21: Vector Field Line Integrals - Valuable Vector Calculus

[@iyadindia862]

A new way of introducing the line integral to me..Thanks for your efforts

[@hashtagaroma7778]

watching right before my exam. you're the GOAT

[@grouchyfluff]

Fantastic stuff. Going into my 4th year of aerospace engineering, I took Calc 3 ages ago and haven't seen ""line integrals"" like this since way back then. First 7 minutes of this video made more conceptual sense than my entire recollection of the latter half of that class. I'll definitely be checking out more videos in this series for a deeper understanding and a refresher on this topic. cheers!

[@austinu]

This was very good, same as your other video on scalar line integrals. Thank you!

[@samuelhafer7516]

Well taught. Wish I had this when I took cal 3.

[@briandwi2504]

Another very clear and concise lesson. Thanks!

[@abdeljalilpr2033]

Very very good choise of subjects and i like tge way of analysis and simplification..ma shaa Allah!! Respect to you

[@jatingarg1375]

Thank u sir ..love from India and much respect to you..♥️👍

[@saqlainahmed4197]

Sir awesome work 👌👌

[@domenscreativemaking3942]

Thank u so much sir

[@GauravKumar-np2qm]

amazing.

[@moondxstq4297]

Can you make a video about solving Circulation involving finding Potential function?

[@denisovtenetova6125]

amazing!

[@nathanisbored]

Hadn’t heard it called a ‘scalar field line integral’ before. my teacher said he didn’t like the way the book explained it and wrote his own notes, and in his notes he called it the ‘tangential integral’ (the other one being called the flux integral). The difference between the two is that one is F dot T and the other is F dot n, where n is normal to the curve

[@arkamninguno8446]

Hola, como ya sé que hablas español. Te quería decir que este tema de integrales de línea está muy interesante, también lo estoy estudiando y espero que llegues a la parte de teorema de stokes porque hay una cosa que no entiendo sobre ese teorema. Gracias y saludos por tu esfuerzo. 👍👍😎

[@A_Box]

Better than my old University notes. LOL

[@danw6406]

Hi, i have a question depending on the difference about scalar to vecotr line integral. In scalar line integration we use magnitude of r-prime * dt, because we are interested in the small change along the curve, which i understand. But in vectorfield line integration we take r-prime * dt without the magnitude. Like i understand you, is that we want to sum the projection of our vectorfiled at a given point on the curve onto the tangential unit vector of the curve. Now to my question shouldnt the r-prime here be a unit vector as well, so we can measure this projection correctly? Because you even mentioned that big T needs to be a unit vector. To my understandig r-prime gives us the speed on a given point on the curve which is not always of the length one. Sorry for my english, im from Germany.

[@MuPrimeMath]

The formulation of the integrand as F(r(t)) ⋅ r'(t) dt is equal to F ⋅ T ds. They're both the exact same integral. We often use the form F(r(t)) ⋅ r'(t) dt simply because it is easier to compute for a given problem. We do the same thing for the scalar field line integral as well. The unit tangent vector T points in the direction of r', so r' = |r'| * T. It's the same conversion in both cases. Whether r' is shorter or longer than T is accounted for by how fast the parametrization r traverses the curve. If r' is larger then the parametrization will traverse faster, thus assigning less weight in the integral. That cancels out the fact that F ⋅ r' is larger if r' is larger.

[@danw6406]

@@MuPrimeMath That makes total sense to me now. Thanks for answering so fast. Just a tangent, would it be a right thought to have, that my r´t will be always of unit length 1 if i parametrize my curve after curve length? Im just curious. Since with a parametrized curve c(t) for example the output matches the input. Like c(8)=8, and so on. So my r´t which consists of dc/dt will be always 1. Im sorry if I monopolize your time. But the help in my university is not as good as you just did. My professor is shy and my tutors for math are worse than me. 😞Thanks again!

[@MuPrimeMath]

Yes - in fact a curve r(t) is arclength parametrized if and only if |r'(t)| = 1.

[@danw6406]

@@MuPrimeMath Ty very very much!