Scalar field line integral independent of path direction | Multivariable Calculus | Khan Academy

[@hedonism13]

Love it when there are no dislikes on a video, especially one as awesome as this.

[@MyNameIssaSimon]

after 9 years, there are 5 dislikes

[@maximlavrenko1164]

@@MyNameIssaSimon 0 now

[@owlattack1971]

I wish my only problem with parameterization was pronouncing it....

[@gamed01]

if anybody had doubts over sign of ds then, it is because ds was calculated usin ds sqr equals dx sqr plus dy sqr that is pythagoras theorem so squaring makes plus dx minus dx same

[@paul_tee]

Rephrasing what you've said, a line integral of a scalar field is essentially a (direct generalization of) the unsigned Riemann integral via parametrization*. By definition, however, parametrization comes with a cost of losing orientation. You lose the data of the sign because you always take the positive value of the square root when you integrate against ds. Given this, it's essentially clear that scalar field line integrals are independent of path direction. *I say generalization in the sense that parametrizating a curve C(x,y) turns the curve into a function of one variable t, i.e. C(x(t),y(t)). Thus, our scalar field also goes from f(x,y) to f(x(t),y(t)) by substitution, a function of one variable. Thus, taking the line integral after parametrization becomes taking an ordinary Riemann integral.

[@infinite9724]

12:08 when t is b, USA! whoop,whoop!

[@imegatrone]

I Really Like The Video Showing that the line integral of a scalar field is independent of path direction From Your

[@ManojKumar-cj7oj]

R(t)=x(t)i+y(t)j a

[@ManojKumar-cj7oj]

x=cos(t), y=sin(t) 0

[@xx_faze_rolf_xxl4029]

In regular integration, dx is negative if u go backwards right? So why is ds not negative in line integrals?? Im confused about this

[@harry_page]

Because in the line integral, the tiny distance you move along the curve is ds, which by Pythagoras can be written sqrt(dx^2 + dy^2), so if either of dx or dy are negative, it doesn't matter because they get squared and become positive, and we strictly take the positive square root I presume. So if you go in the opposite direction along the curve and swap the signs of dx and dy, it doesn't change anything for ds.

[@hilmar5385]

I think the reason it is only positive here is because he defines it to be at 6:33 in the video (without pointing it out) by not defining a positive and negative direction for the line when taking the step size. The x and y axes have directions associated with them, but the line does not, yet he could define one if he chose to do so, but I suppose the logic would be a bit weird to follow either way since "it is negative here because I say so" might not be a satisfactory explanation either

[@hilmar5385]

to continue. It is a bit strange to define the line to be positive in both directions especially for certain cases. Like say the line goes purely along the x axis, when doing an integral along the x axis in opposite directions you would switch signs, but now since the line is defined to be positive in both directions you get equal signs. I think the actual problem is that there isn't an elegant way to express the direction of the line since the magnitude of the line is one directional and the x and y space is two dimensional, so x and y have 4 possible different signs, but the magnitude of the function has only 2 possible signs, For the next video for vector fields he finds a way to do this though. I think this is because the dimensions of the output of the underlying function now matches the input so he can say that the direction of the line at any point is just equivalent to where it is pointing. Not sure if this makes sense.But yeah what I'm getting at is that I don't think it is correct to say that the direction doesn't matter here because it is a scalar field, but rather because the sign of the line is always positive, which it is because it is difficult to represent the sign mathematically so here we defined it to be always positive, and it's difficult to represent the sign mathematically because it is a line in a 2d scalar field.

[@TheJunkieBox]

5:35 Turn on subtitles.

[@altuber99_athlete]

"Let's let" hahaha

[@levi4328]

lmao this phone call 14:24

[@vikraal6974]

It's Grant

[@Cyclingdino]

You're a master at explaining but there's just one thing I didn't understand in your demonstration. When you made the substitution u=a+b-t, you related the variable u with t, but in your final product (the integral in terms of u), you said that u could be equal to t. Can someone please explain?

[@ameyajatinshah458]

u is not equal to t when originally defined as u=a+b-t. But in the last part where he changes the sign of the integral by reversing the limits of the integral, u gets redefined and varies from a to b. This is due to one of the properties of integration that he mentions.

[@VinothKumar-qo7ry]

I'm also having same doubt

[@John-hc6mo]

@@ameyajatinshah458 So when the path is moving from b to a the value of u needs to be as originally defined, u=a+b-t, but once the limits are reversed, equivalent to integrating a path in the opposite direction, you need u=t as explained in the preceeding video. (I think).

[@arjunmandavkar8597]

Why a scaler line integral doesn't changes its sign while simple definite Integral changes, if path is reversed?

[@SohamChakraborty42069]

Scalar line integral gives the algebraic area of the curtain wall. As long as z=f(x,y) does not change its sign, changing the direction in which you move along the curve would not change the sign of the area.

[@ManojKumar-cj7oj]

O my god,I just learned to reverse the parametric curve

[@patrinos13]

There is something i dont understand and i would really appreciate your help. Isnt f(x,y) a Surface and not a scalar field? I thought a scalar field has the coordinates as variables giving us a scalar value as a result but not a coordinate. When a fucntion has coordinate variables and the result is a coordinate isn't that a surface like z=f(x,y)?

[@theflaggeddragon9472]

patrinos13 they can be modeled as both. A scalar field can be written as a surface where the height (f(x,y)) is the value, or it can be a flat plane where at any given point if you select a point, instead of giving it a height it's value is directly computed. This means a scalar field f(x,y,z) can either be a 4 dimensional "surface" or a three dimensional array of points. Hope that helps

[@trollfootballgeneration8645]

Just pause and ponder on your own question,imagine a field in a x,y,z plane and compare it with a surface in the same plane

[@TheJunkieBox]

the TV is equal to white pride lmao at subtitles

[@christucker5760]

I can't understand how (-x(a+b))^2 is equal to x(a+b)^2 ..... these are not mathematical equal

[@MyNameIssaSimon]

Might be late but x is a function

[@richarddeonath]

because du=-dt , u=-t. he was just showing how similar the two results were not dat they are equal