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Line Integral - Stokes theorem

AwesomeTrains

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1. The problem statement, all variables and given/known data
Hello
I was given the vector field: [itex]\vec A (\vec r) =(−y(x^2+y^2),x(x^2+y^2),xyz)[/itex] and had to calculate the line integral [itex] \oint \vec A \cdot d \vec r [/itex] over a circle centered at the origin in the [itex]xy[/itex]-plane, with radius [itex] R [/itex], by using the theorem of Stokes.

Another thing, when calculating [itex]∇× \vec A [/itex] I used Cartesian coordinates but when doing the integration I changed to polar coordinates, is that okay? Or do I have to do it all in Cartesian coordinates, because I did curl in Cartesian coordinates?

2. Relevant equations
Stokes equation:
5f65e93751487f9350c194aa5f2bb8de.png


3. The attempt at a solution
My result is: [itex]2R^4π[/itex] Is that correct ? I evaluated both sides and got the same.

Kind regards Alex
 

vela

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Another thing, when calculating [itex]∇× \vec A [/itex] I used Cartesian coordinates but when doing the integration I changed to polar coordinates, is that okay? Or do I have to do it all in Cartesian coordinates, because I did curl in Cartesian coordinates?
It's fine to change coordinates to make evaluating the integral easier whenever you want.

My result is: [itex]2R^4π[/itex] Is that correct ? I evaluated both sides and got the same.
I didn't actually do the integral on paper so I might have missed something, but the answer looks right to me.
 

AwesomeTrains

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Okay, thanks for the reply :)
 

rude man

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I got 4πR4.

del x A = 4R2 k
dA = R2/2 k
∫2R2R2dθ from 0 to 2π = 4πR4.
 
Last edited:

AwesomeTrains

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Thanks for doing the calculation but why is [itex] dA=\frac{R^2}{2} d \theta [/itex]. My curl vector is the same.
I was using [itex]dxdy=rdrd \theta[/itex]
 

rude man

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Thanks for doing the calculation but why is [itex] dA=\frac{R^2}{2} d \theta [/itex]. My curl vector is the same.
I was using [itex]dxdy=rdrd \theta[/itex]
I took the differential area as that of a triangle of area 1/2 bh = 1/2 R Rrdθ = 1/2 R2
b = base
h = height
Proof: A = ∫dA = 1/2 R2∫dθ from 0 to 2π = πR2. :smile:

You can do it your way too! Just requires integrating twice instead of once:
dA = r dr dθ
A = ∫∫r dr dθ= R2/2 (2θ) = πR2.
So how about showing how you got 2πR4? One of us must have stumbled a bit in the dA department.
 

vela

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I got 4πR4.

del x A = 4R2 k
dA = R2/2 k
∫2R2R2dθ from 0 to 2π = 4πR4.
The curl is ##\nabla\times\vec{A} = 4r^2 \hat{k}##. It depends on variable ##r##, not constant ##R##, so you can't use that area element.
 

AwesomeTrains

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Thanks for the replies guys. I'll write out my calculation then you can see where the mistake lies.
[itex] \nabla \times \vec A = (xz - 0)i + (0 - yz)j + (3x^2+y^2 - (-x^2 - 3y^2))k [/itex]
Chose a vektor normal to the circle:
[itex] d \vec f = 0i+0j+k [/itex]
The dot product is then:
[itex] \nabla x \vec A \cdot d \vec f = 4x^2+4y^2 = 4r^2[/itex]
Then I integrated over the circle, and changed to spherical coordinates where I used the area element: [itex]dA =rdrd \phi[/itex]:
[itex] \int_{0}^{2\pi} \int_{0}^R 4r^2rdrd \phi = \int_{0}^{2Pi} \int_{0}^R 4r^3drd \phi = 2R^4\pi[/itex]
 
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rude man

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I agree, I should not have made r constant in the curl expression.
 

AwesomeTrains

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Is [itex] 2R^4\pi [/itex] then the correct result?
 

vela

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Yes.
 

AwesomeTrains

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Okay, thanks for the help, will be back with some more problems :)
 

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