ques

- anonymous

ques

- Stacey Warren - Expert brainly.com

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- schrodinger

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- anonymous

what is the question

- anonymous

From 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

C is the closed curve which is the boundary of that surface
|dw:1443886311804:dw|

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## More answers

- anonymous

It isn't really enclosing the surface though??

- anonymous

Oh, it must be bounded by C that's the only condition

- anonymous

Suppose 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

the line integral and the surface integral are the same, that is what Stokes' theorem says

- IrishBoy123

you know the [deformable] butterfly net analogy?

- anonymous

Nope, but is my step correct for finding the surface integral of curl F through the whole surface(including circular top) |dw:1443887053149:dw|

- IrishBoy123

yes
but |dw:1443887366725:dw|

- anonymous

Hm, 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

and 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

Wait does that mean for total surface we have,
\[2 \times \oint_\limits{C} \vec F . d \vec r\]
??

- IrishBoy123

at the risk of sounding agricultural, it comes in the hoop and goes out the net.

- IrishBoy123

i can post an example that can be done several ways

- anonymous

So it will be 2 times the line integral??? for total surface??

- anonymous

If it's equal for the surfaces, since they have the same C

- anonymous

then it just becomes 2 times for total surface?

- IrishBoy123

no, they should net out to zero
so take a sphere
|dw:1443888358128:dw|

- IrishBoy123

you should be able to do both surfaces using the loop so they are the same but they cancel out
|dw:1443888386190:dw||dw:1443888398064:dw|

- anonymous

|dw:1443888517480:dw|

- anonymous

Applying 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

Why do they cancel out instead of adding up??

- IrishBoy123

because 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

Hmmm, I remember something similar from school
\[e.m.f=-\frac{d \Phi_{B}}{dt}\]
Faraday's law of electromagnetic induction

- anonymous

Your equation looks different

- IrishBoy123

yours 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

scroll down here to equation 7
http://www.maxwells-equations.com/faraday/faradays-law.php

- anonymous

Shouldn't it be a double integral on the right side of equation 7

- IrishBoy123

there'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

Oh,
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

Electromagnetism 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

that's pretty good!

- anonymous

I 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

try 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 do-able in a few ways, so might be interesting.

##### 1 Attachment

- IrishBoy123

oh, i mean example 1

- anonymous

\[\vec V=4y \hat i+x \hat j+2z \hat k\]
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
\[\iint_{S}=\iint_{R}(\vec \nabla \times \vec V).\hat n \frac{dxdy}{\hat n . \hat k}\]
\[\vec \nabla \times \vec V=-3 \hat k\]
\[\hat n=\frac{x \hat i+y \hat j+z \hat k}{\sqrt{x^2+y^2+z^2}}\]\[(\vec \nabla \times \vec V).\hat n=-\frac{3z}{\sqrt{x^2+y^2+z^2}}\]\[\frac{1}{\hat n . \hat k}=\frac{\sqrt{x^2+y^2+z^2}}{z}\]
\[\iint_\limits{S}=\iint_\limits{R}-\frac{3z}{\sqrt{x^2+y^2+z^2}}.\frac{\sqrt{x^2+y^2+z^2}}{z}dxdy=-3\iint_\limits{R}dxdy=-3\pi a^2\]

- anonymous

@IrishBoy123

- IrishBoy123

yes!
i think this is a really good book, you will know the maths already:
http://www.amazon.co.uk/A-Students-Guide-Maxwells-Equations/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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