# Verification of Stoke's Theorem for a Cylinder

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1. Oct 30, 2014

### NuclearMeerkat

1. The problem statement, all variables and given/known data

2. Relevant equations
Stoke's Theorem:

3. The attempt at a solution
∇×A = (3x,-y,-2(z+y))

I have parametric equation for wall and bottom:
Wall: x(θ,z) = acosθ ; y(θ,z) = asinθ ; z(θ,z) = z [0≤θ≤2π],[0≤z≤h]
Bottom: x(θ,r) = rcosθ ; y(θ,r) = rsinθ ; z(θ,r) = 0 [0≤θ≤2π],[0≤r≤a]

And then I get lost, other examples I have tried to understand the method from are going straight over my head, any help much appreciated.

2. Oct 30, 2014

### vela

Staff Emeritus
Take a stab at writing down the integral on the lefthand side. Surely you can at least write down $\nabla\times\vec{A}$ in terms of the cylindrical coordinates now.

3. Oct 30, 2014

### NuclearMeerkat

∇×A in cylindrical coordinates is (3acosθ,-asinθ,-2(z+asinθ)) and then I integrate w.r.t. θ and z?

4. Oct 30, 2014

### vela

Staff Emeritus
It depends on which surface you're integrating over. The integral over $\Sigma$ will be the sum of the integral over the bottom and the integral over the wall.

5. Oct 30, 2014

### LCKurtz

For the lateral side you have $\vec R(\theta,z) = \langle a\cos\theta,a\sin\theta,z\rangle$. Now you need to calculate$$d\Sigma=\pm \vec R_z\times \vec R_\theta$$with the sign chosen to agree with the orientation. Then you will be ready to integrate$$\iint_\Sigma \nabla \times \vec F\cdot d\Sigma$$

6. Oct 31, 2014

### NuclearMeerkat

Rz = (0,0,1) and Rθ = (-asinθ,acosθ,0)
use -Rz because it the bottom face

giving dΣ = (-acosθ,asinθ,0)

therefore the integral is ∫dz (3x,-y,-2(z+y))⋅(-acosθ,asinθ,0)

7. Oct 31, 2014

### LCKurtz

You have two surfaces, the lateral surface and the bottom face. Don't mix them up or combine them.

I assume you are doing the lateral surface here. Your $d\Sigma$ has a sign error and it needs $d\theta dz$. Also your integral must be in terms of those variables.