I absolutely love your videos. I know you're struggling with viewership but please don't leave youtube, you're an amazing TH-camr and are of so much help! You're the reason I've been able to learn calculus, vectors, trigonometry and everything else in 9th grade. I love you man!
@@lidyasolomon5557 So true, classes as we know them bout to be obsolete. I've used youtube for pretty much all of my classes in order to get decent grades in a fraction of the time the rest of the class are spending
Update: the master's had way more niche topics and youtube either had 0 videos on it or if lucky some video with less than 5k views from some weird small conference
You might not see this Professor Dave, but I literally spent days trying to understand this topic, literally almost no video could help explain it for a person like me. You explained the theory and how to apply this knowledge with such simplicity. This just made my day! Thank you so much Professor dave. You are amazing!
there is a mistake in the question, i put it into claude ai and it changed Z=t^2 to Z=T^4 an then solved to get the right answer, the working iout it did: We're given that C is defined by x = t², y = t³, z = t⁴, where t goes from 0 to 1. The integral we need to evaluate is ∫C (z dx + x dy + y dz). According to Green's theorem, this is equal to the surface integral: ∬S (∂y/∂x - ∂x/∂y) dxdy + (∂z/∂x - ∂x/∂z) dxdz + (∂z/∂y - ∂y/∂z) dydz We need to find these partial derivatives: ∂y/∂x = 3t/2 ∂x/∂y = 2/(3t) ∂z/∂x = 2t ∂x/∂z = 1/(2t²) ∂z/∂y = 4t/3 ∂y/∂z = 3/(4t) Substituting these into the surface integral: ∬S (3t/2 - 2/(3t)) dxdy + (2t - 1/(2t²)) dxdz + (4t/3 - 3/(4t)) dydz Now, we need to change the variables from x, y, z to t. The Jacobian of this transformation is: J = |dx/dt dy/dt dz/dt| = |2t 3t² 4t³| = 24t⁵ So our integral becomes: ∫0¹ [(3t/2 - 2/(3t)) + (2t - 1/(2t²)) + (4t/3 - 3/(4t))] * 24t⁵ dt Simplifying: 24 ∫0¹ (3t⁶/2 - 2t⁴/3 + 2t⁶ - t³/2 + 4t⁶/3 - 3t⁴/4) dt Integrating: 24 [(3t⁷/14) - (2t⁵/15) + (2t⁷/7) - (t⁴/8) + (4t⁷/21) - (3t⁵/20)]0¹ Evaluating from 0 to 1: 24 [(3/14) - (2/15) + (2/7) - (1/8) + (4/21) - (3/20)] Simplifying this fraction gives us 33/20.
I found myself, due to unforeseen circumstances, teaching an AP Calculus AB class, and liked doing it so much I decided to take Calculus III. I'm not sure if I could get through vector calculus without your help! Thanks!
This was really really helpful. My university notes introduced it way way more complicated than this. Now that I've seen this I feel like I can now move onto the more complicated stuff that's in my uni notes, thanks!
Thank you Prof Dave! You not only explain the how but the what and why which are important.Now as I do the calculations it's not just numbers infront of me but I know what it is and why I'm doing what I'm doing.
0:29 This literally saved me. I was so confused over what the line integral was because every other person uses mathematical terms to describe it. Thanks so much!
It's 1:45am. I have a lecture in less than 8 hours where we'll be doing examples on line integrals. I missed the lecture on their theory. So here I am. Thank you professor Dave.
Professor Dave, my daughter learned so much from you and I am forever grateful. Just a correction here, the curve C is, by definition, the curve on the x-y plane, NOT on the surface defined by f(x,y). For each small segment dS on the curve C (which is on the x-y plane), we pick a point (x,y), then evaluate f(x,y), and multiply it by dS. We then sum them all up. BTW, I have seen the error in other videos too, e.g., TH-cam video taught by Michel Van Biezen. Last, I wish I could watch videos like yours when I was growing up.
As a special case, when the curve C is chosen to be a straight line along the x-axis (or y-axis), the line integral reduces to the traditional one-dimensional integral along the x-axes (or y-axis).
The parametrics x(t) and y(t) are dependent upon variable "t" to make C! Kind of like drawing with an etch-a-sketch! Over time, you move your x and y dependently but they make a new curve/line/squiggly.
Hey, when you are doing U-sub and writing out the integral, the notation for the lower and upper bounds should be changed to reflect what U would be when t=0, t=6. Pretty sure it should be the integral of u^1/2 from 0.25 to 144.25. It looks like you were implying that it will get changed back to the t variable, I just thought I would mention this for students watching this video. I remember getting marked down from some of my professors over this.
I can see why an integral under a 2d curve is useful to calculate that area if extruded for instance, but what are some practical applications of a line integral over a surface? I initially thought this would give the length of the curve over the surface. How do you find that? Thanks!
you're awesome mister thank you so much for this, it's actually VERY difficult to find calc 3 material on this level. alot of the material is outdated or just uses typical mathematical nomenclature
There is a minor mistake in 9:58 you have written P instead of Q and R in the line integral derivation. But you are the best teacher I have ever seen. Thanks for your great effort!
Hey professor, quick question on comprehension problem two: I get 3/2, not 33/20. I have asked some of my peers about the problem too and they get the same answer as myself. I am not sure what I am doing wrong. Is there a walkthrough of the integral somewhere? any help would be appreciated
error at 6:38 corrected at 6:52 the integration limits are shown as those for t (but not labelled thus) and not as the limits for u (which is the integrating variable)
THANK YOU FOR THIS AMAZING VIDEO. I basically learned Calc 3 in less than a week cause I didn't pay attention all summer. HAHAHA. Now I get to take the exam with only knowing this stuff for 3 days. LMAO. I should be fine though. I'm probably gonna end the class with a C; that's passing, and at this point I'm too far into my college degree to think about it too much.
I also got 3/2... I'm confused, I would assume the video was wrong. I differentiated x,y and z with respect to t, solved for dx, dy, dz and simply plugged it in. Then integrated with respect to t and got (t^5 + 1/2*t^4) from t=0 to t=1. Which gave the answer 3/2.
I have a question, imagine that I have the four dimensioal spacetime, the minkowski spacetime, then I calculate the line integral considering a four vector field (considering that the infinitesimal spacetime invertal is ds²= (c dt)² - dx² - dy² - dz²) and my question is, what should the resulting area mean? What should it be?
The physical significance of the integration is not really clear to me, but going by dimensional analysis the resultant integration will have a dimension of (length)^3(time) which doesnt reprsent any physical quantity as far as i know
Hey, at 9:53 shouldn't we still have P, Q and R instead of 3 P's ? Also, in the 2nd comprehension problem are you sure about the result ? I get 3/2...not 33/20. Could it be a typo? Other than that, thank you for all the videos! Really great.
@@ProfessorDaveExplains If it makes you feel better, I didn't see it the 1st time I watched the video (and I think neither did most of your viewers). If it wasn't for the comprehension I would never go back and catch it. This seems a lot like the "The Monkey Business Illusion" where a gorilla goes unnoticed until you are told about it.
I love you. You're my favourite youtuber that keeps me sane as a chemistry student since you are able to simplify very difficult topics so I get a picture in my mind before I start studying them further. 😊
I think... there was a slight error om the U sub. The U substitution didn't change the lower and upper bounds of the integral which were still in terms of t from 0 to 6. The lower would be from u = 1/4 + 4(0^2) = 1/4 and the upper bound u = 1/4 + 4(6^2) = 577/4 (if I did that right :D) So the answer would then plug those values on for U. :)
sir,in ordinary integration we usually get area under a curve..and when we double integrate it we get volume... But what i have learned from my university teacher she said we get area for a double integral and volume for triple integral...im totally confused....😐 And things got confused when im solving in vector integration..
@@femboy1164 well i think i havent heared anything wrong...since am doing problems in vector integration( double integration is an vital part in this vector portion).... But im not been able to visualise...
I'll just use this notation in this comment: $ as the integral symbol $(a, b) as the integral from a to b where a is the lower bound and b is the upper bound. Think of it this way: the single integral, say $dx in an interval (a, b), is like adding all the infinitesimal lengths of a line from a to b which is equal to the length b - a. You could also express the area under a curve y = f(x) in an interval (a, b) as a region bound by y = 0 and y = f(x) and, x = a and x = b. Therefore the area under the curve is $(a, b) $(0,y) dy dx Simplifying the innermost integral, you have, $(a, b) (y - 0) dx or $(a, b) y dx Substituting f(x) for y because y = f(x), $(a, b) f(x) dx And there you have it! The area under a curve. Technically the area under a curve is a double integral but we forgo that additional step above and instead just integrate the product of f(x) and dx over an interval so that it becomes a single integral.
We can straighten both into straight line by time and speed of sound or light or convert drawing or photography is simpler and much real then guessing the lazy eay to use z components then more simpler by slice of 3 dimension otherwise just 2 dimention is less data not sufficient
here is just my summary: Line integration will give us an area, this area is under a surface along a particular path within that surface( it's projection is C curve in the xy-plane). or just keep in mind that the base is C and whose height is function f(x,y). does it make sense?
In modern mathematics how "infinitesmall" is defined that is a very serious question. You said that "ds", "dt" are infinitesmall, but according to the modern interpretation of differentials, these are principle part of change not infinitesmall. if they are, we cannot do arithmatic with these, just like we cannot calculate infinity+1 or infinity-1, its impossible
Whoops! At 9:53 I accidentally put three P's instead of P, Q, R. Sorry!
And the answer to comprehension question 2 is 3/2 also, right?
I get 3/2 for question two as well
i was scratching my head like "thats not what i saw in class" XDD
Good job!!! Better then my professor could do...
Unacceptable. Your punishment shall be me skipping your ads.
I absolutely love your videos. I know you're struggling with viewership but please don't leave youtube, you're an amazing TH-camr and are of so much help! You're the reason I've been able to learn calculus, vectors, trigonometry and everything else in 9th grade. I love you man!
Honestly. Your explanations are so clear cut and understandable
i agree you are the best explainer of calculus and chemistry on youtube
Your a genius if your olready doing this in ninth grade.
You’ve cleared an entire semesters worth of work up for me in 10 minutes. Thank you so much!
exactlyyyyy bruhhh I didn't pay attention in class, but it's like who even needs class with videos like this lmfao.
@@lidyasolomon5557 So true, classes as we know them bout to be obsolete. I've used youtube for pretty much all of my classes in order to get decent grades in a fraction of the time the rest of the class are spending
@@Shannxy yup. Waste of my time. 😧
6 min for me
Update: the master's had way more niche topics and youtube either had 0 videos on it or if lucky some video with less than 5k views from some weird small conference
You might not see this Professor Dave, but I literally spent days trying to understand this topic, literally almost no video could help explain it for a person like me. You explained the theory and how to apply this knowledge with such simplicity. This just made my day! Thank you so much Professor dave. You are amazing!
A lot of it covered in such a short time. The best series for Multivariable calculus.
The first part of video really helped with the intuition. It’s the clearest explanation yet on what line integral is.
I've also got 3/2 for the second Integral
same idk if im right tho
wait same, idk why??? Im gonna put it into Claude and Wolfram Alpha, lets see
there is a mistake in the question, i put it into claude ai and it changed Z=t^2 to Z=T^4 an then solved to get the right answer,
the working iout it did:
We're given that C is defined by x = t², y = t³, z = t⁴, where t goes from 0 to 1.
The integral we need to evaluate is ∫C (z dx + x dy + y dz).
According to Green's theorem, this is equal to the surface integral:
∬S (∂y/∂x - ∂x/∂y) dxdy + (∂z/∂x - ∂x/∂z) dxdz + (∂z/∂y - ∂y/∂z) dydz
We need to find these partial derivatives:
∂y/∂x = 3t/2
∂x/∂y = 2/(3t)
∂z/∂x = 2t
∂x/∂z = 1/(2t²)
∂z/∂y = 4t/3
∂y/∂z = 3/(4t)
Substituting these into the surface integral:
∬S (3t/2 - 2/(3t)) dxdy + (2t - 1/(2t²)) dxdz + (4t/3 - 3/(4t)) dydz
Now, we need to change the variables from x, y, z to t. The Jacobian of this transformation is:
J = |dx/dt dy/dt dz/dt| = |2t 3t² 4t³| = 24t⁵
So our integral becomes:
∫0¹ [(3t/2 - 2/(3t)) + (2t - 1/(2t²)) + (4t/3 - 3/(4t))] * 24t⁵ dt
Simplifying:
24 ∫0¹ (3t⁶/2 - 2t⁴/3 + 2t⁶ - t³/2 + 4t⁶/3 - 3t⁴/4) dt
Integrating:
24 [(3t⁷/14) - (2t⁵/15) + (2t⁷/7) - (t⁴/8) + (4t⁷/21) - (3t⁵/20)]0¹
Evaluating from 0 to 1:
24 [(3/14) - (2/15) + (2/7) - (1/8) + (4/21) - (3/20)]
Simplifying this fraction gives us 33/20.
I found myself, due to unforeseen circumstances, teaching an AP Calculus AB class, and liked doing it so much I decided to take Calculus III. I'm not sure if I could get through vector calculus without your help! Thanks!
This was really really helpful. My university notes introduced it way way more complicated than this. Now that I've seen this I feel like I can now move onto the more complicated stuff that's in my uni notes, thanks!
Best explanation on youtube hands down. Thank you for helping us all pass ❤️
you're blowing my mind rn
Thank you Prof Dave! You not only explain the how but the what and why which are important.Now as I do the calculations it's not just numbers infront of me but I know what it is and why I'm doing what I'm doing.
The second comprehension questions answer should be 3/2
ok thank god I thought I'm going insane
Really? Not our mistake?
@@killuaenthusiast SAME
It's definitely 3/2.
@@Fernando-me2ie How did he mess that up. I also panicked. It's not the first time he gives the wrong example solution.
0:29 This literally saved me. I was so confused over what the line integral was because every other person uses mathematical terms to describe it. Thanks so much!
thank you professor dave! you just simplified an entire concept that i thought i’d never be able to understand well
This has given me so much insight into how this works and was really easy to follow
This video explains line integrals so simply. Thanks a lot!
It's 1:45am. I have a lecture in less than 8 hours where we'll be doing examples on line integrals. I missed the lecture on their theory. So here I am. Thank you professor Dave.
You have answered every question which was in my mind
Powerful, This video has really helped me. God bless you Prof.
That was surprisingly clear. Thank you
Very well explained professor!
Professor Dave, my daughter learned so much from you and I am forever grateful. Just a correction here, the curve C is, by definition, the curve on the x-y plane, NOT on the surface defined by f(x,y). For each small segment dS on the curve C (which is on the x-y plane), we pick a point (x,y), then evaluate f(x,y), and multiply it by dS. We then sum them all up. BTW, I have seen the error in other videos too, e.g., TH-cam video taught by Michel Van Biezen. Last, I wish I could watch videos like yours when I was growing up.
As a special case, when the curve C is chosen to be a straight line along the x-axis (or y-axis), the line integral reduces to the traditional one-dimensional integral along the x-axes (or y-axis).
Very good remark! Thank you.
Great explanation. Thank you very much sir!
The parametrics x(t) and y(t) are dependent upon variable "t" to make C!
Kind of like drawing with an etch-a-sketch! Over time, you move your x and y dependently but they make a new curve/line/squiggly.
Hey, when you are doing U-sub and writing out the integral, the notation for the lower and upper bounds should be changed to reflect what U would be when t=0, t=6. Pretty sure it should be the integral of u^1/2 from 0.25 to 144.25. It looks like you were implying that it will get changed back to the t variable, I just thought I would mention this for students watching this video. I remember getting marked down from some of my professors over this.
yeah It should be like that
agree with u,
Thank you so much prof Dave
God bless you . You are such a great simplifier
The most amazing classes ever 🥳
I understood the lesson and solved the two examples . thx so much🧡🧡
Very good content, thank you!
pls post more to this series!
I can see why an integral under a 2d curve is useful to calculate that area if extruded for instance, but what are some practical applications of a line integral over a surface? I initially thought this would give the length of the curve over the surface. How do you find that? Thanks!
you're awesome mister thank you so much for this, it's actually VERY difficult to find calc 3 material on this level. alot of the material is outdated or just uses typical mathematical nomenclature
can someone show me the work for the 2nd comprehension problem? I didn't quite get it.
The answer I got for question 1 was the same as Professor Dave's however, for question 2 I keep getting 3/2 as a solution.
Master video on line integrals.
There is a minor mistake in 9:58 you have written P instead of Q and R in the line integral derivation. But you are the best teacher I have ever seen. Thanks for your great effort!
i love your videos sir!!
Thanks, you saved my day again!
With me, I have nothing to say but only THANK YOU DAVE. Iscored a super A in my Calculas exam.Actually it was amongst the so few As scored.
Why do I even pay for university with content like this lol. Thank you so much!
Math is beautiful!
uh at the end do you find f(x,y) by subbing in x(t) and x(y) in
Oh my god your videos are so helpful THANK YOU
How do I even parametrize (y^2/x) in the "comprehension" section? Do I just plug in the x=4t and y=3t into it?
Exactly.
I understood how to get the answer for the first exercise but not the second with the vector. can someone explain?
love you so much David
Can u make physical interpretation for residues and singularity in complex analysis
Love uhhh sir ..I m really thankful to uh for this great lecture 😍😍❤
@8:38 Request you to show with a graph that what difference will it make when dx and dy are used instead of ds
Hey professor, quick question on comprehension problem two: I get 3/2, not 33/20. I have asked some of my peers about the problem too and they get the same answer as myself. I am not sure what I am doing wrong.
Is there a walkthrough of the integral somewhere? any help would be appreciated
Its a typo, you did nothing wrong
Thank you❤
error at 6:38 corrected at 6:52
the integration limits are shown as those for t (but not labelled thus) and not as the limits for u (which is the integrating variable)
t as a common parameter for x and y; ds=sqrt[(dx)^2+(dy)^2]=sqrt[(dx/dt)^2+(dy/dt)^2]dt
just out of curiosity, how did you get t = 2x from x(t) = t/2?
Can you please have a tutorial on how to solve sample problems of line integral?. Thanks!!
Proff Dave can u please confirm that the answer to ur second question is 3/2 not 33/20?
tnx very much
I have a quiz next class in this is a life saver
Good explanation
Good and enough explanation....
Pls, how did you solve the second question?
This one is gold
Does anyone know how to do the second question on the comprehension check?
THANK YOU FOR THIS AMAZING VIDEO. I basically learned Calc 3 in less than a week cause I didn't pay attention all summer. HAHAHA. Now I get to take the exam with only knowing this stuff for 3 days. LMAO. I should be fine though. I'm probably gonna end the class with a C; that's passing, and at this point I'm too far into my college degree to think about it too much.
6:44 Did you make a mistake with the intervals of the integrals? Or am I mistaken?
thanks alot sir
Is comprehension question 2 3/2 ??????
Ur video is really amazing
why i get 3/2 for second question >-
3/2 is the answer ... 33/20 is wrong
so am I
@@JeffReams are you sure im also getting 3/2
I also got 3/2... I'm confused, I would assume the video was wrong. I differentiated x,y and z with respect to t, solved for dx, dy, dz and simply plugged it in. Then integrated with respect to t and got (t^5 + 1/2*t^4) from t=0 to t=1. Which gave the answer 3/2.
@@HDitzzDH Same! :(
I have a question, imagine that I have the four dimensioal spacetime, the minkowski spacetime, then I calculate the line integral considering a four vector field (considering that the infinitesimal spacetime invertal is ds²= (c dt)² - dx² - dy² - dz²) and my question is, what should the resulting area mean? What should it be?
The physical significance of the integration is not really clear to me, but going by dimensional analysis the resultant integration will have a dimension of (length)^3(time)
which doesnt reprsent any physical quantity as far as i know
Hey, at 9:53 shouldn't we still have P, Q and R instead of 3 P's ?
Also, in the 2nd comprehension problem are you sure about the result ? I get 3/2...not 33/20. Could it be a typo?
Other than that, thank you for all the videos! Really great.
oh man what a dumb error with the P's! i pinned a comment. as for the comprehension i'll have to double check
I also got 3/2 I thought I did something wrong
@@ProfessorDaveExplains
If it makes you feel better, I didn't see it the 1st time I watched the video (and I think neither did most of your viewers).
If it wasn't for the comprehension I would never go back and catch it.
This seems a lot like the "The Monkey Business Illusion" where a gorilla goes unnoticed until you are told about it.
matan guedj I got 3/2 too so it might be a typo I’m not sure
why doesn't ∫f(x,y)sqrt((dy/dx)^2+1)dx work?
It is awesome!!!!!
Hi loved your video is it possible to get in contact for private tutoring?
love your tutorials♥️ regards from Argentina
can someone tell me how they did comprehension question 1? I keep getting 15/2
can anyone explain how to do the second comprehension question ??
What did you get?
Not sure what op got but I got 3/2
@@sw3aterCS_ Yeah buddy, It should be 3/2
Coolio. Thanks for checking
@@sw3aterCS_ oh thank jesus I also got 3/2
I love you. You're my favourite youtuber that keeps me sane as a chemistry student since you are able to simplify very difficult topics so I get a picture in my mind before I start studying them further. 😊
I think... there was a slight error om the U sub. The U substitution didn't change the lower and upper bounds of the integral which were still in terms of t from 0 to 6.
The lower would be from u = 1/4 + 4(0^2) = 1/4 and the upper bound u = 1/4 + 4(6^2) = 577/4 (if I did that right :D) So the answer would then plug those values on for U. :)
Yes, I thought that too. The integration limits do change with the substitution 'u'.
He converts back from u to t before plugging in the limits, so don’t worry
sir,in ordinary integration we usually get area under a curve..and when we double integrate it we get volume...
But what i have learned from my university teacher she said we get area for a double integral and volume for triple integral...im totally confused....😐
And things got confused when im solving in vector integration..
You heard wrong
@@femboy1164 well i think i havent heared anything wrong...since am doing problems in vector integration( double integration is an vital part in this vector portion)....
But im not been able to visualise...
I'll just use this notation in this comment:
$ as the integral symbol
$(a, b) as the integral from a to b where a is the lower bound and b is the upper bound.
Think of it this way: the single integral, say $dx in an interval (a, b), is like adding all the infinitesimal lengths of a line from a to b which is equal to the length b - a.
You could also express the area under a curve y = f(x) in an interval (a, b) as a region bound by y = 0 and y = f(x) and, x = a and x = b. Therefore the area under the curve is
$(a, b) $(0,y) dy dx
Simplifying the innermost integral, you have,
$(a, b) (y - 0) dx or
$(a, b) y dx
Substituting f(x) for y because y = f(x),
$(a, b) f(x) dx
And there you have it! The area under a curve. Technically the area under a curve is a double integral but we forgo that additional step above and instead just integrate the product of f(x) and dx over an interval so that it becomes a single integral.
could you upload the slides somewhere?
These are not slides, it's a video file.
Is this last video of playlist ? ☺️
more coming soon
Hey Prof you should do something on tensors...😀😄
literally the goat
🔥🔥🔥🔥🙏👍👍👍 thanks sir
The answer for the second question should have been 3/2.But your video was really helpful tanx.
i got the same answer, but we're probably doing it wrong
got the 2nd problem, 3/2 you have to separately find the line integral for each term then sum them all because dx,dy,dz are different from each other.
We can straighten both into straight line by time and speed of sound or light or convert drawing or photography is simpler and much real then guessing the lazy eay to use z components then more simpler by slice of 3 dimension otherwise just 2 dimention is less data not sufficient
here is just my summary: Line integration will give us an area, this area is under a surface along a particular path within that surface( it's projection is C curve in the xy-plane). or just keep in mind that the base is C and whose height is function f(x,y).
does it make sense?
10 mins of your video = hours of lectures
Hi Professor Dave, do you have an email?
I believe you forgot to write the bounds of integration in terms of u at 6:38
♥️♥️♥️♥️♥️♥️♥️ for you and for your videos as well
correct me if i'm wrong but I think this is scalar line integral I'm looking for vector line integrals anyways thanks
bahuth acha hai sir
How x=t/2 and y=t²
for 2) i got 3/2 instead of 33/20
Same
Guys, just a reminder. Please go back to all the previous videos and smash the like button.
I still didn't get it 😭.
I got the first answer wrong 🤦 what do i do? 🥺
In modern mathematics how "infinitesmall" is defined that is a very serious question. You said that "ds", "dt" are infinitesmall, but according to the modern interpretation of differentials, these are principle part of change not infinitesmall. if they are, we cannot do arithmatic with these, just like we cannot calculate infinity+1 or infinity-1, its impossible