Speed down the path on a half of a sphere

What is the speed at any point of a path [math] \theta = \theta(\phi)[/math] on a rough sphere. The forces that affect the moving of a point particle are the force of gravity and the force of friction.

I tried this way but I am not sure if the normal vector is right:

[math]

\mathbf{T}=\frac{dx}{ds}\mathbf{i}+\frac{dy}{ds}\mathbf{j}+\frac{dz}{ds}\mathbf{k}

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[math]

\mathbf{N}=(-\frac{dz}{ds}-\frac{dy}{ds})\mathbf{i}+(-\frac{dz}{ds}+\frac{dx}{ds})\mathbf{j}+(\frac{dx}{ ds}+\frac{dy}{ds})\mathbf{k}

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[math]

F_g = -mg \mathbf{k}

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[math]

F_n=|\textbf{F}_g\cdot\textbf{N}|=|-mg (\frac{dx}{ds} +\frac{dy}{ds}) \textbf{k}|=mg (\frac{dx}{ds} +\frac{dy}{ds})

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[math]

\textbf{F}_{friction}=-\mu F_n\textbf{T}=-\mu mg (\frac{dx}{ds} +\frac{dy}{ds}) \textbf{T}

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[math]

ma=F_g -F_{riction}

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\frac{1}{2}\frac{d(v^2)}{ds}=g \frac{dz}{ds}-\mu g (\frac{dx}{ds} +\frac{dy}{ds})

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After integrating and converting to spherical coordinates i get:

[math]

v=\sqrt{2g((\cos{\theta_0}-\cos \theta)-\mu((\cos \phi_0 \sin \theta_0 - \cos \phi \ \sin \theta)+(\sin \phi_0 \sin \theta_0 - \sin \phi \sin \theta)) )}

[/math]

What do you think about the formulas about. Have I done it wrong? I think it would be better to go in spherical coordinates but I didn't know how.

Merged post follows:

I found how to solve it.