- #1

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Verify Stokes' theorem by explicit calcluation where S is the circle of radius a in the z = 0 plane.

You may like to employ the ideas that [tex]\vec{dS} = \rho d\phi dz \hat{\rho} + dz d\rho \hat{\phi} + \rho d\rho d\phi \hat{z}[/tex] and [tex]\vec{dl} = d\rho \hat{\rho} + \rho d\phi \hat{\phi} + dz \hat{z}[/tex]

Stokes' theorem: [tex]\int_{S} \vec{dS} \cdot (\nabla \times \vec{A}) = \int_{C} \vec{dl} \cdot \vec{A}[/tex]

I found [tex]\nabla \times \vec{A} = \cos \phi \hat{\rho} - \sin \phi \hat{\phi} + 3\rho^2 \hat{z}[/tex]

So, [tex]\vec{dS} \cdot (\nabla \times \vec{A}) = \cos \phi \rho d\phi dz - \sin \phi dz d\rho + 3\rho^2 d\rho d\phi[/tex]

Now, I said dz = 0 because the circle is in the z = 0 plane, so z = constant and dz = 0. Is that right?

[tex]\int_{S} \vec{dS} \cdot (\nabla \times \vec{A}) = \int_0^{2\pi} d\phi \int_0^a 3\rho^2 d\rho = 2\pi a^3[/tex]

Now, [tex]\vec{dl} \cdot \vec{A} = \rho^3 d\phi[/tex] (since there's no [itex]\hat{\rho}[/itex] component and again dz = 0)

[tex]\int_C \vec{dl} \cdot \vec{A} = \int_0^{2\pi} \rho^3 d\phi = [\rho^3 \phi]_{0}^{2\pi} = 2\pi \rho^3 = 2\pi a^3[/tex]

This is what's supposed to happen, but I'm not sure if I've actually done the calculations right or got them right by luck. If someone could tell me where (if anywhere) I've gone wrong, thanks.