Line Integral - Stokes theorem

In summary, Alex was given a vector field and had to integrate it over a circle. He used Cartesian coordinates to calculate the line integral and then changed to polar coordinates when doing the integration. He found that the curl is ##\nabla\times\vec{A} = 4r^2 \hat{k}##.
  • #1
AwesomeTrains
116
3

Homework Statement


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?

Homework Equations


Stokes equation:
5f65e93751487f9350c194aa5f2bb8de.png


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
 
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  • #2
AwesomeTrains said:
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.
 
  • #3
Okay, thanks for the reply :)
 
  • #4
I got 4πR4.

del x A = 4R2 k
dA = R2/2 k
∫2R2R2dθ from 0 to 2π = 4πR4.
 
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  • #5
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]
 
  • #6
AwesomeTrains said:
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.
 
  • #7
rude man said:
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.
 
  • #8
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]
 
Last edited:
  • #9
I agree, I should not have made r constant in the curl expression.
 
  • #10
Is [itex] 2R^4\pi [/itex] then the correct result?
 
  • #11
Yes.
 
  • #12
AwesomeTrains said:
Is [itex] 2R^4\pi [/itex] then the correct result?
Yes, I say it is, and you were right to use r dr dθ as the elemental area.
 
  • #13
Okay, thanks for the help, will be back with some more problems :)
 
  • #14
AwesomeTrains said:
Okay, thanks for the help, will be back with some more problems :)
Please do! I'll try not to mess up next time ... o:)
 

1. What is the Line Integral - Stokes theorem?

The Line Integral - Stokes theorem is a mathematical theorem that describes the relationship between a line integral of a vector field over a closed curve and a surface integral of the vector field over the surface bounded by the curve. It is a fundamental tool in vector calculus and is often used to solve problems in electromagnetism, fluid mechanics, and other fields.

2. How is the Line Integral - Stokes theorem different from the Fundamental Theorem of Calculus?

The Fundamental Theorem of Calculus is used to evaluate the definite integral of a function over an interval, while the Line Integral - Stokes theorem is used to relate the line integral of a vector field to the surface integral of the same vector field. In other words, the Fundamental Theorem of Calculus deals with one-dimensional integrals, while the Line Integral - Stokes theorem deals with two-dimensional integrals.

3. What are the applications of the Line Integral - Stokes theorem?

The Line Integral - Stokes theorem has many applications in physics, engineering, and other fields. It is used to calculate work done by a force, fluid flow through a surface, and circulation of a vector field. It is also used in the study of electromagnetic fields and fluid mechanics.

4. How is the Line Integral - Stokes theorem related to Green's theorem?

Green's theorem is a special case of the Line Integral - Stokes theorem for two-dimensional vector fields in the plane. Green's theorem relates a line integral around a closed curve to a double integral over the region enclosed by the curve. The Line Integral - Stokes theorem extends this concept to three-dimensional vector fields and surfaces.

5. What are some common misunderstandings about the Line Integral - Stokes theorem?

One common misunderstanding about the Line Integral - Stokes theorem is that it only applies to conservative vector fields. In reality, the theorem can be applied to any continuous vector field. Another misconception is that the curve and surface in the theorem have to be simple and smooth, when in fact they can be quite complex. Additionally, the orientation of the curve and surface must be taken into account when applying the theorem, which can be confusing for some students.

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