Stokes theorem in a cylindrical co-ordinates, vector field

[Ratpigeon]

Homework Statement

given a vector field v[/B=]Kθ/s θ (which is a two dimensional vector field in the direction of the angle, θ with a distance s from the origin) find the curl of the field and verify stokes theorem applies to this field, using a circle of radius R around the origin

Homework Equations

Stokes Theorem is:

[itex]\int[/itex]∇×v.da=[itex]\oint[/itex]v . dl
and the curl in cylindrical co-ordinates is:
1/s (∂vz/∂θ-∂vθ/∂z) s+(∂vs/∂z-∂vz/∂s) θ+1/s(∂/∂s (s vθ)-∂vs/∂θ) z
Where vz=0; vs=0; vθ=kθ/s

The Attempt at a Solution

IN cylindrical co-ords; dl=ds s +s dθ θ+dz z

The line integral is hence equal to

∫kθdθ with θ runing from 0 to 2[itex]\pi[/itex]

Which has a solution of 2k [itex]\pi[/itex]²
However, the curl is zero except for at the centre, where 1/s goes to infinity; so the integral on the other side has a delta function, and the integral will come out at 2k[itex]\pi[/itex] ²meaning the integral will be something like:

∫∫∂(s)kθdθds which evaluates to 2k [itex]\pi[/itex]² as required;

But I'm not sure how to get that integral...

 

[anathema]

To find the curl of the vector field, we can use the formula given in the problem:

curl(v) = 1/s (∂vz/∂θ-∂vθ/∂z) s+(∂vs/∂z-∂vz/∂s) θ+1/s(∂/∂s (s vθ)-∂vs/∂θ) z

Since vz=0 and vs=0, the formula simplifies to:

curl(v) = 1/s (∂vθ/∂z) s + 1/s(∂/∂s (s vθ)-∂vs/∂θ) z

Substituting in the given values of vθ=kθ/s, we get:

curl(v) = 1/s (∂/∂z (kθ)-∂kθ/∂θ) s + 1/s(∂/∂s (s kθ)-∂(0)/∂θ) z

= 1/s (0-0) s + 1/s(kθ-0) z

= kθ/z

Now, to verify Stokes' Theorem, we need to calculate the line integral of the vector field around a circle of radius R around the origin. Using the formula given in the problem for dl, we get:

dl = Rdθ θ + Rdz z

The line integral can be written as:

∫v . dl = ∫kθ/s (Rdθ θ + Rdz z)

= ∫kR dθ

= kRθ|θ=0 to θ=2π

= 2kπ

This is equal to the integral of the curl of the vector field over the surface enclosed by the circle, which is:

∫∫curl(v) . da = ∫∫kθ/z (Rdz dz dθ)

= kRθ|θ=0 to θ=2π

= 2kπ

Thus, we have verified that Stokes' Theorem holds for this vector field.