Stokes Theorem in cylindrical coordinates

[SDas]

Homework Statement

A vector field A is in cylindrical coordinates is given.

A circle S of radius ρ is defined.

The line integral \intA∙dl and the surface integral \int∇×A.dS are different.

Homework Equations

Field: A = ρcos(φ/2)u_ρ+ρ² sin(φ/4) u_φ+(1+z)u_z (1)

The Attempt at a Solution

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

\intA∙dl =\int_0^{2\pi}ρ³sin(φ/4)dφ = 4ρ³ (2)

The surface integral over the area of the circle is

\int∇×A.dS = \int_0^{2\pi} \int_0^\rho(3ρsin(φ/4)+(1/2)sin(φ/2))ρdρdφ = \int_0^{2\pi}(ρ³sin(φ/4)+ρ²/4 sin(φ/2))dφ=4ρ³+ρ² (3)

Observing more closely, in the surface integral (3), we get an additional term \int_0^{2\pi}(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?

 

[gabbagabbahey]

Hi SDas, welcome to PF!

Hint: what are the conditions \textbf{A} must satisfy for Stokes theorem to be true on a given region?

 

[SDas]

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?

 

[Dick]

Is your field even continuous? What happens when the angular coordinate passes through 2pi?

 

[SDas]

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.