For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

### Line Integrals, Part 3

_{1}+ C

_{2}+ C

_{3}+ C

_{4}.

- 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 C
_{1}(t) = (0,0) + t(1,0) = (t,0) and so ∫_{C1}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}(2t,3t) ×(1,0)dt = ∫_{0}^{1}2tdt = 1 - C
_{2}(t) = (1,0) + t(0,1) = (1,t) and so ∫_{C2}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}(2 − t,3 + 2t) ×(0,1)dt = ∫_{0}^{1}(3 + 2t)dt = 4 - C
_{3}(t) = (1,1) + t( − 1,0) = (1 − t,1) and so ∫_{C3}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}(1 − 2t,5 − 3t) ×( − 1,0)dt = ∫_{0}^{1}(2t − 1)dt = 0 - C
_{4}(t) = (0,1) + t(0, − 1) = (0,1 − t) and so ∫_{C4}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}(t − 1,2 − 2t) ×(0, − 1)dt = ∫_{0}^{1}(2t − 2)dt = − 1

_{C}F = ∫

_{C1}F + ∫

_{C2}F + ∫

_{C3}F + ∫

_{C4}F = 1 + 4 + 0 + ( − 1) = 4.

^{2}− y

^{2},xy − 1 ) along C = C

_{1}+ C

_{2}+ C

_{3}.

- 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 C
_{1}(t) = ( − 1,0) + t(2,0) = ( − 1 + 2t,0) and so ∫_{C1}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}( [1/2] − 2t + 2t^{2}, − 1 ) ×(2,0)dt = ∫_{0}^{1}(1 − 4t + 4t^{2})dt = [1/3] - C
_{2}(t) = (1,0) + t( − 1,1) = (1 − t,t) and so ∫_{C2}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}( − [1/2]t^{2}− t + [1/2], − t^{2}) ×( − 1,1)dt = ∫_{0}^{1}( − [1/2]t^{2}+ t − [1/2] )dt = − [1/6] - C
_{3}(t) = (0,1) + t( − 1, − 1) = ( − t,1 − t) and so ∫_{C3}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}( − [1/2]t^{2}+ 2t − 1,t^{2}− t − 1 ) ×( − 1, − 1)dt = ∫_{0}^{1}( − [1/2]t^{2}− t + 2 )dt = [4/3]

_{C}F = ∫

_{C1}F + ∫

_{C2}F + ∫

_{C3}F = [1/3] − [1/6] + [4/3] = [3/2].

^{2}+ y

^{2},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 = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{p}( 2 + 2cos(t),1 ) ×( − sin(t),cos(t))dt . Note that in our seimicircle 0 ≤ t ≤ p.

_{C}F = ∫

_{0}

^{p}( − 2sin(t) − 2sin(t)cos(t) + cos(t) ) dt = (2cos(t)) |

_{0}

^{p}− (sin

^{2}(t)) |

_{0}

^{p}+ (sin(t)) |

_{0}

^{p}= − 4.

^{x},e

^{y}) along C = C

_{1}+ C

_{2}, where C

_{2}is the curve represented by the parabola y = − x

^{2}+ 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 = − t^{2}+ 1. - We must make sure the direction of the path is conserved, we will compute ∫
_{C2}F from 1 ≤ t ≤ − 1. - Then ∫
_{C1}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}( e^{ − 1 + 2t},1 ) ×(2,0)dt = ∫_{0}^{1}( 2e^{ − 1 + 2t})dt = e − e^{ − 1}. - Now C
_{2}(t) = (t, − t^{2}+ 1) and so ∫_{C2}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{1}^{ − 1}( e^{t},e^{ − t2 + 1}) ×(1, − 2t)dt = ∫_{1}^{ − 1}( e^{t}− 2te^{ − t2 + 1})dt = − ∫_{ − 1}^{1}( e^{t}− 2te^{ − t2 + 1})dt = − e + e^{ − 1}.

_{C}F = ∫

_{C1}F + ∫

_{C2}F = e − e

^{ − 1}+ ( − e) + e

^{ − 1}= 0.

^{ − 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).

^{ − 1}(t) = (cos(t), − 2sin(t)).

^{2}ln(y))/2],[(y

^{2}ln(x))/2] ), x = t and y = t

^{2}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,t
^{2}) 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 + t^{2}) and so C^{ − 1}(t) = (1 − t,1 − 2t + t^{2}).

^{ − 1}(t)) = ( [((1 − t)

^{2}ln(1 − 2t + t

^{2}))/2],[((1 − 2t + t

^{2})

^{2}ln(1 − t))/2] ).

_{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

_{C1}F = 2, by substitution 2 = ∫

_{C1}F = − ∫

_{C2}F and ∫

_{C2}F = − 2. Similarly 2 = ∫

_{C1}F = − ∫

_{C3}F and ∫

_{C3}F = − 2.

_{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

_{C−}F = − ∫

_{C}F . It follows that ∫

_{C1−}F = − 2, ∫

_{C2−}F = 2 and ∫

_{C3−}F = 2.

_{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

_{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.

_{C1}F ∫

_{C2}F , given that F(x,y) = ( xy,xy ), C

_{1}(t) = (t,t

^{2}) and C

_{2}(t) = (t,t) for 0 ≤ t ≤ 1.

- We compute ∫
_{C1}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}( t^{3},t^{3}) ×( 1,2t ) dt = ∫_{0}^{1}( t^{3}+ 2t^{4}) dt = [13/20]

_{C2}F = ∫

_{a}

^{b}F(C(t)) ×C′(t) dt = ∫

_{0}

^{1}( t

^{2},t

^{2}) ×( 1,1 ) dt = ∫

_{0}

^{1}( 2t

^{2}) dt = [2/3]. Hence ∫

_{C1}F ∫

_{C2}F .

_{C1}F = ∫

_{C2}F , given that F(x,y) = ( x,y ), C

_{1}(t) = (t,t

^{2}) and C

_{2}(t) = (t,t) for 0 ≤ t ≤ 1.

- We compute ∫
_{C1}F = ∫_{a}^{b}F(C(t)) ×C′(t) dt = ∫_{0}^{1}( t,t^{2}) ×( 1,2t ) dt = ∫_{0}^{1}( t + 2t^{3}) dt = 1

_{C2}F = ∫

_{a}

^{b}F(C(t)) ×C′(t) dt = ∫

_{0}

^{1}( t,t ) ×( 1,1 ) dt = ∫

_{0}

^{1}( 2t ) dt = 1. Hence ∫

_{C1}F = ∫

_{C2}F , note the role a vector field plays while computing path integrals.

_{C}f given f(x,y,z) = x + y + z where x = t, y = − t and z = t

^{2}from 0 ≤ t ≤ 1. Do not integrate.

- To find the path integral ∫
_{C}f of a function f, we compute ∫_{C}f = ∫_{a}^{b}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,t
^{2}). Then f(C(t)) = t^{2}, C′(t) = (1, − 1,2t) and || C′(t) || = √{1 + 1 + 4t^{4}} = ( 2 + 4t^{4})^{1 \mathord/ phantom 1 2 2}.

_{C}f = ∫

_{a}

^{b}f(C(t))|| C′(t) || dt = ∫

_{0}

^{1}t

^{2}( 2 + 4t

^{2})

^{1 \mathord/ phantom 1 2 2}dt.

_{C}f given f(x,y,z) = xyz where x = cos(t), y = e

^{t}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 = ∫_{a}^{b}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),e
^{t},t). Then f(C(t)) = te^{t}cos(t), C′(t) = ( − sin(t),e^{t},1) and || C′(t) || = √{sin^{2}(t) + e^{2t}+ 1} = ( sin^{2}(t) + e^{2t}+ 1 )^{1 \mathord/ phantom 1 2 2}.

_{C}f = ∫

_{a}

^{b}f(C(t))|| C′(t) || dt = ∫

_{0}

^{1}te

^{t}cos(t)( sin

^{2}(t) + e

^{2t}+ 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.

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

### Multivariable Calculus

### 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 c _{n}... 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 c _{n}, the last curve, is the same as the beginning point of c_{1}, 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) = x ^{2},xy^{2}.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 cos ^{3}(t), then cos(t)sin^{2}(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)cos ^{3}(t) + cos^{2}(t)sin^{2}(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)cos ^{3}(t) + cos^{2}(t)sin^{2}(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 x ^{3}, and I think it was going to be xy^{2}, 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 t ^{3} and t × 1-t^{2}.*0661

*Then we solve for this one, this is going to equal t ^{3}, we expand this out and multiply it out. We are going to get t - 2t^{2} + 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 + 2t ^{2} - t^{3}, which is equal to 2t^{2} - t^{3}, 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 2t ^{2} - t^{3} dt, and this one of course you can do by hand, so this becomes 2t^{3} over 3 - t^{4}/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

*Again, what is nice about this is you do not necessarily have to change the parameterization.*1237

*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, x ^{2},y,z^{3}.*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 cos ^{2}(t) × y which is sin(t) × z^{3}, so × t^{3}, so this equals, in general we put the variable first, so t^{3}cos^{2}(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, sin ^{2} + cos^{2} + 1 under the radical. And sin^{2} + cos^{2} 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 t ^{3}cos^{2}(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 + 9/28pi ^{3} - 41/27 + 15/32pi^{2}.*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

2 answers

Last reply by: michael Boocher

Fri Aug 2, 2013 8:11 AM

Post by Josh Winfield on May 11, 2013

Example 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 answer

Last reply by: Professor Hovasapian

Sat May 4, 2013 4:13 PM

Post by Eunhee Kim on May 4, 2013

Why is c^-(t)=(a+b-t)?