We use stokes and greens theorem a lot in electromagnetism. Another way to determine orientation is using right hand rule. Point your right thumb in the direction of the outward normal vector, which ever direction your fingers curl in is the direction you integrate.
dr peyam, although i've convinced myself that your answer to the given problem is indeed correct, i believe, that your reasoning concerning the normal vectors of the barrel is not. specifically, since we're told that |z| < 4, the barrel has no lids, and therefore there should be no top lid and bottom lid normal vectors, only the side ones. for if there were, the surface would become closed and by theorem the curl of the vector field over this surface would have to be zero for the following reason: we could cut this surface anywhere and anyhow with a plane obtaining two surfaces with a common boundary, but with the opposite orientation of this boundary resulting in them cancelling each other out.
I've been brushing up on my vector calculus recently and this was a great practice question for Stokes' Theorem! Interestingly, when deciding the orientation of the top and bottom paths, the way I pictured it was making a cut vertically along the side of the barrel. You then pull it apart an infinitesimal distance, giving you essentially a rolled up rectangle as the path C. You can then very easily see the direction of the paths using your method or the RH rule. Since you only pulled the two cut edges ever so slightly apart, the paths along these two edges have equal and opposite contributions to the total line integral, so cancel out to leave only the top and bottom rings with the correct orientation.
Wow, I think I discovered something extraordinary. You can write cos(x) as 1/2 [e^ix + e^-ix] But one solution of this differential equation f(x)² - f'(x)² = 1 is: f(x) = 1/2 [e^x + e^-x] Without 'i' in the exponent! The interesting thing is, like sin(x) and cos(x) are related over derivatives(slope)/integrals(area), this function is related over the length of its graph. But I'm just hobby-mathematician. I don't have the education/skills to explore such things more precise. Sorry for my bad english. Greetings from Germany.
This can be done even easier. F = . Curl(F) = . To find the flux of the curl through the barrel (including top and bottom) we use divergence theorem: Div(Curl(F)) = 1 + 1 - 2 = 0, so the flux is zero! Now, to find the flux through the side of the barrel, we need to subtract the flux through the top and the bottom from 0. Curl(Top) = , the normal to the top is , the area is pi*3^2, so top flux is -8*1*9*pi = -72pi. Similarly for the bottom. So the answer is 0 - (-72pi) - (-72pi) = 144pi.
Hello, nice video! Just a question since im dumb: I don't really understand the part in which you specify orientation, shouldn't the direction be counter-clockwise for c1 (upper part) and clockwise for c2 (lower part)? Would the orientation of c1 be counter-clockwise if the bottom part was completely closed? Because to me it still feels like the surface is to the right if you move clockwise on c1. So in short, I guess I don't understand why the right-hand rule fails to determine the direction in which the man travels around the boundary.
i think the more fitting surface to use instead would be an Ellipsoid that is stretched over the z axis so it does't look so oval, great video overall though, recently finished my calc 3 class and i understand this.
Dr Peyam would u mind showing us how to find a tangent plane to a 3D surface using vectors and partial derivatives please?? Keep up the good videos and thanks so much for the help!!
Hi, it's provable if you know that you can factorise (1-x^n) into (1-x) \times \sum_{k=0}^{n-1} 1^{n-k} x^k and then it's not so hard calculus. I can write it in LaTeX if you wanna take a look at the proof ^^. Hoping this was useful :), voilà. P.S : I'm just using the fact that you can factorise (a^n-b^n) into (a-b) \times \sum_{k=0}^{n-1} a^{n-k} b^k
You mean Hn are the partial sums of the harmonic series. This is pretty easy if you know that (x^n - 1)/(x -1) = x^(n-1) + x^(n-2) + ... + 1 (You can calculate (x -1)*(x^(n-1) + x^(n-2) + ... 1) and that becomes a telescoping sum) Now the antiderivative is just 1/n*x^n + 1/(n-1)*x^(n-1) ... + 1*x + c Evaluating at zero gives just c and evaluating at 1 gives 1/n + 1/(n-1) + ... + 1 + c which is just Hn + c So your integral is Hn
We use stokes and greens theorem a lot in electromagnetism. Another way to determine orientation is using right hand rule. Point your right thumb in the direction of the outward normal vector, which ever direction your fingers curl in is the direction you integrate.
dr peyam, although i've convinced myself that your answer to the given problem is indeed correct, i believe, that your reasoning concerning the normal vectors of the barrel is not. specifically, since we're told that |z| < 4, the barrel has no lids, and therefore there should be no top lid and bottom lid normal vectors, only the side ones. for if there were, the surface would become closed and by theorem the curl of the vector field over this surface would have to be zero for the following reason: we could cut this surface anywhere and anyhow with a plane obtaining two surfaces with a common boundary, but with the opposite orientation of this boundary resulting in them cancelling each other out.
I've been brushing up on my vector calculus recently and this was a great practice question for Stokes' Theorem!
Interestingly, when deciding the orientation of the top and bottom paths, the way I pictured it was making a cut vertically along the side of the barrel. You then pull it apart an infinitesimal distance, giving you essentially a rolled up rectangle as the path C. You can then very easily see the direction of the paths using your method or the RH rule. Since you only pulled the two cut edges ever so slightly apart, the paths along these two edges have equal and opposite contributions to the total line integral, so cancel out to leave only the top and bottom rings with the correct orientation.
Wow, I think I discovered something extraordinary. You can write cos(x) as 1/2 [e^ix + e^-ix]
But one solution of this differential equation f(x)² - f'(x)² = 1 is: f(x) = 1/2 [e^x + e^-x]
Without 'i' in the exponent! The interesting thing is, like sin(x) and cos(x) are related over derivatives(slope)/integrals(area), this function is related over the length of its graph.
But I'm just hobby-mathematician. I don't have the education/skills to explore such things more precise. Sorry for my bad english. Greetings from Germany.
This can be done even easier. F = . Curl(F) = .
To find the flux of the curl through the barrel (including top and bottom) we use divergence theorem:
Div(Curl(F)) = 1 + 1 - 2 = 0, so the flux is zero! Now, to find the flux through the side of the barrel,
we need to subtract the flux through the top and the bottom from 0. Curl(Top) = ,
the normal to the top is , the area is pi*3^2, so top flux is -8*1*9*pi = -72pi. Similarly for the bottom.
So the answer is 0 - (-72pi) - (-72pi) = 144pi.
Ganz toll Dr. p! Jetzt könnnen wir uns auf Deutsch verständigen!! Ich freue mich immerauf diese Serie
Stokes' and Green's are my favorite part of multivariable. Brings back great memories of my brain becoming jello.
Hello Dr.Peyam, watching here from Australia.
great warrick
Hello, nice video! Just a question since im dumb:
I don't really understand the part in which you specify orientation, shouldn't the direction be counter-clockwise for c1 (upper part) and clockwise for c2 (lower part)? Would the orientation of c1 be counter-clockwise if the bottom part was completely closed? Because to me it still feels like the surface is to the right if you move clockwise on c1. So in short, I guess I don't understand why the right-hand rule fails to determine the direction in which the man travels around the boundary.
The surface has to be to your left when you walk around it, that determines the orientation, not the right hand rule
8:52 The direction of the arrows should be reverse??? considering the guy’s left hand?
Oh, the surface is S, not C.
Thank you!!
nicely done- had to go back and rewatch, because my mind went straight to cylinder+napkin ring, forgot all about F
I’d like to see this problem solved the “silly” way ;)
Berkeley still hosts some Math 53 past papers on their Math Department site. What year did you take your Math 53 final exam?
2006 :)
Doing this integral is like doing a barrel roll.
Video on generalized stokes theorem?
Hi Peyam, can you please explain why to 'walk left' while choosing orientation?
Because we need a positive orientation, and that’s just a convention so that everyone has the same answer
@@drpeyam I think it is like a right hand rule, sorry to all the left-handers including Dr Peyam :)
Your chalk writes invisible letters.Thank you.
i think the more fitting surface to use instead would be an Ellipsoid that is stretched over the z axis so it does't look so oval, great video overall though, recently finished my calc 3 class and i understand this.
Dr Peyam would u mind showing us how to find a tangent plane to a 3D surface using vectors and partial derivatives please?? Keep up the good videos and thanks so much for the help!!
Great idea!
Dr.payem will u plizz solve this question Hn=integral from 0 to 1 of (1-xⁿ)/(1-x)
Hi, it's provable if you know that you can factorise (1-x^n) into (1-x) \times \sum_{k=0}^{n-1} 1^{n-k} x^k and then it's not so hard calculus. I can write it in LaTeX if you wanna take a look at the proof ^^. Hoping this was useful :), voilà.
P.S : I'm just using the fact that you can factorise (a^n-b^n) into (a-b) \times \sum_{k=0}^{n-1} a^{n-k} b^k
You mean Hn are the partial sums of the harmonic series.
This is pretty easy if you know that
(x^n - 1)/(x -1) = x^(n-1) + x^(n-2) + ... + 1
(You can calculate (x -1)*(x^(n-1) + x^(n-2) + ... 1) and that becomes a telescoping sum)
Now the antiderivative is just 1/n*x^n + 1/(n-1)*x^(n-1) ... + 1*x + c
Evaluating at zero gives just c and evaluating at 1 gives 1/n + 1/(n-1) + ... + 1 + c which is just Hn + c
So your integral is Hn
GEORGE EULER BAITE notice that: 1 + x + x^2 + ... + x^n-1= (1-x^n)/(1-x) if you now Integrate both sides from 0 to 1 you get your identity.
Thank you sir
In physics class, we remembered the orientation as being another part of the "right hand rule".
Yeah this mountain stuff is so weird. If your thumb is along the normal, it moves in the direction you curl your fingers
@@madhuragrawal5685 which hand?
@@alvinlepik5265 the right, like always
72pi + 72pi?? thats a gross pi
Where Hn=harmonic series