. Okay.0086

This just -- the integrand -- is just fg · dx/dy, so it actually has the advantage of sort of representing some dot product already.0106

Now, it is very, very important to remember of course that when we use notations, these are short-hand, so fdx + gdy is short for the following: let me write it up here, well, no, let me write it down here. It is short for f(x(t),y(t)) × dx/dt + g(x(t),y(t)) dy/dt dt.0128

When we multiply all of this out, this fdx + gdy is actually just short for that, which is equivalent to our more theoretical definition f(c(t)) · c'(t). Again, this is just a shorthand notation.0184

This is what it is when it is fully expanded, so, once again I am going to write this... so, again, the integral of f along c which is equal to the integral of f(c(t)) · dc/dt, c', dc/dt, just different notations... dt, forgot that little dt there... is equal to fdx + gdy.0207

This is the notation that is really, really important, but again it is just a question of notation as long as you remember what this means. This believe it or not is actually the important one. It is the definition of the line integral so that is the one that we are going to keep using.0254

We will use this one more and more as we move towards Green's Theorem. Okay.0268

Often, a curve will be given in a non-parameterized way, so, a non-parameterized way.0276

In other words, they might say integrate this function or this vector field over the parabola as the parabola moves from this point to this point.0300

Well, you are going to have to come up with a parameterization because the line integral... the definition of the line integral involves the parameterization c(t).0309

So, you are going to have to find a way to take that curve -- which is expressed usually as a function of (x,y) equals some function of x -- and find a way to express it as a single parameter t.0317

In general, for y = f(x), let x = t and y = f(t).0328

This way, you will always get the parameterization that you need. Now that does not necessarily mean that this is the only parameterization, that is the thing about parameterizations. There is usually more than the one.0345

The nice thing about it is that the integral itself is actually independent of the parameterization. Some parameterizations will make the problem easier, some will make it harder. There is more than one parameterization, so those are not unique.0354

Okay. Just some general examples for the circle of radius r, x = rcos(t) and y = rsin(t). That is the parameterization for a circle. For a straight line segment between p and q, between the points p and q, the parameterization is as follow: c(t) = the point p + t × q - p.0369

We will actually see an application of this a little bit later -- where t runs from 0 to 1.0430

Again, the integral... this is the most important part... the integral is independent of the parameterization.0438

You know what, I do not like this capital letters, I will just stick to writing it out -- is independent of the parameterization chosen.0455

Profoundly important. So, two different parameterizations, same curve, same direction, you are going to get the same integral.0478

Now, when you choose a parameterization, you are specifying x a direction of traversal.0488

We will say more about this in the next lesson when we actually talk about reversing the path, integrating in one direction, integrating along another direction. You are specifying a direction of traversal.0514

In other words, if you were to parameterize that curve moving in this direction from here to here would be one parameterization, but if you wanted to move in this direction, if you wanted to move like this from here to here, it actually changes the parameterization.0528

There is a formal mathematical way of doing that and again we will see that in the next lesson when we see some examples, but for the most part what is important is that the parameterization that you use does not matter, the integral will be the same, that is what is important.0545

Let us do an example. Okay, so example 1. Find the integral of the vector field f(x,y) = (xy2,x2y) on the ellipse centered at the origin with major radius 4 and minor radius 2 from θ = pi/6 to θ = 3pi/4.0559

Okay, so find the integral of the function on the ellipse, so we are integrating along the ellipse, centered at the origin, the major radius is 4 the minor radius is 2 and we want to go from the point on the ellipse represented by pi/6 to 3pi/4.0630

Let us go ahead and draw what this looks like, and then we will do the parameterization of the ellipse.0646

We have that, major radius is 4 so this distance from there to there is 4, minor radius is 2, so something like this -- sort of like that, I am not the best ellipse drawer but something like this -- and θ = pi/6, that means this point and 3pi/4 is the 45 degree angle to this point.0653

What we are doing is we are integrating from here to here, so let me do this in red. We are integrating along this path. Here to here. This function.0678

Let us go ahead and write our, so we have f(x), let us go ahead and write our c(t). As it turns out, you parameterize an ellipse sort of the same way you parameterize a circle.0692

You have rcos(t),rsin(t) but now the two radii are different, so it is just going to be this radius cos(t), and this radius sin(t).0705

What we have is 4cos(t) and 2sin(t), now, this particular parameterization, by choosing this parameterization I am actually moving in this direction.0715

This is what we mean when we say any time you choose a parameterization you are specifying a direction. I am specifying a counterclockwise movement along this path.0728

If I had gone, let us say this way, let us say I wanted to parameterize from this point to this point, the parameterization would actually have to change.0736

So, when you choose a parameterization, you are actually specifying a particular direction and it is very important that you know what that direction is.0745

You have to make sure that you understand all of the particulars of the objects that you are working with. Okay, so let us go ahead and flip the page. I am going to rewrite the parameterization of the function and then I am going to start working with it and integrate it.0753

Let me go back to blue here. So, we have f(x,y) = (xy2,x2y), and c(t) = 4cos(t)2sin(t), and t is going to run from pi/6 to 3pi/4.0765

Okay. Let us form, as always, we form f(c(t)) · dc/dt. O0801

Okay, so, f(c(t)) = let us put -- this is x, this is y -- we put those in here, we get 4cos(t),sin2(t)... uh, no, sorry, that is not right... 4cos(t) × 2sin(t)2, and we get 4cos(t)2 × 2sin(t).0820

This is going to equal 2 × 2 is 4, this is going to be 16 × cos(t) sin2(t), and this is going to be 32cos2(t)sin(t).0853

Okay. Now we have that. Now let us form dc/dt. dc/dt = the derivative of this thing equals -4 × sin(t) and we have 2 × cos(t).0874

Now, when we form the f(c(t)) · dc/dt, it is this × that + this × that. We end up with -64cos(t)sin3(t) + this × that which is 64cos3(t)sin(t).0890

So, our integral of f over c is equal to the integral from pi/6 to 3pi/4 of this expression, -64cos(t)sin3(t) + 64cos3(t)sin(t) dt.0926

When we use our mathematical software, we end up with the number 2. That is it. I am certainly hoping that you are not trying to do this manually, it is not worth it.0951

One of the nice things about having -- in fact, the best hting about having math software is now you are no longer confined to examples that simply for logistical reasons are simple so that you can get through them.0960

Now you can choose any function that you want, any parameterization, and boom software gives you the answer. This is what is important, being able to set this up. This a computer can do. This is what you want to understand.0974

So let us go ahead and do another example here. A very, very important example.0987

So, example 2. Let us see here. Find the integral of the vector field f(x,y) = -y/x2 + y2 x/x2 + y2, around the circle of radius 4 from (4,0) to the point (2,2sqrt(3)).0995

In this particular case, they are actually giving us two specific points, from this point to this point.1068

So, what we are going to have to do is, in order to find the parameterization, we are going to have to find the lower and upper limits of integration.1073

We are going to have to convert this point and this point to t values based on the parameterization that we choose. Okay.1080

Let us go ahead and do a little geometry here just to see what this looks like, so that we make sure we understand what we are doing.1086

So, we have this circle of radius 4, right? This is 4, that is 4, we know how to parameterize that, that is going to be 4cos(t), 4sin(t).1092

What we are doing is we are going from this point, the point (4,0) all the way to this point right here (2,2sqrt(3)) and we are moving in this direction, that direction, that is the parameterization.1102

Our c(t) is going to be the 4cos(t), 4sin(t)... now, when we do the integration, we need values of t because we are integrating along the curve so the t values are going to be the lower and upper limits of integration.1122

So, this point is represented by t = 0, so at the point (4,0), t = 0, so that is our lower limit of integration. Well, how about here, what is t going to be when we get to here?1143

Well, at the point (2, 2sqrt(3)), well, let us go ahead and solve, 4cos(t) is the x value here, this is the y value, let us just go ahead and solve for t.1161

So, 4cos(t) = 2, cos(t) = 2/4 = 1/2, and since we are in the first quadrant, t = pi/3.1177

So, we are going to be integrating from 0 to pi/3, that is our lower limit of integration, that is our upper limit of integration.1194

Notice that in this case they did not give us the parameterization and the t values that we are going to traverse, they gave us 2 points and they gave us just the geometric on this unit circle.1201

We came up with a parameterization, that should not be a problem. We need to find out what the t value is so we have to evaluate one of these to find the t values.1212

t goes to 0 as t moves up to pi/3, that covers that segment of the circle.1221

Okay. So, t is going to run from 0 all the way to pi/3. Alright, let us go ahead and solve the integral. Well, f(c(t)) when I put this parameterization into there, I end up with the following.1231

I end up with -4sin(t)/16cos2(t) + 16sin2(t), and I end up with 4cos(t)/16cos2(t) + 16sin2(t).1255

Okay. This of course is equal to -4sin(t)/16, cos2(t), sin2(t) = 1, so that factors out, and I have 4cos(t)/16. Let us get rid of these random little lines that tend to show up at the bottom of the page.1284

This is going to equal sin(t) - sin(t)/4, and... I am sorry, -4... oh, this is ridiculous, let us see... -4sin(t)/4, and 4... I have already cancelled, now I am the one that is messing up arithmetic. 4 and 16, 4 and 16, so I get -sin(t)/4 and I get cos(t)/4.1312

How is that? That is good. Okay. Now, let us do dc/dt.1351

So, dc/dt, or c'(t), that is going to equal -4sin(t) and 4cos(t), so, f(c(t)) · dc/dt is going to equal -- let me rewrite the f(c(t)) again -- f(c(t)) is going to equal (-sin(t)/4,cos(t)/4), so it is the dot product of this thing and this thing.1359

When I do that, I am going to get sin2(t) + cos2(t) and that is going to equal 1, so our final integral, the integral of f over this particular curve is going to be the integral from 0 to pi/3 of just plain old dt, 1dt.1404

Of course, that equals pi/3.1427

Here we are busy trying to solve this line integral, and we end up with notice, 0 to pi/3 and we end up with an answer for this integral is pi/3. This is not a coincidence.1433

Okay. This is a very, very important result and a very important vector field, especially for those of you in physics and engineering. You are going to see this vector field over and over and over again.1443

Let us just discuss this real quickly. So, this is a very important result and a very important vector field.1465

Here is the result. Here is why it is important: if you ever integrate this vector field along any curve, along any... and this one I definitely will put in capitals... along any curve, from point p1 to p2, and op1 and op2 make angles of θ1 and θ2, respectively, with the x-axis. I will draw this out in just a minute so that you see what happens... respectively with the x-axis, then the integral of f along that particular curve is equal to θ2 - θ1, or δθ.1481

Here is what we mean by that. So, let us say we have this coordinate plane and let us say we have this curve that is going like this, okay?1565

If this is point p1 and this is point p2, p1 and p2, if this is θ1 and this is -- this angle is θ1 and this angle is θ2, remember? we always measure our θs from the positive x-axis moving in a counterclockwise direction.1578

Then the integral of this particular vector field, f(x,y) -y/x2 + y2, x/x2 + y2, the integral of this vector field over any curve, no matter what curve you trace, whether it is like this or this or it could be all whacky and end up here, the difference in the angle, the integral of this vector field is always going to be the difference between θ2 and θ1, no matter what it is.1604

It does not have to start at the origin by the way, notice here this is one angle, this is the other angle. In our example, we start from the origin and we integrate up to pi/3.1636

The angle difference was pi/3, the integral was pi/3. Here, same thing. If this were 3pi/4 and this were pi/4, well, we would have 3pi/4 - pi/4, 2pi/4, the integral would be pi/2.1646

This is a profoundly important vector field. You will see it over and over and over again.1659

It will save you a lot of time if you recognize that as you integrate this vector field over absolutely any path, where this angle arrangement is involved, you can just take the different of the angles and that is the integral.1665

Thank you for joining us here at educator.com, we will see you next time for a further discussion of line integrals. Take care, bye-bye.1678