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$\begingroup$

So I am given a Vector $A$ where $$A=r^2\hat{e}\psi$$

This vector is already considered to be in spherical coordinates.

I am confused by the fact that I am not given $\hat{r} \hat{\theta}$ and instead am given $\hat{e}$, which makes no sense at it is not an angle.

Can anyone explain to me why $\hat{e}$ is part of the spherical coordinate system? Especially considering it is not an angle.

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    What's your convention used? $r(\sin \theta \cos \psi, \sin \theta \sin \psi,\cos \theta)$ or $r(\sin \psi \cos \theta, \sin \psi\sin \theta,\cos \psi)$ or else? They are different from $r(\sin \theta \cos \phi, \sin \theta \sin \phi,\cos \theta)$ which I commonly use. The bases are usually written as $\mathbf{e}_r$, $\mathbf{e}_\theta$, $\mathbf{e}_\phi$ or $\hat{r}$ $\hat{\theta}$, $\hat{\phi}$. – 2017-02-07
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    @NgChungTak The $\hat{e}$ has no bases. I am asked to use this to find the value of an integral of a hemispherical surface. I am given the value of $r=1$ for $z >= 0$. I am quite stumped by the fact that there is a $\hat{e}$ in the vector in spherical coordinates. Never encountered anything like this before and can't find any examples online. – 2017-02-07
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    It think that might be a typo mistake - just means $\hat{e}_{\psi}$ Do check your convention. In my convention used $\mathbf{e}_r=(\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta)$, $\mathbf{e}_{\theta}=(\cos \theta \cos \phi, \cos \theta \sin \phi, -\sin \theta)$, $\mathbf{e}_{\phi}=(-\sin \theta \sin \phi, \sin \theta \cos \phi,0)$ – 2017-02-07
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    @NgChungTak Oooohh I see, so If you were going to write out the full vector equation would it be: $A=0\hat{er} + r^2\hat{e\psi} + 0\hat{e\theta}$? – 2017-02-07
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    Don't forget to write the subscript as "\hat{e}_{r}" ,etc. – 2017-02-07
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    @NgChungTak Ahh wasn't sure how to do that :) thank you! Just to comfirm, was my full vector equation correct? I want to make sure for went I calculate the curl. – 2017-02-07
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    It seems to be OK! – 2017-02-07

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