The Line Integral, A Visual Introduction

[@vcubingx]

Follow-up on Green's Theorem! -> kzbin.info/www/bejne/boTafHdrlLt_a9E Support me on Patreon: patreon.com/vcubingx Here's a small intuition correction: When I said "projecting the integral", that isn't particularly correct. It's more of a "stretching out" the curve so it becomes straight, without distorting its length. This is why we have the factor of sqrt(x'^2 + y'^2) in the conversion formula. (Credit to Dr. Steve Trettel for pointing this out!) MISTAKES! At 4:11, the problem should be (1,-1). I read it wrong and typed it wrong. Wow. Also in the first example the integral's function should be 3(3-2t)^2-2(6-7t). ALSO IN 8:15 I MISSED A MINUS SIGN HOLY

[@md.rafiqurrahman792]

vcubingx You put the parametric equation such as 3(3-2t)^2-2(6-7t)^2 where the main function is (3x^2-2y). I think you put a square by mistake.

[@vcubingx]

@@md.rafiqurrahman792 sheesh you're right

[@AdamsAppleseed]

Just wanted to add onto this that if you calculate the problem at 4:11 to be to (1,1) then you should get an answer of 6 * (sqrt(29)) At least that's what I got lol

[@theInternet633]

This kinda looks like 3Blue1Brown content. Not to say this isn't awsome still. That's more of a compliment than anything else but i'll be damned if you aren't an avid viewer of him.

[@vcubingx]

Haha yeah I use the animation engine he created for his videos, which is why it looks so similar

[@borekworek69]

When I saw the first seconds of the video my first thought was "Yes! More Manim users!" I think Manim should be a standard for math-related videos. It's such a great library and it's a shame that not that many people know about it. Sorry for my poor English :)

[@zakirreshi6737]

Same here...and i love these types of videos

[@bostangpalaguna8467]

you should check out: manim.

[@quitest4850]

I am a simple man, I see videos helping people to understand the concept of math, I hit like

[@mahatmawisesa9706]

8:14 isn't that integral should be -3(7+5t)-5(33+39t) ?

[@johnstone6848]

your parameterization for the curve C at 4:25 is incorrect. It should be 6-5t not 6-7t

[@wduandy]

How to do this parametrization?

[@mrdiin.dev_]

@@wduandy n order to parametrize a line, you need to know at least one point on the line, and the direction of the line. If you know two points on the line, you can find its direction. The parametrization of a line is r(t) = u + tv, where u is a point on the line and v is a vector parallel to the line. There are lots of possible such vectors u and v. To find one such vector v, find the difference between any two points on the line. In this example, we have A(3, 6) and B(1, 1) or I guess he meant to say B(1, -1) hence his result. So using the latter, the direction of the line will be AB = B - A = (1, -1) - (3, 6) = (-2, -7). Thus us our v. The line can then be written as r(t) = A + tv or B + tv, it doesn't really matter, but he used r(t) = A + tv. So r(t) = (3, 6) + t(-2, -7) which means x = 3 - 2t and y = 6 - 7t.

[@onemadscientist7305]

SInce we're working with a segment here, one better way to think about this is in terms of barycenters between the two endpoints with coefficients between 0 and 1. If A and B are the endpoints, then a point P inside the segment can be written as follows : P=t*A+(1-t)*B where 0

[@onemadscientist7305]

Also, the parameterization in the video is definitely wrong and this actually changes the value of the integral ! For a start, , it should be ds=sqrt(29)*dt and not ds=sqrt(53)*dt...

[@gd3xsilon783]

I think he made a mistake in the definition of C. He said in the video “C is the line segment from (3,6) to (-1,1)”.

[@Miguelocod]

As an electronics engineering I needed to refresh this 6 years after taking Multivariate calculus, and this was an excellent explanation. Thanks!

[@steventhijs6921]

There is a minor mistake at 4:50, is should just be '-2(6-7t)', not '-2(6-7t)^2'. The answer of 8 × sqrt(53) is correct though if you meant a line segment to (1, -1)

[@michaeljuergen4326]

I was thinking the same

[@andrewshain899]

@@michaeljuergen4326 I was second guessing myself The two mistakes on one screen was really getting to me.

[@DougChemist-oj4ev]

How is 8 x sqrt(53) correct given the corrected '-2(6-7t)' term and the (1, -1) endpoint? Evaluating the eqn at bottom of 4:50 (with the correct term substituted) at t=1 gives 5 x sqrt(53), and for t=0 one gets -15 x sqrt(53), so isn't the correct answer -10 x sqrt(53)?

[@vashon100]

@@DougChemist-oj4ev Because 6 -5 = 1, not 6-7 = 1 . (3,6) to (1,1)

[@DougChemist-oj4ev]

Sorry I forgot to take the antiderivative when evaluating the final integral; the given answer of 8 times the square root of 53 is correct@@vashon100

[@isaacaguirre1881]

I am a first year EE student studying for my vector calc exam . This is the best channel i have found this year regarding this topic. Keep it up, you're the best.

[@Harry-ub2fv]

This encouraged me to learn Vector Calculus.

[@tandlose]

This is great! I like the balance you have between questions and explanation, 3blue1brown never uses questions which is a shame because it is a nice way to apply what you've learnt and get a breather from the learning. On the other hand the examples are quick and you don't waste time explaining basic calculus so it doesn't feel dragged out and uninformative like many other math videos who uses two thirds of the time for examples. Great work and I can't wait to see more!

[@understandtheuniverse2199]

Bro you don't know value of basic calculus and we need to think of it every time

[@Pavan_Gaonkar_abc]

ತುಂಬು ಹೃದಯದ ಧನ್ಯವಾದಗಳು... ಇದೊಂದು ಅದ್ಭುತ ಪ್ರದರ್ಶನ... ಸಂಪೂರ್ಣವಾಗಿ ಅರ್ಥವಾಯಿತು...

[@martinsanchez-hw4fi]

If C was from (1,1) to (3,6), shouldn't the parametrization be [3-2t, 6-5t]?

[@vashon100]

Yes, vid is wrong.

[@lad4694]

​@@vashon100 He forgot a minus sign. Rather than (1,1) it should have been (1,-1)

[@gururajmundargi7939]

sir you have started to teach difficult math using animation it really helps to me.iam a engineering student from india. please dont stop uploding this kind of vedio

[@quantumleap7964]

the animations really helped solidify the concept, great job.

[@vcubingx]

Thanks! Glad you liked it!

[@fyu1945]

Really good but I think you make the same mistake 3Blue1Brown made when he was starting. Talk a little bit slower. But otherwise great video!

[@Michallote]

Is it weird to watch this videos on 1.5x?

[@agrajyadav2951]

@@Michallote yes

[@howmathematicianscreatemat9226]

The downside of being gifted means not being able to see what others cannot see.

[@ahlecksplodes]

@@howmathematicianscreatemat9226shut yo corny ahh up

[@utkarshsaxena3511]

Thank you for giving such details. I don't know why these are not written in books clearly, and we have to search the internet for better understanding, especially for mathematics.

[@zprogress4004]

3:52 the line integral is supposed to be the area of the orange colored region right? the arrow of the integral is pointing on the projected shadow on the xz-plane?

[@saiprasadmohanty8339]

Even i have the same doubt so can you please help me incase if you have got your doubt cleared. Thanks

[@gibranayala3903]

underrated content very good

[@sidimohamedbenelmalih7133]

This is the first time when I started to understand the integrals, and it was great, like a schematization to what's Waiting for me, and be prepared to it

[@CKFOKAY]

I think your parametric representation of the line should be . Am I wrong ?

[@myonlynick]

4:24 it would help us a lot from confusion if you take the extra step and correcting the point (1,1). It should say (1,-1). Maybe embed an annotation in the current video?

[@EngineerNick]

Really good video thankyou so much :) I am confused at 6:26 - I think the red and yellow vector arrows should be in the same direction? I think the red arrow in the circle is facing the wrong way; for the dot product to be negative there must be an obtuse angle between the vectors? Still the animation is amazing well done :)

[@alexdeac2951]

such a great video! concise, intuitive and covers all the relevant information

[@bostangpalaguna8467]

dude... your visualization is crystal clear! thank's a lot !

[@thevegg3275]

Thanks! What is the precise method of parameterizing C? Looks easy to figure out for a straight line but a wavy curve?...can't be done without a specific method. Can you define that method in detail please?

[@vcubingx]

You can't always parametrize the curve by hand. For example if I draw any squiggle on a paper, you can't always parameterize it. But, when solving problems by hand you would probably get curves that can be easily parameterized, like circles, ellipses, lines, quadratics, etc. You can just Google "parametric curves" and that should answer your question if you're curious on how to parameterize those curves. Thanks for watching!

[@andrewshain899]

kzbin.info/www/bejne/iqjToH94lt5-j68 This video here does a great job of explaining starting at 1:32.

[@erickperez5930]

Your videos are great! Please keep the great job you have been doing so far.

[@vcubingx]

Thank you so much!

[@morgengabe1]

The instrumentals in this video are something else lol. Stylishly done!

[@prathmeshpatel5657]

Dude this is great. It made everything so much clearer. Thank you for making this

[@visbellum]

It's my pleasure to watch it!! Thank's for doing it!!!

[@vcubingx]

Thanks!

[@thehighground583]

THANK YOU for helping for giving me a visual intuition of where the formula for arc length comes from at the start of the video!! That was something that I had previously lacked. Also love the background music!

[@mariogalindoq]

There are a mistake: the line segment from (3,6) to (1,1) is not C=(3-2t, 6-7t) 0

[@DiegoMathemagician]

0:28 just pointing out this little mistake i did noticed f(x)=x^2+y^2; it should be f(x.y)=x^2+y^2. Thanks for doing such a great video, by the way.

[@vcubingx]

Thanks for pointing that out!

[@saadsaoud5185]

top class fantastic best leacture of all about introductionary line integeration

[@vikramnagarjuna3549]

Hi, I am from India, that was great video sir... 👏👏👏👍

[@jayy7842]

At 4:32, how did you parametrize the curve from 3x^2 -2y to become < 3 - 2t, 6 - 7t > ?

[@dylan1160]

it's wrong. check the line starts at (1,-1) not (1,1)

[@mnada72]

3:55 I didn't get what is the projection pointed to by the arrow.?

[@fruxo2000]

When the animation starts playing at 7:16, isn't the red arrow pointing both in the wrong direction and with the wrong magnitude? Shouldn't the red arrow, which is meant to be r'(t), be pointing in the same direction as the yellow arrow in the graph, which is also meant to be r'(t). The yellow arrow appears correct. And, since the dot product of two vectors that are more than 90d apart is negative, it would only make sense if the red arrow was on the opposite side of the green arrow, until after C crosses origo, when the green arrow should swap sides (as it correctly does) but the red arrow should remain pointing up-right (which it currently only does for the latter half of the animation). Am I mistaken?

[@photon2724]

indians are finally making quality content.

[@krishnabharadwaj4638]

Yes finally something other than jee and neet

[@ritishbehera4711]

Is he indian? I thought its a foreign youtube channel

[@robertmegee9052]

Ok, I'm stuck at about 2:10 into the video. You reduce your delta-x and delta-y to be derivitives but where did the delta-t come from? No explanation was given other than you can multiply and devide by the equivalent of 1. Please tell what it actually means. Where is it on the coordinate axis?

[@sonu7642]

Finally i understood the concept..........thankyou 😎😎😎😎

[@lok7396]

Nicely made.

[@Matheus-yx6qb]

There's an error in 7:15. The red vector inside the circle was supposed to be the (vec r) derivative, but it seems that the red vector was (vec r) itself

[@predatoryanimal6397]

the lo-fi track is a nice touch

[@ignassablinskas9175]

Hello, you wrongly parametrized the curve C at 4:25 , it should be < 3-2t , 6-5t >

[@1951split]

0:28 you mean f(x,y) = x^2 + y^2 instead of "f(x)" 4:24 line segment is wrong. you drew (3,6) to (1,-1) instead of (1,1)

[@aalexren]

přesně tak

[@kalaiselvanrajasekaran8352]

One suggestion pal,if you could have used a normal curvy surface like a parabolic roof top kind and explain the line integral concept then it would be made easily more comprehensive than now.But it is a good explanation pal.Impressive one.

[@sheepphic]

You can make parentheses scale to the size of their contents in LaTeX by using \left( and ight) instead of plain ( and ); this might work better for things like your derivatives at 2:07

[@LagAttacktoSlay]

Finally, people are recognizing 3b1b's Pythonlib lmao. Good video!

[@szilike_10]

3:57 how could the projected area be tha same as the area under the graph? I mean if we'd stretch that curve to a line, the area of it would be much bigger. Isn't it like saying that given a line y = mx, on the OXY plane, and a segment a to b on the X axis, the distance between (a, y(a)) and (b, y(b)) is the same as d(a, b)? The pythagorean theorem says the exact opposite. I know it is not the same, I reduced the number of dimensions, but I'm still curious if I'm not noticing something. However great video, keep up the work.

[@szilike_10]

Also, if I have a sheet of paper its shadow can be either a line or the shape of the paper depending how I keep it in the air. The area of the paper (the line integral) won't change however the area of the shadow (projection area) will.

[@probablyapproximatelyok8146]

I don’t think the area of the projection is the same as the area under the curve. I’m very much new to these topics from multivariable calculus but my best guess as to why it is being projected onto the OXZ plane is b/c the integral is being taken ‘with respect to’ x (with the dx at the end) so it is only multiplying the height of each rectangle by the x component of its respective ‘step’, and when the integral has the ds at the end it is equal to the area under the curve b/c it multiplies the rectangles by the actual magnitude of their respective ‘step’ instead of a projected version of it. Maybe if it were dy instead of a dx it would be projected onto the OZY plane. Hope this helps! (Sorry if it doesn’t or if I misunderstood your question)

[@sarvagyagupta1744]

At 1:52, how did you get 𝚫t? What does that represent?

[@vcubingx]

Delta t represents a change in time. We're moving to a parametric form

[@sarvagyagupta1744]

@@vcubingx So it could any other variable right? It just needs to be consistent with other axes.

[@professoreisenoxid9882]

Why are F(r(t)l and r'(t) at 7:16 aligned? Shouldn't r'(t) be orientied in the other direction because the displayed dot product is mit negative-it would be with a negative orientation. Therefore the (yellow) vector r'(t) in the vectorfield doesn't show the direction as the red one. Somehow I am confused 😅

[@jagsproduction]

Dope animation, it helps in understanding better

[@harshans7712]

thanks a lot for this explanation but I have a doubt, how did u get the last question's limits as 0 to 1?

[@ulyssesfewl1059]

At 4:15 you say "to -1,1", yet on the screen the end point is shown as (1,1). Which is correct?

[@DougChemist-oj4ev]

Neither is correct - the endpoint should be (1, -1) as shown on the graph, to give the answer dS = sqrt53dt. Also note the eqn at 4:50 of 3(3-2t)^2-2(6-7t)^2 should instead be 3(3-2t)^2-2(6-7t), to give the final answer given of 8 times sqrt53, These things have been pointed out by others. Please can you keep this pinned at the top?

[@wgwandawg]

This is so well explained. I also liked the way you played the animation again so that we could get a clearer picture of whats going on. Thank you!

[@kaavyabanerjee]

How you parametrize that equation?

[@sanjayganesh6760]

How did you parametrize the curve?

[@Mirabel151]

Thank you so much for this video! it helped it to better understand line integrals for my finals!

[@lad4694]

At 4:24, for the parametrization to be correct, you missed a minus sign. Rather than (1,1) it should have been (1,-1)

[@Linaiz]

Thank you for all your helpful videos. I think you're doing an amazing work. If I could suggest one thing, once your channel expands more, you could hire a professional voice-over guy. I think that would facilitate understanding, because visuals are already top notch, and I only ever get lost in the audio explanation. Best of luck with your channel!

[@uwaispatel4352]

theres a mistake in the parametrization at 4:29

[@Iris-zr1rf]

I like to think about the line integral as being the total elevation graph as you take a hike (Curve C) through a mountainous region (f(xy))

[@filomenamendes5349]

Incredibly helpful! Thank you so much

[@dordavid3032]

4:25 the parametrization is incorrect! should be C=(3-2t,6-5t) the correct answer is sqrt of 1044

[@crimsonarmor.]

Excellent explanation and awesome animations! Subscribed :)

[@ashishratnawat2711]

Excellent Video! Just one thing, around 6:20 it should be "dot product indicate how aligned two vectors are" right ?

[@guitsgarcia5446]

i think he meant 'how parallel two vectors are' instead of perpendicular

[@SimpMaker]

Thank you for the visualizer

[@alexn.2702]

@8:15 when calculating the final product, wouldn't you add 3(7+5t)^2 with 5(33+39t) since its the dot product of the two

[@vcubingx]

yeah I missed a minus sign there. I put a list of all the mistakes in a pinned comment. Sorry about that!

[@alexn.2702]

vcubingx oh okay didn’t see that, thank you!

[@126sivgucsivanshgupta2]

I really like the song choice in the background ... And offcourse the mathematics

[@omjoglekar3677]

3:53 why did we squash it onto X-Z plane ?

[@nagesh1805]

Awesome!. I'm wondering if you made these videos using manim(3b1b) library.

[@DevPatel-nf5mp]

Great work 🔥🔥🔥

[@pastorofmuppets7654]

Nicely done! Can you suggest some good Manim tutorials?

[@muhammadyaseen8923]

Yep. Moar! manim tutorials.

[@vcubingx]

Try Todd Zimmerman's. But the thing that works best for me was going through Grant's projects.

[@muhammadyaseen8923]

@@vcubingx thanks for suggesting, can you link? I'm not sure which "Todd Zimmerman" you are referring to.

[@vcubingx]

talkingphysics.wordpress.com/2019/01/08/getting-started-animating-with-manim-and-python-3-7/

[@Efrain0465]

Que programa usaste? Gracias por el video

[@karlaserit4523]

Hello. How to parameterize a curve into a vector? Thank you for your answer ^_^

[@Michallote]

I have the same question when dealing with more elaborate curves. But to parametrize a line is actually pretty simple. I'll try to be as clear as i can: First main idea is to use a variable t which conviniently for us goes from 0 to 1 So let's say we wanted to get from point a to b which are fixed using t as the parameter. Our parametric function will be f(t) So when t=0 the f(t=0) = a t=1, f(t=1) = b so you can think a bit about it and you can intuitively get to: f(t) = a * (1-t) + b * t So that those conditions stay true. The beautiful thing about it is that a and b can be anything, could be scalar valued, 2D vectors: f(t)= (ax,ay) * (t -1) + (bx,by) * t 3D vectors. The parametrization still linearly gets you from one to the next. More formally it is called linear interpolation look it up on Wikipedia it's awesome.

[@mayankchhipa503]

Can you make a series of complex variable analysis!??

[@vcubingx]

Noted! I definitely want to cover topics within complex analysis, probably not an entire series though

[@mayankchhipa503]

@@vcubingx thanks for consideration now can't wait to see complex analysis visually!!

[@harrypotter1155]

3:53 so the curvy line integral of the scalar field is as the same as the projected one?

[@vcubingx]

Yep!

[@mrdiin.dev_]

Chillhop music is my favourite and made me love this video more! What software did you use to animate though?

[@vcubingx]

I used manim: github.com/3b1b/manim. Glad you enjoyed!

[@gby9861]

Thank you for the help! Cemented my understanding!

[@vcubingx]

Glad it helped!

[@conzpicuouz]

Awesome video. Thank you very much for your help.

[@vcubingx]

Glad it helped!

[@bugi6606]

Thank you for that video! Do more such videos !!!

[@arrabax]

great work!

[@creatorofanimation]

What music did you use in this video?

[@BoZhaoengineering]

It is a better interpretation than textbook. But I would suggest go to the text book first after getting the concept of line integral and back to the animation video. The idea of line integral in a vector field becomes much more clear! Thank you for your nice video and interpreting

[@vcubingx]

I agree! I think I've improved in my explanations over time, might revisit this topic in the future!

[@BoZhaoengineering]

vcubingx cheers. Today’s learning resource and landscape has changed to the better than before. Digital tutorial video becomes indispensable as a learning companion. Before the rise of Khan academy , it seemed nobody predict that while learning ‘stem’, the help could be made from enthusiasts by digital technologies.

[@ismailrachdy]

cann Someone explain to me why is between 1 and 0 «

[@edissonstevencastanomesa4841]

Nice video! Just a comment, for better notation, it should be used a "dt" for last shown equation at minute 8.11.

[@rayyan21d]

Thank you so much, this was really helpful!

[@mohammedkhan4990]

Excellent presentations. Your videos helped me do much. Thank you.

[@vcubingx]

Thanks for watching

[@harendrasingh2319]

Great work keep going,,👍

[@zakirreshi6737]

Thanks for these beautiful videos..

[@aalexren]

Sorry, but on 4:34 you made a mistake. Parametrisation supposed to be , not the 6-7t. Maybe I didn't get the point, then could you please fix my issue. UPD. OK, got the pinned comment

[@ingridp8233]

such a great explanation, thank you!

[@GubeTube19]

How do u parametrize curves like at 4:28?

[@vcubingx]

r(t) = (1-t) + t

[@studiosbonsoir]

Yeah I'm also confused about this

[@rogeliomontelongo1398]

@@vcubingx So, why is -7t in the second part? isn't -5t?

[@quark67000]

@@rogeliomontelongo1398 There's probably an error in the text: if you compare with the graph near it, the line segment C is from (3,6) to (1,-1), not to (1,1). So the parametrisation with -7t is correct.

[@TZ-nd1cm]

is the answer to the first question is 11sqrt29?

[@abhinaamrawat1680]

Just have one word to say "exquisite"

[@bharathsundar2378]

great work mate...

[@DOPEBEATZBOYS]

the first example was wrong wasn't it? the parameterization seems off

[@DOPEBEATZBOYS]

and in the animation, you had a square for the y and it wasn't in the problem, not sure if you used that for the final answer

[@vcubingx]

Yeah both the examples have problems. I addressed all of them in a pinned comment. Thanks for pointing them out!

[@DOPEBEATZBOYS]

@@vcubingx oof sorry, i should have looked for that first

[@ayooluwakolawole]

Great vid, really helped me build intuition! Only critique is some of the coordinates of the curves didn't match the graphs you put

[@shahrukhmechanical]

Was it -549.5 or am I wrong?

[@AlePreludioFinal]

Hey! A great video. Thank you

[@vcubingx]

Thanks! You're welcome!