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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0From Stoke's Theorem we have \[\iint_\limits S (\vec \nabla \times \vec F). \hat n \space ds=\oint_\limits C \vec F.d \vec r\] Where S is any surface and C is a closed curve around the surface I can see how this works out in the case of a 2D surface dw:1443885818098:dw But if we consider the surface of a 3D object, we have dw:1443885892309:dw Here S is the lateral surface of that figure What do we take for our C here??

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1C is the closed curve which is the boundary of that surface dw:1443886311804:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It isn't really enclosing the surface though??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, it must be bounded by C that's the only condition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Suppose we want to find the surface integral of curl F both through the lateral surface and the circular surface, Would that be done something like.. \[\oint_\limits C \vec F.d \vec r+\iint_\limits R (\vec \nabla \times \vec F).\hat k \space dxdy\]dw:1443886644120:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3the line integral and the surface integral are the same, that is what Stokes' theorem says

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3you know the [deformable] butterfly net analogy?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope, but is my step correct for finding the surface integral of curl F through the whole surface(including circular top) dw:1443887053149:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3yes but dw:1443887366725:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hm, that's easy to understand, because the curves bounding these surfaces are the same, so the line integral through these curves should be the same for a particular vector field!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3and the flat net enclosed by the ring = the puffed out nets because all are enclosed by same ring.....and covered by the same line integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wait does that mean for total surface we have, \[2 \times \oint_\limits{C} \vec F . d \vec r\] ??

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3at the risk of sounding agricultural, it comes in the hoop and goes out the net.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3i can post an example that can be done several ways

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it will be 2 times the line integral??? for total surface??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If it's equal for the surfaces, since they have the same C

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then it just becomes 2 times for total surface?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3no, they should net out to zero so take a sphere dw:1443888358128:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3you should be able to do both surfaces using the loop so they are the same but they cancel out dw:1443888386190:dwdw:1443888398064:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443888517480:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Applying stokes theorem on R and S, they convert to line integrals of the same curve, so just 2 times the line integral for total??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why do they cancel out instead of adding up??

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3because in the physical sense you are typically looking at the flux (like flow) of the curl of some field. so say from Maxwell you have \(\nabla \times \mathbf{E} = \frac{\partial \mathbf{B}} {\partial t}\) relating the curl of an eletric field with a magnetic field and its flux.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm, I remember something similar from school \[e.m.f=\frac{d \Phi_{B}}{dt}\] Faraday's law of electromagnetic induction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your equation looks different

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3yours is faraday's in simplest form  induced voltage from a changing magnetic field i quoted maxwell #3 in its differential form which is the basis of what you quoted my point though is that when we look at surfaces we are usually concerned with the rate at which something is flowing through it, be it a fluid or a magnetic field.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3scroll down here to equation 7 http://www.maxwellsequations.com/faraday/faradayslaw.php

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Shouldn't it be a double integral on the right side of equation 7

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3there's a lot of variation in notation. i think the \(\int_S \dots dS \) is enough for some people to signify a double integral

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, cuz I thought about it like this \[\vec \nabla \times \vec E=\frac{\partial \vec B}{\partial t}\] \[(\vec \nabla \times \vec E).d \vec A=\frac{\partial \vec B}{\partial t}.d \vec A\]\[\iint (\vec \nabla \times \vec E). \space d \vec A=\iint \frac{\partial \vec B}{\partial t}. \space d \vec A\]\[\oint \vec E.d \vec l=\iint \frac{\partial \vec B}{\partial t}.d \vec A\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Electromagnetism is fascinating, I can't wait to study it in later semesters, the moment when you realize ahh you can apply the math you've learned so beautifully

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3that's pretty good!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I'll leave discussion about finding total flux over that surface, it still looks a bit confusing and I doubt I'd use it right now, most of the problems require me to find the flux over the lateral surfaces (they mainly want to see how u can apply the theorems etc.) so my exams will be more theorems(statements) and problems based. My main doubt was how to find C for a 3D figure I quote from my syllabus "The emphasis of course is on applications in solving problems of interest to physicists. The students are to be examined entirely on the basis of problems, seen and unseen."

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3try this. i could really do with the practise too but i've got a lot of rugby to watch this evening:) so i'll have a go later.... it is set up to be doable in a few ways, so might be interesting.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3oh, i mean example 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\vec V=4y \hat i+x \hat j+2z \hat k\] S is \[x^2+y^2+z^2=a^2\] \[z \ge 0\] dw:1443891158189:dw Hmm one way we can do is convert into line integral using stokes' theorem! \[\iint_\limits{S}(\vec \nabla \times \vec V).\hat n \space d \sigma=\oint_\limits C \vec V.d \vec r\] C is the circular loop in the xy plane therefore z=0, dz=0 \[\iint_\limits S=\oint_\limits{C}4ydx+xdy=\int\limits_{0}^{2\pi} (4a^2\sin^2(\theta)+a^2\cos^2(\theta)) d \theta\]\[\iint_\limits{S}=\frac{a^2}{2}\int\limits_{0}^{2\pi}(5\cos(2\theta)3)d \theta=\frac{a^2}{2}[\frac{5\sin(2 \theta)}{2}3 \theta]_{0}^{2\pi}=3 \pi a^2\] 2nd we can find curl then take the projection on xy plane \[\iint_{S}=\iint_{R}(\vec \nabla \times \vec V).\hat n \frac{dxdy}{\hat n . \hat k}\] \[\vec \nabla \times \vec V=3 \hat k\] \[\hat n=\frac{x \hat i+y \hat j+z \hat k}{\sqrt{x^2+y^2+z^2}}\]\[(\vec \nabla \times \vec V).\hat n=\frac{3z}{\sqrt{x^2+y^2+z^2}}\]\[\frac{1}{\hat n . \hat k}=\frac{\sqrt{x^2+y^2+z^2}}{z}\] \[\iint_\limits{S}=\iint_\limits{R}\frac{3z}{\sqrt{x^2+y^2+z^2}}.\frac{\sqrt{x^2+y^2+z^2}}{z}dxdy=3\iint_\limits{R}dxdy=3\pi a^2\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.3yes! i think this is a really good book, you will know the maths already: http://www.amazon.co.uk/AStudentsGuideMaxwellsEquations/dp/0521701473 there is a pdf somewhere on the net so you can try it first and see if you like it. by the time you finish it you will see how Maxwell worked out the speed of light whilst sitting at his desk doing some maths! amazing stuff really.
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