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 2 answersLast reply by: michael BoocherFri Aug 2, 2013 8:11 AMPost by Josh Winfield on May 11, 2013Example 1: t^3 cancels when taking the dot product of f(c2(t)) with c'(t). The t term was canceled but t^3 remained, i think the t^3 cancels and the t reamains. 1 answerLast reply by: Professor HovasapianSat May 4, 2013 4:13 PMPost by Eunhee Kim on May 4, 2013Why is c^-(t)=(a+b-t)?

### Line Integrals, Part 3

Find the path integral of F(x,y) = (2x − y,3x + 2y) along C = C1 + C2 + C3 + C4.
• To find the path integral ∫C F , we can find the sum of all path integrals that make up C. So ∫C F = ∫C1 F + ∫C2 F + ∫C3 F + ∫C4 F .
• Our four paths are lines, which can be represented parametrically by C(t) = P + t(Q − P) where 0 ≤ t ≤ 1 and P and Q are points on the line.
• We must make sure the direction of the path is conserved, we will let the endpoints of each line, Q, be the point the arrow is on.
• Then C1(t) = (0,0) + t(1,0) = (t,0) and so ∫C1 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 (2t,3t) ×(1,0)dt = ∫01 2tdt = 1
• C2(t) = (1,0) + t(0,1) = (1,t) and so ∫C2 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 (2 − t,3 + 2t) ×(0,1)dt = ∫01 (3 + 2t)dt = 4
• C3(t) = (1,1) + t( − 1,0) = (1 − t,1) and so ∫C3 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 (1 − 2t,5 − 3t) ×( − 1,0)dt = ∫01 (2t − 1)dt = 0
• C4(t) = (0,1) + t(0, − 1) = (0,1 − t) and so ∫C4 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 (t − 1,2 − 2t) ×(0, − 1)dt = ∫01 (2t − 2)dt = − 1
Hence ∫C F = ∫C1 F + ∫C2 F + ∫C3 F + ∫C4 F = 1 + 4 + 0 + ( − 1) = 4.
Find the path integral of F(x,y) = ( [1/2]x2 − y2,xy − 1 ) along C = C1 + C2 + C3.
• To find the path integral ∫C F , we can find the sum of all path integrals that make up C. So ∫C F = ∫C1 F + ∫C2 F + ∫C3 F .
• Our three paths are lines, which can be represented parametrically by C(t) = P + t(Q − P) where 0 ≤ t ≤ 1 and P and Q are points on the line.
• We must make sure the direction of the path is conserved, we will let the endpoints of each line, Q, be the point the arrow is on.
• Then C1(t) = ( − 1,0) + t(2,0) = ( − 1 + 2t,0) and so ∫C1 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( [1/2] − 2t + 2t2, − 1 ) ×(2,0)dt = ∫01 (1 − 4t + 4t2)dt = [1/3]
• C2(t) = (1,0) + t( − 1,1) = (1 − t,t) and so ∫C2 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( − [1/2]t2 − t + [1/2], − t2 ) ×( − 1,1)dt = ∫01 ( − [1/2]t2 + t − [1/2] )dt = − [1/6]
• C3(t) = (0,1) + t( − 1, − 1) = ( − t,1 − t) and so ∫C3 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( − [1/2]t2 + 2t − 1,t2 − t − 1 ) ×( − 1, − 1)dt = ∫01 ( − [1/2]t2 − t + 2 )dt = [4/3]
Hence ∫C F = ∫C1 F + ∫C2 F + ∫C3 F = [1/3] − [1/6] + [4/3] = [3/2].
Find the path integral of F(x,y) = ( x2 + y2,1 ) along C.
• To find the path integral ∫C F , we parametrize our half circle utilizing x = rcos(t), y = rsin(t) and r being the radius, 0 ≤ t ≤ 2p.
• Note that our circle is not centered at (0,0), to compensate for the change of center we add 1 to our x - axis so that x = 1 + rcos(t).
• We have C(t) = (1 + cos(t),sin(t)), then ∫C F = ∫ab F(C(t)) ×C′(t) dt = ∫0p ( 2 + 2cos(t),1 ) ×( − sin(t),cos(t))dt . Note that in our seimicircle 0 ≤ t ≤ p.
Thus ∫C F = ∫0p ( − 2sin(t) − 2sin(t)cos(t) + cos(t) ) dt = (2cos(t)) |0p − (sin2(t)) |0p + (sin(t)) |0p = − 4.
Find the path integral of F(x,y) = ( ex,ey ) along C = C1 + C2, where C2 is the curve represented by the parabola y = − x2 + 1.
• To find the path integral ∫C F , we can find the sum of all path integrals that make up C. So ∫C F = ∫C1 F + ∫C2 F .
• Note that ∫C1 F is a line that is parametrized by C(t) = ( − 1 + 2t,0) with 0 ≤ t ≤ 1 and ∫C2 F can be parametrized by letting x = t and so y = − t2 + 1.
• We must make sure the direction of the path is conserved, we will compute ∫C2 F from 1 ≤ t ≤ − 1.
• Then ∫C1 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( e − 1 + 2t,1 ) ×(2,0)dt = ∫01 ( 2e − 1 + 2t )dt = e − e − 1.
• Now C2(t) = (t, − t2 + 1) and so ∫C2 F = ∫ab F(C(t)) ×C′(t) dt = ∫1 − 1 ( et,e − t2 + 1 ) ×(1, − 2t)dt = ∫1 − 1 ( et − 2te − t2 + 1 )dt = − ∫ − 11 ( et − 2te − t2 + 1 )dt = − e + e − 1.
Hence ∫C F = ∫C1 F + ∫C2 F = e − e − 1 + ( − e) + e − 1 = 0.
Given C(t) = (cos(t),2sin(t)), − [p/2] ≤ t ≤ [p/2] find C − 1.
• To change the orientation of a curve, we change the parameter C(t) to C − 1(t) = C(a + b − t) where t ∈ [a,b] is our interval.
• We have a = − [p/2], b = [p/2] so C − 1(t) = C( − [p/2] + [p/2] − t ) = C( − t).
By substitution we obtain C( − t) = (cos( − t),2sin( − t)) = (cos(t), − 2sin(t)) and so C − 1(t) = (cos(t), − 2sin(t)).
Given F(x,y) = ( [(x2ln(y))/2],[(y2ln(x))/2] ), x = t and y = t2 find F(C − 1(t)) from (0,0) to (1,1).
• To change the orientation of a curve, we change the parameter C(t) to C − 1(t) = C(a + b − t) where t ∈ [a,b] is our interval.
• Since x and y are parametrized, we have C(t) = (t,t2) with 0 ≤ t ≤ 1, then a = 0, b = 1 so C − 1(t) = C( 0 + 1 − t ) = C(1 − t).
• By substitution we obtain C(1 − t) = (1 − t,(1 − t)2) = (1 − t,1 − 2t + t2) and so C − 1(t) = (1 − t,1 − 2t + t2).
Hence F(C − 1(t)) = ( [((1 − t)2ln(1 − 2t + t2))/2],[((1 − 2t + t2)2ln(1 − t))/2] ).
Let ∫C F = ∫C1 F + ∫C2 F + ∫C3 F , where ∫C1 F = 2, ∫C1 F = − ∫C2 F and ∫C1 F = − ∫C3 F .
i) Find ∫C3 F and ∫C2 F
Note that ∫C1 F = 2, by substitution 2 = ∫C1 F = − ∫C2 F and ∫C2 F = − 2. Similarly 2 = ∫C1 F = − ∫C3 F and ∫C3 F = − 2.
Let ∫C F = ∫C1 F + ∫C2 F + ∫C3 F , where ∫C1 F = 2, ∫C1 F = − ∫C2 F and ∫C1 F = − ∫C3 F .
ii) Find ∫C1 F , ∫C2 F and ∫C3 F
Recall that ∫C F = − ∫C F . It follows that ∫C1 F = − 2, ∫C2 F = 2 and ∫C3 F = 2.
Let ∫C F = ∫C1 F + ∫C2 F + ∫C3 F , where ∫C1 F = 2, ∫C1 F = − ∫C2 F and ∫C1 F = − ∫C3 F .
iii) Find the value of ∫C F and ∫C F = ∫C1 F + ∫C3 F + ∫C2 F
Since ∫C F = ∫C1 F + ∫C2 F + ∫C3 F we substitute our values to obtain ∫C F = 2 − 2 − 2 = − 2. Similarly ∫C F = ∫C1 F + ∫C3 F + ∫C2 F = − 2 + 2 + 2 = 2.
Verify that ∫C1 F ∫C2 F , given that F(x,y) = ( xy,xy ), C1(t) = (t,t2) and C2(t) = (t,t) for 0 ≤ t ≤ 1.
• We compute ∫C1 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( t3,t3 ) ×( 1,2t ) dt = ∫01 ( t3 + 2t4 ) dt = [13/20]
Also ∫C2 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( t2,t2 ) ×( 1,1 ) dt = ∫01 ( 2t2 ) dt = [2/3]. Hence ∫C1 F ∫C2 F .
Verify that ∫C1 F = ∫C2 F , given that F(x,y) = ( x,y ), C1(t) = (t,t2) and C2(t) = (t,t) for 0 ≤ t ≤ 1.
• We compute ∫C1 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( t,t2 ) ×( 1,2t ) dt = ∫01 ( t + 2t3 ) dt = 1
Also ∫C2 F = ∫ab F(C(t)) ×C′(t) dt = ∫01 ( t,t ) ×( 1,1 ) dt = ∫01 ( 2t ) dt = 1. Hence ∫C1 F = ∫C2 F , note the role a vector field plays while computing path integrals.
Find ∫C f given f(x,y,z) = x + y + z where x = t, y = − t and z = t2 from 0 ≤ t ≤ 1. Do not integrate.
• To find the path integral ∫C f of a function f, we compute ∫C f = ∫ab f(C(t))|| C′(t) || dt. Note that this is our usual multiplication.
• Since x, y and z are parametrized we obtain C(t) = (t, − t,t2). Then f(C(t)) = t2, C′(t) = (1, − 1,2t) and || C′(t) || = √{1 + 1 + 4t4} = ( 2 + 4t4 )1 \mathord/ phantom 1 2 2.
Hence ∫C f = ∫ab f(C(t))|| C′(t) || dt = ∫01 t2( 2 + 4t2 )1 \mathord/ phantom 1 2 2 dt.
Find ∫C f given f(x,y,z) = xyz where x = cos(t), y = et and z = t from 0 ≤ t ≤ 1. Do not integrate.
• To find the path integral ∫C f of a function f, we compute ∫C f = ∫ab f(C(t))|| C′(t) || dt. Note that this is our usual multiplication.
• Since x, y and z are parametrized we obtain C(t) = (cos(t),et,t). Then f(C(t)) = tetcos(t), C′(t) = ( − sin(t),et,1) and || C′(t) || = √{sin2(t) + e2t + 1} = ( sin2(t) + e2t + 1 )1 \mathord/ phantom 1 2 2.
Hence ∫C f = ∫ab f(C(t))|| C′(t) || dt = ∫01 tetcos(t)( sin2(t) + e2t + 1 )1 \mathord/ phantom 1 2 2 dt.

*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.

### Line Integrals, Part 3

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 0:12
• Piecewise Continuous Path
• Closed Path
• Example 1: Find the Integral
• The Reverse Path
• Theorem 1
• Parameterization for the Reverse Path
• Example 2
• Line Integrals of Functions on ℝn
• Example 3

### Transcription: Line Integrals, Part 3

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

The last few lessons, we have been talking about line integrals and we are going to continue our discussion of line integrals and we are going to discuss closed paths today. Let us go ahead and get started.0004

Okay. So, first of all, we will go ahead and talk about a piece wise continuous path and then we will go into closed paths.0014

If a path is given in a piecewise continuous fashion -- and I will draw this out in just a minute -- in a piecewise continuous form.0024

In other words if you had something like that, like this, and maybe this little half circle like that, and let us say, this is curve 1, this is curve 2, this little circle here is curve 3 and this one is curve 4 and the points of continuity are here, so it is continuous, it is just called piecewise continuous.0049

Here, if this were the case, then the integral of f along c... c is the entire thing, well it is just equal to the integral of f along c1 + the integral along c2 + ... the integral of f along cn... however many pieces there are in this curve, that is it.0074

Essentially what you are going to be doing is you are going to be parameterizing each segment and then just adding up the integrals. Completely intuitive. Okay.0100

Now, let us talk about closed paths. So, we say a path is closed if the end point of cn, the last curve, is the same as the beginning point of c1, the first.0109

Let us take a couple of points like this. Let us take p1 over here, and let us take p2 over here.0150

So, we might have, let me go ahead and draw some dots so that we know what we are talking about. So, we might have something that goes like this and we are parameterizing it in this direction.0160

That is one path or we might have another one, we might have one here and there, something like that.0174

So, let us say we start over here and we move in this direction, so this will be -- let us just call this p3, let us call this p4, these are also point, so you know we start over here... so this is our start, we come this way, we come over here and then we go this way, then that way, then this way.0191

Then let me make sure to put these arrows here so that we know where we are going, and then we end up here where we started, so this is a closed path.0215

In other words, p1 and p5 are the same. That is it, a closed path. That is exactly what you think it is. It is just a closed path.0222

Okay. So, let us go ahead and do an example here of a line integral on a closed path, so example 1.0231

We want to find the integral along the path c, which is equal to c1 + c2, and we will draw this out in just a minute for the following function.0243

So, we are going to draw it out, and from the drawing of the actual path, we are going to find a parameterization for it. We are not given the curve, we are actually going to have to come up with the parameterization ourselves for the following function.0267

f(x,y) = x2,xy2.The particular path we are going to be following is this one.0282

Okay. So, this is going to be the point (1,0), and this is going to be the point (0,1), so we are going to follow... our path is going to be like that, and then it is going to be like this.0298

So, in a counterclockwise fashion we are going to follow the circle from here to here, and then we are going to follow the line from here to here.0312

We want to know what the integral of this function is along this closed path. Let us see what we can find out.0320

Let me go ahead and mark some points so this is... oops, got that reversed, we are on the x-axis here... so this is the point (1,0), and this is the point (0,1), so we will go ahead and call this c1, and this is c2. Okay.0329

Well, the integral of f along c is just going to be the integral of f along c1 + the integral of f along c2. We are going to have to solve 2 integrals here.0344

We are going to parameterize this one, we are going to parameterize this one, we are going to do the integrals separately and that is it. So, let us go ahead and parameterize these curves.0356

So, c1, let me do this in blue actually. There we go, so c1 of t, well, c1 of t is just the quarter circle from this point to this point, so the parameterization is going to be cos(t) and sin(t), and t is going to run from 0 to pi/2. That is it.0365

That is the standard parameterization of the circle. That is it.0398

okay... yes, and c2, well c2(t), this is a line segment, a line segment from here to here. How we parameterize a line segment is we take the beginning point + t × the ending point, I am going to call it... let us call this p1, and let us call this p2.0403

So, p2 - p1, well p1 is the point (0,1), and notice I am writing it as a column vector simply because it makes it a little easier for me to see. I think about it this way.0428

So, (0,1) + t × p2, which is the column vector (1,0) - (0,1), so let us go ahead and solve this.0443

We get (0,1) + t ×, well 1-0 is 1 and 0-1 is -1, so now let us go ahead and put this together.0459

That is fine, I will do it over here... so this is going to be... so the top is going to be just t, and this is going to be 1 + t × -1, so it is going to be 1 - t. That is our parameterization. If you want to see it in terms of a row vector like this, it is going to be (t, 1-t), so we actually found the parameterization for this particular line segment.0473

So, we are going to do this line integral, we are going to do this line integral for the function, and then add them together.0502

So let us go ahead and do f(c1(t)), so f(c1(t)... well we put this into these right here. This is x, this is y, so we put them in.0508

We end up with cos3(t), then cos(t)sin2(t), so this is going to be our f(c1(t)).0525

Now let us do our c1'(t), or I will write it as dc1/dt, in keeping with standard single variable notation.0538

So, the derivative of c1(t) is -sin(t), and then we have cos(t), there we go, and then when we take the dot product of this and this, we are going to end up with -sin(t)cos3(t) + cos2(t)sin2(t).0546

This is what we are actually going to integrate, so now the integral of... should I do it here, or should I do it on the next... that is fine, I can do it here... you know what, no, I can do it on the next page. I do not want these stray lines.0576

The integral of f along c1 is going to be the integral from 0 to pi/2 of the expression that we just had, -sin(t)cos3(t) + cos2(t)sin2(t) dt.0593

Then, when you go ahead and put this into some math software, because I am hoping that you are not actually going to do this by hand, you are going to end up with -1/4 + pi/16.0616

Just a number, that is it. That is what an integral is. Just a number. Okay. So that is our first curve.0625

Now, let us go ahead and do our second curve. So, now for c2. Again, I will go ahead and write out the parameterization. It is t and 1-t, and let me go ahead and just write f again just so we have it on this page.0632

It is going to be, I think it was going to be x3, and I think it was going to be xy2, is that correct? Yes, that is correct, so we have f.0650

Now, let us go ahead and form f(c2(t)), so let us go ahead and put these values, this in for x and this for y, and we go ahead and work it out so you are going to end up with t3 and t × 1-t2.0661

Then we solve for this one, this is going to equal t3, we expand this out and multiply it out. We are going to get t - 2t2 + t 3, right?0680

So, expand this out, binomial, multiply, so there you go, now you have f(c(t)), or c2(t), now let us go ahead and do dc2/dt, so c2'.0696

Well that is nice and easy, that is just going to equal 1 and -1. That is nice and simple.0708

So, when we go ahead and we take the dot product of this and this, which is going to be our integrand, we end up with t - t + 2t2 - t3, which is equal to 2t2 - t3, so this is our integrand.0715

This is the thing that is going to be under the integral sign, so, the integral of f along c2 is going to equal the integral from 0 to 1, remember the parameterization for a line is really, really... for a line segment is easy because you are always going from 0 to 1. That never changes.0742

0 to 1... of our 2t2 - t3 dt, and this one of course you can do by hand, so this becomes 2t3 over 3 - t4/4 from 0 to 1, and this gives you... I hope I did my arithmetic correct... 2/3 - 1/4.0765

Our final integral of f/c is equal to the integral of f/c1 + integral of f/c2 = -1/4 + pi/16, let me put boxes around the 2 numbers... + 2/3 - 1/4.0792

When you add all of that together, you get -1/2, you just put the 1/4 together, I did not feel like... I am going to leave the 2/3 alone, it is just a number it does not really matter.0818

pi/16, I have never been a big fan of too much simplification myself. I believe that too much simplification actually -- well, in terms of arithmetic it is fine, but in general when you are working with math, you do not want to oversimplify too much just to make it elegant, because then you start to hide things. You want to see what is happening.0828

+ 2/3, and there you go. That is your final answer. Nice and simple, how is that?0846

Now let us talk about the reverse path, so, if we are going to integrate from one point to another along a path, what happens when we actually go from this point to this point. What do you think happens?0855

Well intuition might say it is just the reverse integral. The answer is yes, you are just going to switch signs. Let us just talk about this formally really quick.0867

So, the reverse path... the reverse path, okay... now, let c be a path -- actually I think I am going to go back to black here. I like to do my discussions in black -- the reverse path, so let c be a path traversed in a given direction. That is fine.0878

In a given direction... Then, c with a little negative sign up in the top right is the notation for the path traversed in the reverse direction, in the opposite direction, or reverse direction.0933

So, nothing strange here, just a question of notation, so if I have p and if I have q, and if I have some path where we go this way from p to q, well that is c.0964

The reverse path if we go this way, that is just c-, that is it.0975

Now, let us write out a theorem. Let f be a vector field on an open set... on some open set s... and let c be a curve in s defined on the interval a, b, so the beginning point and the ending point of the particular curve, for the parameterization.0981

Then, the integral of f along the reverse path is equal to -the integral of f along the path in one direction. That is it, you are just going to switch signs.1029

Now, this is what is important. The parameterization for the reverse path is not the same, very, very important... is not the same as for the forward path.1047

Now, formally, there is actually a way to convert a... if you have a parameterization for a particular curve, to actually get a parameterization for the reverse path, and it would go like this.1082

So, formally the parameterization of c(t) = c(a+b) = t, so let us just do an example.1095

Fortunately we are not going to have to worry too much about reverse paths and actually coming up with parameterizations for reverse paths.1110

Generally we just come up with a parameterization for a particular path that works, and if we need to go in the reverse direction, we just switch the sign of the integral. That is what is nice.1117

You do not have to actually reparameterize it and do it that way, just switch the sign. That is what this lets you do.1125

Let us just do an example for the heck of it. So, example 2.1131

Let c(t) equal our standard cos(t)sin(t), parameterize our circle t is in 0 to 2pi.1140

Find the parameterization for traversing that circle in the opposite direction, so clockwise instead of counterclockwise.1155

So by our definition, c-t = c(a+b) - t.1168

Well, that is equal to c of, this is a, this is b, 0 + 2pi - 2 = c(2pi-t) and c(2pi-t) = cos(2pi-t)sin(2pi-t). That is it Nice and simple.1178

Basically you have this circle that you have traversed in the counter clockwise direction. That is the normal cos(t)sin(t).1210

Now if you use this parameterization, you are traversing it in this direction. If you integrate along this way, it is just the reverse of integrating along this way, you just change the sign of the integral. Good.1219

When you are given a particular curve, one parameterization might just be sort of natural. If you look at it you might say "oh, let me just do this."1241

But perhaps you are traversing it in the opposite direction. Well, you do not have to change the parameterization, just take the integral with a parameterization that you already have and just switch the sign. That is it.1249

So, the integral of f, c- = the integral from b to a of f(c(t)) · c'(t) dt.1260

That equals negative the integral of a to b. What we have been doing of f(c(t)) this is just everything written out. Let us see c't/dt, that is it. Just change the sign.1278

If you are going to integrate from a to b, but you go from b to a, just change the sign.1292

So, now let us move on to another topic. It is still line integrals but up to this point, we have defined line integrals with respect to vector fields.1298

We said that there is some vector field, some function which is a mapping from R2 to R2, R3 to R3, well, it is possible but we have also been dealing with functions of several variables where we are mapping R2 to R, R3 to R, where you have just one function instead of a vector field.1311

Is it possible to actually integrate a function over a path, a contour the same way that we did for single variable calculus. The answer is yes.1327

Let us go ahead and define what we mean by the line integral of a function of several variables and then we will do an example.1337

So, this is going to be line integrals of functions on RN, and remember, functions again, that is the word that we set aside to use when our arrival space happens to be the real numbers, not R2 or R3, when it is just R, the real numbers.1346

So, now let us define, so let us define the integral of f along c and notice I have used a small f so generally we will us a capital f for -- oops, wow, that is a crazy line, look at that -- so generally we will use a small f for functions, a capital F for vector fields, at least I will, unless there is something in the problem where we have to do something else.1377

So, the integral of a function over the curve is equal to the integral from a to b of f(c(t)), so this part is the same except for instead of a dot product because we are not dealing with a vector is we are going to take the norm of c'(t) dt.1411

With the definition of a line integral, what we need to do is f(c(t)) · c'(t), but now we are dealing with a function not a vector field, so it is f(c(t)) still, except now you are just multiplying it by the norm of the derivative of the actual curve itself.1438

Let us just go ahead and do an example. Example 3.1456

Now, let us go ahead and let f(x,y,z), we will do a function of 3 variables, x2,y,z3.1471

So you see we plug in a vector, a three vector and we get out a number, it is a mapping from R3 to R.1483

Well, let us see, let us go ahead and let c(t), the curve along which we parameterize, let it be the spiral in 3 space defined by cos(t)sin(t) and t, and let us go ahead and let t go from 0 to 3pi/4.1491

Just to let you know what this looks like, I am going to draw out my 3-dimensional... okay, and I ill go ahead and go back that way, this way, so this thing that is coming out is the x-axis. This thing is the y-axis, and this is the z.1510

The z and the y are the ones that are in the plane, and the x-axis is the one coming out at you.1528

This spiral, cos(t),sin(t),t, if it is just cos(t) and sin(t), it is the circle, the unit circle in the x,y plane, however with this t, now we have this... so what you are going to get is it is going to be from 0 to 3pi/4, so we are going to move along 135 degrees.1534

We are actually going to end up coming up, so if I were to take the x,y plane and look at it just from the x,y plane, I would see this curve that is going up like this. Sort of like a ramp of a parking lot that works in a spiral.1558

That is exactly what it is. It is basically a spiral because now you are also, you are not just moving along x and y, you are also moving along z for every step that you take. So, you are going like this.1576

We are going to just -- let us see what we can do with this line integral here -- okay.1588

So, f(c(t)), so let us find f(c(t)), that is going to equal -- actually I do not need this because I am not talking about a vector.1595

We have cos2(t) × y which is sin(t) × z3, so × t3, so this equals, in general we put the variable first, so t3cos2(t)sin(t).1608

Now let us go ahead and find c'(t), well c'(t) is -sin(t), cos(t) and 1.1630

Now let us go ahead and find the norm of c'(t). The length, in other words, of the tangent vector at any given moment, that is going to equal this squared + this squared + this squared all under the radical.1642

So, sin2 + cos2 + 1 under the radical. And sin2 + cos2 is 1, 1 + 1 is 2, so we have sqrt(2).1661

The integral of f along this particular contour is equal to the integral from 0 to 3pi/4 of t3cos2(t)sin(t) × sqrt(2) dt.1673

Then when we put this into mathematical software because again I am not going to do this by hand, we get the following answer: -13pi/12 + 9/28pi3 - 41/27 + 15/32pi2.1695

Again, it is just a number expressed in terms of pi and other numbers, nothing simplified, put it in a calculator and you are just going to get a number.1721

It is possible to actually have a function of several variables and integrate that function just like it is possible, what we have been doing, you know, it is what we started off with, integrating a vector-field over a certain path.1729

That is it, everything is exactly the same except the definition is slightly different. Instead of doing a dot product, you are just multiplying by the norm of the tangent vector which is just dotting it with the vector itself.1747

That is it for line integrals here. Thank you for joining us here at educator.com and we will see you next time for a further discussion of potential functions. Bye-bye.1760