.Dr. Peyam, at the point at which you parameterized the surface, could you have used polar coordinates and the dS factor and avoided having to do the cross product? I saw that at the end you used polar coordinates anyway.
Do you ever plan on doing some analysis on manifolds? I notice you have been gearing towards analysis on the real line and a little bit of topology. Maybe some differential forms/the generalized stokes theorem?
sir can you maybe include an example on how stoke's theorem can be used with gauss's theorem to calculate open loops, or any other example as such. a my professor taught it, but I wasn't too sure, and your videos are superrrrrr clear. thanks so much
@@drpeyam an example of question would be (integration sign{c})(a.dr) where a =(-y/(x^2+y^2) ,x/(x^2+y^2),1) where c in the first octant is given by : x^2 + y^2 =1 , x+2y-z=1 it starts from (1,0,0) to (0,1,1) ans. (pi/2) +1
Peyam joon, please make series of vedio explain all vector fields from beginning till this green and stokes theorems for uneducated ones like me to get the point. Merci
Dear Dr Payem, may you help me solve this problem, please ? Give a,b are real numbers. Evaluating the integral of x.e^(-x^2) dx, from a to b, by letting t=x^2, with 3 conditions : 1. 0 ≤ a ≤ b 2. a ≤ 0 ≤ b 3. a ≤ b ≤ 0 And I also wonder if we can use Stokes Theorem in this problem, sir. Thank you sir
I don't believe that you can have any arbitrary surface. Your surface can't extend beyond the outermost line curve defined by dropping perpendicular down from every point on the surface.
You can integrate over the circular area or over the hemisphere, correct? Which one will be simpler? (Asked very early in viewing, prior to computation of the determinant)... compute one component & permute the variables
This was the only video I could find that went over finding normal vector and parameterization. Thank you for posting
thank you so much Dr. Peyam for this walkthrough! it helped me get a better understanding for the idea!
U lost me at “let’s”
On a serious note, r u teaching this to ur calc 3 students already? 😱
No. we stop at surface integrals 😭😭😭
@@drpeyam because of limited time issues right?
@@drpeyam is this part of calc 3 as we were taught this is calc2
The dark side of the Stokes theorem is a pathway to many abilities some consider to be unnatural.
is it possible to learn this power?
I misread the title as “The Dark Side of Strokes” and I was all “wait, there’s a light side to those?”
Love your videos! Both so fun and educational
Initially learning stoke's theorem nearly gave me a stroke.
Claro que si amigo!
This video is absolutely fantastic
Thank you :)
Very clear, thank you
.Dr. Peyam, at the point at which you parameterized the surface, could you have used polar coordinates and the dS factor and avoided having to do the cross product? I saw that at the end you used polar coordinates anyway.
Doing that would make dS messy. It's a lot cleaner this way.
Do you ever plan on doing some analysis on manifolds? I notice you have been gearing towards analysis on the real line and a little bit of topology. Maybe some differential forms/the generalized stokes theorem?
Even I thought this
Probably not
Does it not interest you as much?
@@luna9200 he did a PhD on pde
One question, is a transcendental number the integral from 2 to 3 of the zeta function?
No idea
Very good lecture
3:09 - I've always felt that is a really hard way to take a cross-product/determinate of a 3x3 matrix
thank youuu!
sir can you maybe include an example on how stoke's theorem can be used with gauss's theorem to calculate open loops, or any other example as such. a my professor taught it, but I wasn't too sure, and your videos are superrrrrr clear. thanks so much
Never heard of it
@@drpeyam an example of question would be (integration sign{c})(a.dr) where a =(-y/(x^2+y^2) ,x/(x^2+y^2),1)
where c in the first octant is given by :
x^2 + y^2 =1 , x+2y-z=1
it starts from (1,0,0) to (0,1,1)
ans. (pi/2) +1
I actually subsequently did this integral without parametrizing and without the cross product, and got the same answer.
Good for you
Peyam joon, please make series of vedio explain all vector fields from beginning till this green and stokes theorems for uneducated ones like me to get the point. Merci
Already done
Love you
Dear Dr Payem, may you help me solve this problem, please ?
Give a,b are real numbers. Evaluating the integral of x.e^(-x^2) dx, from a to b, by letting t=x^2,
with 3 conditions :
1. 0 ≤ a ≤ b
2. a ≤ 0 ≤ b
3. a ≤ b ≤ 0
And I also wonder if we can use Stokes Theorem in this problem, sir.
Thank you sir
This has nothing to do with stokes, it’s a single variable integral
Dr. Peyam, how do you know the parametruzation of "S"? How I know That is r(x, y) = (x, y, 1)? Someone ecuation?
Check out my video on parametric surfaces
@@drpeyam ohhh, that's right. Jajajaj thank you, I see. 😊
Dang, now i feel better bombing vector calculus knowing that Dr. peyam struggled with stoke’s theorem too lol
so funny and amazing to follow😇
thanks
I actually used the dark side more than the other side
I don't believe that you can have any arbitrary surface. Your surface can't extend beyond the outermost line curve defined by dropping perpendicular down from every point on the surface.
You can integrate over the circular area or over the hemisphere, correct? Which one will be simpler? (Asked very early in viewing, prior to computation of the determinant)... compute one component & permute the variables
Please is there any where you can make your lessons a bit simpler? Although I love your lessons but I usually get lost at some point
That’s the simplest way to present this topic. Also check out the playlist
@@drpeyam Thank Dr Peyam..I am hoping to be as good as you someday in mathe😩
धन्यवाद ।
Que el redultado sea cero, no significa que no sea interesante.
The solution is zero does not mean it is not interesting!
I think I had this excact ssme problem on my calc 3 final
Im cooked😢
Turns out manifolds and exterior derivatives are important
Bye
easy of hard = hard of easy
Hi
Stokes ist der beste Freund von Tom crawford . Den hättest du einladen müssen
GOAT
Ok, do you think you'd be able to prove it or umprove it?
wow is very simple video...thanks...F(x,y,z)=z^2 i+2xj+y^2 k
S:z=1-x^2-y^2,z≥0
This is too dark
Horse shit, a length is not equal to a surface area. Why do you omit the units of the integrals?