2 answers

What is phi-hat cross s-hat in cylindrical coordinates

Question:

what is phi-hat cross s-hat in cylindrical coordinates?

Answers

phi-hat=-sin(phi) x-hat+ cos(phi) y-hat
s-hat=cos(phi) x-hat+ sin(phi) y-hat
phi-hat x s-hat= -sin^2(phi) z-hat -cos^2(phi) z-hat=-1 z-hat

.

A vector field is defined by [tex]vec{A}(vec{r})} = rho^2 hat{phi} + rho sin phi hat{z}[/tex]

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 dphi dz hat{rho} + dz drho hat{phi} + rho drho dphi hat{z}[/tex] and [tex]vec{dl} =drho hat{rho} + rho dphi 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} + 3rho^2 hat{z}[/tex]

So, [tex]vec{dS} cdot (nabla times vec{A}) = cos phi rho dphi dz - sin phi dz drho + 3rho^2 drho dphi[/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^{2pi} dphi int_0^a 3rho^2 drho = 2pi a^3[/tex]

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

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

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