Stokes Theorem in cylindrical coordinates

In summary: Thanks for pointing it out!In summary, SDas was attempting to solve a problem involving a vector field in cylindrical coordinates and the line and surface integrals over a circle of radius ρ. However, upon closer observation, they realized that the field A did not satisfy the necessary conditions for Stokes' theorem to hold. This led to a discrepancy between the line and surface integrals, with an additional term appearing in the latter due to the discontinuity of the field at φ=2π.
  • #1
SDas
3
0

Homework Statement



A vector field A is in cylindrical coordinates is given.

A circle S of radius ρ is defined.

The line integral [tex]\int[/tex]A∙dl and the surface integral [tex]\int[/tex]∇×A.dS are different.

Homework Equations



Field: A = ρcos(φ/2)uρ2 sin(φ/4) uφ+(1+z)uz (1)

The Attempt at a Solution



The line integal of A over the circumference of the circle S is

[tex]\int[/tex]Adl =[tex]\int_0^{2\pi}[/tex]ρ3sin(φ/4)dφ = 4ρ3 (2)

The surface integral over the area of the circle is

[tex]\int[/tex]∇×A.dS = [tex]\int_0^{2\pi} \int_0^\rho[/tex](3ρsin(φ/4)+(1/2)sin(φ/2))ρdρdφ = [tex]\int_0^{2\pi}[/tex](ρ3sin(φ/4)+ρ2/4 sin(φ/2))dφ=4ρ32 (3)

Observing more closely, in the surface integral (3), we get an additional term [tex]\int_0^{2\pi}[/tex](ρ2/4sin(φ/2))dφ which is not present in the line integral (2).
This gives rise to the second term ρ2 in (3) after the integration.
What is the reason this new term appears in the surface integral, but not the line integral?
 
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  • #2
Hi SDas, welcome to PF!:smile:

Hint: what are the conditions [itex]\textbf{A}[/itex] must satisfy for Stokes theorem to be true on a given region?:wink:
 
  • #3
Hi gabbagabbahey, thanks for the welcome.

I've been trying to check the conditions for Stoke's theorem to hold. I could not find them in any textbook on electromagnetics. Intuitively, the field A should
(i) be bounded, and
(ii) have continuous, first order partial derivatives w.r.t. ρ, φ, and z.
Am I missing something?
 
  • #4
Is your field even continuous? What happens when the angular coordinate passes through 2pi?
 
  • #5
Got it! Of course it isn't. I should not have missed that. The sin(φ/4) and the cos(φ/2) should have been sin(4φ) and cos(2φ) to make A be the same at φ=0 and φ=2π. That was the mistake in what I was trying to do.
 

FAQ: Stokes Theorem in cylindrical coordinates

What is Stokes Theorem in cylindrical coordinates?

Stokes Theorem is a mathematical theorem that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field along the boundary of the surface. In cylindrical coordinates, it is used to calculate the circulation of a vector field around a curve that lies on a cylindrical surface.

How is Stokes Theorem related to the curl of a vector field?

Stokes Theorem states that the surface integral of the curl of a vector field over a closed surface is equal to the line integral of the vector field along the boundary of the surface. This means that the circulation of a vector field around a closed curve is equal to the sum of the curls of the vector field over any surface bounded by the curve.

What are the applications of Stokes Theorem in cylindrical coordinates?

Stokes Theorem is used in many areas of physics and engineering, such as fluid dynamics, electromagnetism, and heat transfer. It is also used in mathematical modeling and numerical analysis to solve complex problems involving vector fields in cylindrical coordinates.

How is Stokes Theorem derived in cylindrical coordinates?

Stokes Theorem can be derived using the Divergence Theorem and the fundamental theorem of calculus. In cylindrical coordinates, the divergence theorem is used to convert a surface integral to a volume integral, and then the fundamental theorem of calculus is used to convert the volume integral to a line integral, resulting in the formulation of Stokes Theorem.

What are the limitations of Stokes Theorem in cylindrical coordinates?

Stokes Theorem only applies to smooth surfaces and vector fields. It also assumes that the surface and curve are orientable, which means they have a consistent direction along their boundaries. Additionally, the curve must be closed and the surface must be bounded by the curve in order for Stokes Theorem to be applicable.

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