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Lecture Comments (5)

0 answers

Post by Burhan Akram on March 9, 2015

Hi Prof. Raffi,

At 13:48, I think the (-2,-2) should be in 4th quadrant.

3 answers

Last reply by: Professor Hovasapian
Sat May 18, 2013 3:33 AM

Post by Eunhee Kim on May 4, 2013

is the line integral in multivariable calc. the area under the curve also?

Line Integrals

Sketch the vectors of the following Vector Field at the indicated points.
F(x,y) = (x,y); F( − 1, − 1), F(1,0), F(0,1), F(1,1), F( − 1,0), F(0, − 1)
  • We first find the values of F at the indicated points: F( − 1, − 1) = ( − 1, − 1), F(1,0) = (1,0), F(0,1) = (0,1), F(1,1) = (1,1), F( − 1,0) = ( − 1,0), F(0, − 1) = (0, − 1).
  • To sketch these vectors we start at the point (x,y) and move towards the direction and length F(x,y). That is, we treat (x,y) as the origin for the vector F(x,y).
Sketch the vectors of the following Vector Field at the indicated points.
F(x,y) = (x2, − y); F( − 1, − 1), F(1,1), F(2,2), F( − 1,1), F( − 2,2), F( − 2, − 2)
  • We first find the values of F at the indicated points.
  • Then F( − 1, − 1) = (( − 1)2, − (1)) = (1,1), similarly F(1,1) = (1, − 1), F(2,2) = (4, − 2), F( − 1,1) = (1, − 1), F( − 2,2) = (4, − 2), F( − 2, − 2) = (4,2).
  • To sketch these vectors we start at the point (x,y) and move towards the direction and length F(x,y). That is, we treat (x,y) as the origin for the vector F(x,y).
Sketch the vectors of the following Vector Field at the indicated points.
F(x,y) = (x − y,x + y); F( − 1, − 1), F(1,0), F(2,0), F( − 1,0), F( − 2,0), F(0,1)
  • We first find the values of F at the indicated points.
  • Then F(0, − 1) = ((0) − ( − 1),(0) + ( − 1)) = (1, − 1), similarly F(1,0) = (1,1), F(2,0) = (2,2), F( − 1,0) = ( − 1, − 1), F( − 2,0) = ( − 2, − 2), F(0,1) = ( − 1,1).
  • To sketch these vectors we start at the point (x,y) and move towards the direction and length F(x,y). That is, we treat (x,y) as the origin for the vector F(x,y).
Let F(x,y) = sin(x) + sin(y).
i) Find the gradient of F.
  • Recall that the gradient of F is ∇F(x,y) = ( [dF/dx],[dF/dy] ) .
Then ∇F = (cos(x),cos(y)).
Let F(x,y) = sin(x) + sin(y).
ii) Sketch the vectors of ∇F(x,y) for the following points: ∇F(0,0), ∇F( 0,[p/2] ), F( [p/2],0 )
  • First we compute each vale of ∇F(x,y): ∇F(0,0) = (cos(0),cos(0)) = (1,1), ∇F( 0,[p/2] ) = ( cos(0),cos( [p/2] ) ) = (1,0) and ∇F( [p/2],0 ) = ( cos( [p/2] ),cos(0) ) = (0,1).
  • To sketch these vectors we start at the point (x,y) and move towards the direction and length ∇F(x,y). That is, we treat (x,y) as the origin for the vector ∇F(x,y).
Given F(x,y) = (xy,x + y) and C(t) = (2t,t), find the line integral of F along C at 0 ≤ t ≤ 1.
  • The line integral of a function F along a curve C on t ∈ [a,b] is defined as ∫C F = ∫ab F(C(t)) ×C′(t)dt . Note that this is a scalar product.
  • Now, C′(t) = (2,1) and F(C(t)) is found by letting x = 2t and y = t.
  • So F(C(t)) = ((2t)(t),(2t) + (t)) = (2t2,t).
Hence ∫ab F(C(t)) ×C′(t)dt = ∫01 (2t2,t) × (2,1)dt = ∫01 (4t2 + t)dt = [4/3]t3 | 01 + [1/2]t2 |01 = [11/6]
Given F(x,y) = (cos(x),sin(x)) and C(t) = (t, − t), find the line integral of F along C at 0 ≤ t ≤ p.
  • The line integral of a function F along a curve C on t ∈ [a,b] is defined as ∫C F = ∫ab F(C(t)) ×C′(t)dt . Note that this is a scalar product.
  • Now, C′(t) = (1,1) and F(C(t)) is found by letting x = t and y = − t.
  • So F(C(t)) = (cos(t),sin( − t)) = (cos(t), − sin(t)).
Hence ∫ab F(C(t)) ×C′(t)dt = ∫0p (cos(t), − sin(t)) × (1,1)dt = ∫0p (cos(t) − sin(t))dt = sin(t) | 0p + cos(t) |0p = − 2
Given F(x,y) = ( x2 + y2,[1/(x2 + y2)] ) and C(t) = (cos(t),sin(t)), find the line integral of F along C at 0 ≤ t ≤ [p/2].
  • The line integral of a function F along a curve C on t ∈ [a,b] is defined as ∫C F = ∫ab F(C(t)) ×C′(t)dt . Note that this is a scalar product.
  • Now, C′(t) = ( − sin(t),cos(t)) and F(C(t)) is found by letting x = cos(t) and y = sin(t).
  • So F(C(t)) = ( cos2(t) + sin2(t),[1/(cos2(t) + sin2(t))] ) = (1,1).
Hence ∫ab F(C(t)) ×C′(t)dt = ∫0p \mathord/ phantom p 2 2 (1,1) × ( − sin(t),cos(t))dt = ∫0p \mathord/ phantom p 2 2 ( − sin(t) + cos(t))dt = cos(t) | 0p \mathord/ phantom p 2 2 + sin(t) |0p \mathord/ phantom p 2 2 = 0.
Given F(x,y) = ( √{x2 + y2} , − xy ) and C(t) = (sin(t),cos(t)), find the line integral of F along C at 0 ≤ t ≤ [3p/2].
  • The line integral of a function F along a curve C on t ∈ [a,b] is defined as ∫C F = ∫ab F(C(t)) ×C′(t)dt . Note that this is a scalar product.
  • Now, C′(t) = (cos(t), − sin(t)) and F(C(t)) is found by letting x = sin(t) and y = cos(t).
  • So F(C(t)) = ( √{sin2(t) + cos2(t)} , − (sin(t))(cos(t)) ) = (1, − sin(t)cos(t)).
Hence ∫ab F(C(t)) ×C′(t)dt = ∫03p \mathord/ phantom 3p 4 4 (1, − sin(t)cos(t)) × (cos(t), − sin(t))dt = ∫03p \mathord/ phantom 3p 4 4 (cos(t) + sin2(t)cos(t))dt = sin(t) | 03p \mathord/ phantom 3p 4 4 + [1/3]sin3(t) |03p \mathord/ phantom 3p 4 4 = − [4/3]
Given F(x,y) = ( 2x2 − y2,[1/x] + [1/y] ) and C(t) = ( et,e − t ), find the line integral of F along C at 0 ≤ t ≤ 1.
  • The line integral of a function F along a curve C on t ∈ [a,b] is defined as ∫C F = ∫ab F(C(t)) ×C′(t)dt . Note that this is a scalar product.
  • Now, C′(t) = ( et, − e − t ) and F(C(t)) is found by letting x = et and y = e − t.
  • So F(C(t)) = ( 2( et )2 − ( e − t )2,[1/(et)] + [1/(e − t)] ) = ( 2e2t − e − 2t,e − t + et ).
Hence ∫01 F(C(t)) ×C′(t)dt = ∫01 ( 2e2t − e − 2t,e − t + et ) × ( et, − e − t )dt = ∫01 (2e3t − e − 2t − e − t − 1)dt = [2/3]e3t |01 + [1/2]e − 2t |01 + e − t | 01 − t |01 = [2/3]e3 + [1/2]e − 2 + e − 1 − [19/6]
Given F(x,y) = ( x + y,x − y ) and C(t) = (t2, − 3t), find the line integral of F along C at 1 ≤ t ≤ 3.
  • The line integral of a function F along a curve C on t ∈ [a,b] is defined as ∫C F = ∫ab F(C(t)) ×C′(t)dt . Note that this is a scalar product.
  • Now, C′(t) = (2t, − 3) and F(C(t)) is found by letting x = t2 and y = − 3t.
  • So F(C(t)) = (t2 + ( − 3t),t2 − ( − 3t)) = (t2 − 3t,t2 + 3t).
Hence ∫ab F(C(t)) ×C′(t)dt = ∫13 (t2 − 3t,t2 + 3t) × (2t, − 3)dt = ∫13 (2t3 − 9t2 − 9t)dt = [1/2]t4 |13 − 3t3 | 13 − [9/2]t2 |13 = − 74.

*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

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
  • Line Integrals 0:18
    • Introduction to Line Integrals
    • Definition 1: Vector Field
    • Example 1
    • Example 2: Gradient Operator & Vector Field
    • Example 3
    • Vector Field, Curve in Space & Line Integrals
    • Definition 2: F(C(t)) ∙ C'(t) is a Function of t
    • Example 4
    • Definition 3: Line Integrals
    • Example 5
    • Example 6

Transcription: Line Integrals

Hello, and welcome back to educator.com and multi variable calculus.0000

Today, we are going to introduce the concept of a line integral. Profoundly important concept, absolutely shows up everywhere. It is the heart and soul of math and science.0006

So, with that, let us just jump right on in. Okay.0016

So, this idea of a line integral, it is not only called a line integral. I also want to give you a couple of other terms that you are going to hear it referred to as.0021

It is also called a path integral, or a curve integral, so it does not matter what you call it. Again, the idea is what is going on. 0030

So, path integral, curve integral, line integral... I think curve integral personally actually is the best, because what you are doing is you are actually just integrating along a curve.0045

It makes the most sense to me, but it does not matter what is you call it.0056

Before we actually introduce the definition of the integral, I am going to talk about something called a vector field and give you that definition, give you a couple of examples and then we will go ahead and give the definition of a line integral and do some examples of that.0060

So, we have had functions that go from R to let us say R2. This is just a curve in 2-space.0075

We take a value t, and then we have some function where you actually have a coordinate, a first coordinate function is f(t), well, something like this.0095

An example would be let us say f(t) = let us say cos(t) and sin(t), so what you end up... you take 1 variable and you end up creating a vector from it. That vector traces out a curve in 2-space.0110

Okay. We have also had the following, what we have been dealing with lately. We have had a function, a multivariable function, so R2 to R, or R3 to R, or R4 to R. Some function of several variables.0131

Well, you take several variables, but you spit out an actual single variable.0143

This is just a function in 2-space and an example of that would be f(x,y) = x2 + y4, something like that. You take 2 values, you evaluate the function, and you end up with one thing. Okay.0149

So, we have talked about functions where the departure space and the arrival space are not the same dimension, for example a function f from R2 to R3, or R4 to R2.0172

We have talked about functions where you have some RN going to RM, and now we are going to talk about a special case of that.0189

Now, let us talk about where the departure space and the arrival space happen to be the same dimension.0203

Let us talk about functions f mapping from R2 to R2, or R3 to R3, and again, for our purposes R2 and R3 is what is important, but this is generally functions from RN to RN.0214

So, where the dimension of the departure space is the same of the arrival space. Let us go ahead and formalize this. Let us have a definition.0232

We are going to define something called a vector field. Okay. Let u be an open set -- and I am sure you remember what open set is, it is just a set that does not include its boundary points -- let f be a mapping which to every point of u associates a vector of the same dimension. That is it.0244

Let us say if I am working from R2 to R2, so if I have some mapping... actually you know what, let us just go ahead and do the example.0307

So, in R2, we have the vector field f, which is a mapping from R2 to R2, so in every point in R2, the x, y, plane, we associate a vector emanating from that point.0315

Let us just go ahead and do an example here. Let us start the example on the next page, so example 1.0347

So, f is a mapping from R2 to R2 defined by f(x,y) is equal to 2x and 2y, so we start with a point in 2-space, and we end up with a point in 2-space. This is the first coordinate, this is the second coordinate.0357

Well, when we actually map this, what we end up with is something that looks like this. Let us say we go to the point (1,1). 0383

If I go to the point (1,1) and I put it in here, (1,1), that means f(x,y) is (2,2). That means from here I have a vector in this direction that the x coordinate is 2 and the y coordinate is 2 and there is a vector in that direction. 0390

If I go to the point (-1,-1), okay, so, now I am over here at this point. Well, if I put (-1,-1) into here, I get (-2,-2). That is a vector pointing in that direction.0408

Over here, if I go to this point, I am going to have a vector pointing in that direction, over here I am going to have a vector pointing that way, over here I am going to have a vector pointing that way, that is it.0420

Each point in a given domain, in this case x, y, I associate a vector where the tail of that vector actually starts at that point and goes in a particular direction. 0431

The direction of that vector, well, the x coordinate is the first coordinate of the function and the second coordinate is the other coordinate, that is all that is going on here.0443

So, to each point in R2, there is a 2 vector emanating from that point, and the direction is given by the function itself.0458

Let us do another example, and this is actually a great example and we have actually been dealing with vector fields, believe it or not -- the gradient operator.0487

The gradient operator in other words, when you actually end up taking the gradient of a function, what you are actually doing is creating a vector field because the gradient is a vector.0501

At any given point, all you are doing is you are saying that once you pick a point, let us say point p, (x,y) from there is some vector that is emanating from that point. 0510

Well, the x direction, the x coordinate of that vector is the derivative in the x direction.0523

The y coordinate is the derivative in the y direction. Operator is a vector field.0527

So, given some function f which is let us say a mapping from R2 to R, so some function of 2 variables, excuse me, the gradient of f is a mapping from R2 to R2, and the gradient is defined by -- well, exactly what we have been doing all along -- it is (df/dx,df/dy). 0541

That is it. We are forming a gradient vector, and it is a vector field. It is a mapping from R2 to R2.0577

I am going to keep repeating myself a couple of times here. I hope you will forgive me. 0587

At a given point p in the domain, the gradient vector starts at that point and has df/dx... let me write that a little bit better... it has df/dx evaluated at p for its x value, for its x coordinate I should say, the vector, and df/dy evaluated at p for its y coordinate. 0595

Okay. Now a little bit more, so every point has a vector associated with it under a mapping for a vector field.0663

Plus, the domain looks like it is covered in vectors. That is why they call it a field.0703

Who knows? We have no idea what direction they are going in, it is just that every point has some vector associated with it. You can imagine if you are standing at any given point, you are going to be pushed in that direction. That is what a vector field is.0728

Okay. So, let us do another example just to make sure we have got it. Example 3.0741

f(x,y) = x2 + y2, so this is a mapping from R2 to R2, so we are looking at a vector field in 2-space.0749

Well, let us go ahead and take a look at f(2,3). So, f(2,3) is equal to... when I put 2 in here, 2 × 3 is 6, 6 × 2 is 12, 22 is 4, 32 is 9, so I have my 10, 11, 12, 13, so it is (12,13).0764

At the point (2,3), I have a vector (12,13). That is it. Let us take the point f(-1,0). Well, when I do (-1,0), -1, 0, I am going to end up with 0 for the first coordinate, -12 is 1 and I get (0,1).0787

When I go to the point (-1,0), I have a vector that is (0,1). That is that one. Let us try f(-2,-2).0810

I am going to come over to this point right here. Some are like right about there.0825

-2 × -2 is 4, and 2 × 4 is 8, so the first coordinate is 8. -22, -22, oh this is going to be (8,8), so I have a vector that is that way. That is it, that is all that you are doing. Nice and straight forward. 0831

Now, let us go ahead and make our way over to line integrals. Let me go ahead and draw a curve in space, something like that, and let me go ahead and draw a vector field -- sort of some random vectors here, and I will put a couple actually on the... so this is a vector field, and this right here is a curve in space, that is c(t).0852

Now we can start. Let f(x,y) which is equal to f1(x,y) function 1(x,y), and function 2(x,y) -- those are the coordinate functions -- be a vector field on R2. So, something like that.0886

Now, let c(t) = c1(t) and c2(t), these are just functions of t, and they are the coordinate functions of the curve, be a curve in 2-space.0911

Since c(t) passes through this vector field, because it is a curve in 2-space and we are in R2, so every single point of R2 has some vector associated with it because we have f, right?0938

As it turns out, some of the point on the curve, I can actually form vectors... in other words I can actually form f(c(t)).0955

Since c(t) passes through R2 and the vector field, we can form f(c(t)), right? 0970

A curve passes through the vector field, well there are points on the curve that are in R2, so I can actually form f(c(t)), points along the curve.0995

You know they have certain vectors associated with them which whatever direction they happen to be moving in, we do not know, because f and c are not specified.1007

Well, we can also form c'(t), in other words we can differentiate the curve, that is just the tangent vector, that is c'(t).1016

The tangent vector 2c(t) at c(t), in other words, if I have some point here, I can not only form f(c(t)), but I can also form the tangent vector.1030

So this is c'(t) and this is f(c(t)), that is it. Vector this way and then the tangent vector. The curve is moving along like that.1051

Now, let us go ahead and define if I take f(c(t)) and I dot it with c'(t), this is actually a function of t.1066

Let us do an example of that. Example 4. We will let f(x,y) = ex+y and the second coordinate function will be y3.1091

We will let c(t), the curve, equal to t and let us say cos(t). Okay?1112

Let us form f(c(t)). f(c(t)), in other words I put these 2 in for x and y, and I end up with et + cos(t), and I get cos3t.1119

Well, let us form c'(t), which is also written as dc/dt, you know more the notation that you are used to from single variable calculus.1140

The derivative of this is 1, the derivative of this is -sin(t). So, now we form... that is fine, I guess I can do it down here... f(c(t)) · c'(t).1150

When I do the dot product of this thing and this thing -- let me do it in blue -- I get e(t) + cos(t) + ... or in this case minus... minus this × that.1167

Minus sin(t)cos2(t). That is it. I do not need this little open parentheses.1189

So, that is what is important. This is our definition right here. We will start with this definition.1198

When we are given a vector field and we are given a curve with some parameterization, we can actually form f(c(t)) · c'(t), and it ends up being an actual function of t.1203

Now we get to our definition of the line integral. I am going to move onto the next page for this one.1217

So, we have... definition, if c(t) is defined on the interval where t on the interval a, b, (2,6), (5,9), whatever, then the integral of f along c from a to b is -- this is the symbol, the integral of f -- along the curve c is equal to the integral from a to b of f(c(t)) · c'(t) dt.1227

Or, we can also write it as the integral from a to b of f(c(t)), let me make this c(t) a little bit clearer here... f(c(t)) · dc/dt dt.1296

What we are saying is when you are given a vector field and when you are given a parameterized curve, you want to form f(c(t)), you want to form dc/dt, you want ot take the dot product of that, and then you want to put that in your integrand.1317

This is what is important right here. This right here. Okay? That is your integrand. It is going to end up being a function of t.1332

We can integrate that function of t. This is just this and this is just the form of integration. That is what we surround our integrand by. This is the definition of the line integral.1339

Let us talk about what it is that we are actually doing here. When we were doing integration in single variable calculus, you were actually taking a line integral except it was a very, very special line that you were moving across. It was the x axis.1350

In other words, when you took some integral of some function like this, let us say from this point to this point, what you were doing is you were integrating the function along this line, along the x axis. 1361

Well, the x axis is a very special line. Now what we are saying is well, it does not matter, we do not have to stick to just that line. I can take any line in any dimension. I can take this x axis and I can curve it this way, I can curve it this way, I can twist it this way so now I might have some curve like this.1373

I can take that function and I can integrate it along this curve, along this line, along this path. 1392

I can still do it because integration, all integration is, is allowing to come up with some number which is a function of whatever variable you are dealing with and just multiplying it by some differential and then adding it all up along the length.1398

That is what you are doing. Because f(c(t)) · c'(t) is a function of t, essentially it is just a number at any given value of t, I can add up all of those little numbers. That is all I am doing.1415

I take the function of t, multiply it by dt and I integrate it. That is all an integration is. Just a really, really long addition problem.1428

So, we can do this, and the definition of a line integral is precisely this: When you are given a vector field, when you are given a curve that is parameterized that happens to pass through that vector field, you can take f(c(t)) for that composite, and then you can dot it with the tangent vector c'(t), and you are going to get yourself an integrand.1438

You can put that integrand, you can surround it with the integral formula... you can form this integral... and you can integrate it just like you do single variable calculus. That is what this is. This is just an integral of the single variable, t. 1456

This is really, really fantastic. It is actually amazing that we can do this. So, integration in single variable calculus is insane, it is a line integral but it is a very, very special line, it is just an x-axis, it is just one direction. This is actually the same thing.1471

So, let us go ahead and do an example here. Let us see how this turns out. Let us see... should we do this one? Yeah, we will do a couple of examples. Let me go back to blue. So, example 5.1487

Let f(x,y) = xy2 and x3... and, well, let me put it below... and let c(t) = t and t2.1508

That is it. Essentially what we are doing here, let me go ahead and draw this out, this parameterization (t,t2), that is the parameterization of the standard parabola that passes through the origin y = x2. This is the parameterization.1532

It says that x is t, and y is t2, so this is the parameterization for the parabola. What we are going to be doing is we are going to be integrating this vector field along this parabola.1549

Okay. We want to find the integral of f along c or t in the integral from 0 to 3, so from 0 to 3, it is going ot travel along this path and say come up to here. Something like that. 0 to 3, the square of 3 is 9, so when we integrate the function along this curve, this is what we are going to get.1562

That is what this is, so let us just go ahead and do it. Well, the first thing we are going to do is we are going to find f(c(t)), so f(c(t)), nice and simple, we just plug them in and see what happens.1593

f(c(t)), c(t) is t and t2, so, this is the x value and this is the y value, so we are going to have t × (t2)2, and this is x3, so it is going to be t3, so that is f(c(t)).1606

That equals t3, and now we are going to form dc/dt, or c'(t), however you want to think about it. That equals... the derivative of t is 1, and the derivative of t2 is 2t.1626

When we dot these two, it is this × this + this × that, so we get f(c(t)) dc/dt t5 + 2t4 + 2t4, and that is what we are going to end up integrating. 1645

So, let us go ahead and write this out. The integral of f along c from the interval 0 to 3, it equals the integral from 0 to 3 of t5 + 2t4 dt.1674

This is a simple integration problem. You have been doing this since the first couple of weeks of calculus, so let us go ahead and do that and evaluate it. 1692

We have t6/6 + 2t5/5 + 2t5. We are evaluating it from 0 to 3 and when put these numbers in we get 729/6 + 486 -- oops, it is going to be 486 -- over 5, and that is our answer.1705

You are welcome to leave it like that. I do not know what it is that your teacher actually expects. If he actually expects a full evaluation, it is a perfectly good number, I am not going to go ahead and work that out or if he expects you just to stop at that point.1726

Once you have this, you know, evaluation is the easy part. If you do not want to evaluate it manually you can just use an algebra system like mathematica, maple, something like that, and it will do it for you.1739

This is the important part, not necessarily the evaluation. That is it. Okay. One comment before we actually do our final answer. Clearly our ability to evaluate a line integral depends on our ability to parameterize the curve.1752

Sometimes the parameterization will be given to you like these examples, sometimes you will have to come up with a parameterization yourself, and do not worry, we will be doing some examples in subsequent lessons where we actually have to come up with a parameterization for a given curve.1793

For example, in this last example, they might have said along the parabola y = x2, we would have to come up with a parameterization to see if t = (t,t2), something like that.1807

To parameterize the path, so our ability to evaluate the integral depends on our ability to parameterize the path because the definition is based on the parameterization c(t).1821

Okay, so let us do an example again. Slightly more complicated. Example 6. This time we will let our function f(x,y) = xy and we will do x2 - y3, and we will let our curve, c(t) = cos(t) and the sin(t) and in case you do not recognize that, this is a parameterization for the unit circle moving in the counterclockwise direction.1834

We want to evaluate the integral in the interval 0 to pi/2 so as t moves from 0 to pi/2, well as t moves from 0 to pi/2, we are going to move along this path right here and we are going to stop right there.1871

We are going to integrate this function along this path from here to here. That is what we are doing. Well? Let us go ahead and do it.1889

f(c(t)) = well, this is x, this is y, so we put it in there. So we get cos(t), sin(t)... let me write that a little bit better... cos(t) and sin(t).1901

Then we have x2 which is cos2(t) - y3, which is -sin3(t). Okay.1921

Now let us do c'(t). I am going to write dc/dt, so when I take the derivative of cos(t) I get -sin(t). I take the derivative of sin which is cos(t) and then I go ahead and I form the dot product, so I take this dotted with this, and I end up with the following. 1932

I am going to shorten the notation. I am just going to do f · dc, I do not want write out all the c's and t's and things like that. So f · dc, it is equal to, well, this dotted with that.1950

So, we are going to have this × this, so it is going to be -sin2... hmm, let us write this slowly, because we do not want these random lines to show up... -sin2(t)cos(t) + this × that. + cos3(t) - ... okay, actually you know what, let me go ahead and erase this and let me start on the next page. I want this particular integrant to be very, very clear.1961

So, f · d(c) = -sin2(t)cos(t) + cos3(t) - cos(t)sin3(t). 2001

When you take the dot product of those two, you are going to end up with that. That is your integrand. 2022

So, The integral of f along the particular path that we chose, in the given interval, is going to be the integral from 0 to pi/2 of - sin2(t)cos(t) + cos3(t) - cos(t)sin3(t) dt.2032

This is the important part. Being able to get here. The rest of it is just basically just technique, so I am not going to be able to integrate this manually. What I did when I did this was I just put this into my particular mathematic software. 2055

I used maple myself and it gave me the answer, nice and quick... and the answer is 1/12.2072

There you go. That is a line integral. You are doing the same thing that you did with single variable calculus except now the line is no longer just the x axis. 2078

We have taken that x axis, and now we can just integrate the function along any curve we want. In this particular case we have integrated a vector field, which is a vector from R2 to R2, or R3 to R3, or R4 to R4, where the dimension of the departure space is the dimension of the arrival space. 2089

In the next lesson I am going to talk about a -- so what we have done here is we have integrated a function from R2 to R2. We have integrated a vector field around a curve.2109

In the next lesson we are going to talk about multivariable functions. Let us say an R2 to R, where the arrival space is just the real number line. Dimension 1, what we have been dealing with, functions of several variables. 2125

You can also do line integrals of functions of several variables, not just a vector field, so we will define that in a subsequent lesson.2140

For the time being, all you have to really worry about is this basic definition. Form f(c(t)), form c'(t), take the dot product, you will end up with a function of a single variable t and then you will just integrate it, just like single variable calculus.2148

Thank you very much for joining us here at educator.com. We will see you next time, bye-bye.2164