Does dP_x dP_y = d^2\vec P : Integration scalar / vector var

[binbagsss]

Excuse me if this is a bad question but:
Does $d P_x d P_y = d^2 \vec P$?
I thought not because $P_x$ is a scalar , a component of the vector, whereas $\vec P$ is a vector?
Thanks in advance

 

[haushofer]

I'd say they're equal, but this is merely notation, no?

 

[binbagsss]

mm I'm confused:

Say I have $\int dP_x dP_y e^{-\vec{P}^2}$

if $dP_x dP_y =d\vec P$* then this is $\int d\vec{P}e^{-\vec{P}^2}=\sqrt{\pi}$

Since $\vec P^2$ is a scalar I can do the integral working with the components $dP_x$ and $dP_y$ however had it not been and just been $\vec{P}$ I don't know how you convert the integral between $dP_x dP_y$ and $d\vec P$

So $\int dP_x dP_y e^{-\vec{P}^2}=\int dP_x dP_y e^{-P_x^2-P_y^2}=\int dP_x e^{-P_x^2}\int dP_x e^{-P_y^2}=\sqrt{\pi}\sqrt{\pi}=\pi$

so using * these don't agree?

 

[binbagsss]

mm I'm confused:

Say I have $\int dP_x dP_y e^{-\vec{P}^2}$

if $dP_x dP_y =d\vec P$* then this is $\int d\vec{P}e^{-\vec{P}^2}=\sqrt{\pi}$

Since $\vec P^2$ is a scalar I can do the integral working with the components $dP_x$ and $dP_y$ however had it not been and just been $\vec{P}$ I don't know how you convert the integral between $dP_x dP_y$ and $d\vec P$

So $\int dP_x dP_y e^{-\vec{P}^2}=\int dP_x dP_y e^{-P_x^2-P_y^2}=\int dP_x e^{-P_x^2}\int dP_x e^{-P_y^2}=\sqrt{\pi}\sqrt{\pi}=\pi$

so using * these don't agree?

anyone? What have I done wrong here?
Many thanks