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