# Finding a function given a differential

I'm still pretty new to this differential business and I have a couple of question concerning this problem I've come across. We're given a differential form $\alpha=p_1 dx_1+p_2 dx_2-H(p_1,p_2)dt$ defined on $\mathbb{R}^5=\{(p_1,x_1,p_2,x_2,t)\}$ with $H(p_1,p_2)$ a globally defined smooth function dependent only on $p_1$ and $p_2$. First, I'd like to find $d\alpha$. This is fairly straightforward, but I'm having some notational difficulties. I should have $$d\alpha=dp_1\wedge dx_1+dp_2\wedge dx_2 - [\text{something}]\wedge dt$$ where the "something'' above should be the total derivative of $H(p_1,p_2)$. Is there a good notation for expressing the total derivative of a general function in two variables, namely $H(p_1,p_2)$?

If $\alpha$ is the differential of a globally defined smooth function $f$, then I would have that $d\alpha=0$. Since this isn't the case, $\alpha$ is not the differential of a globally defined smooth function.

Finally, I'd like to consider the restriction of $d\alpha$ to the $3$-dimensional plane $\{p_1=4,p_2=5\}$. Then $$d\alpha = 0$$ since $\alpha = 4dx_1+5dx_2-H(4,5)dt$. Because $d\alpha=0$, it must be the differential of a smooth function defined on this plane. In order to find this function explicitly, I think I need to integrate $\alpha:$ $$\int 4dx_1+\int 5dx_2 - \int H(4,5)dt = 4x_1+5x_2-H(4,5)t$$ Is this correct? Specifically, is the restriction of $d\alpha$ to the plane provided above the differential of $4x_1+5x_2-H(4,5)t$?

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Do you want a notation other than $dH$ for the total derivative? If so, you could just write it out since it isn't that long with only two variables. For the second part. Yes, that function is linear so it is quick to check that $d$ of it is your restricted $\alpha$. –  Matt Oct 17 '12 at 5:18
@Matt It's late and I seem to be overlooking obvious solutions. Thanks. –  chris Oct 17 '12 at 5:23