Integrate dx, dy, dz: Order Matters

[captain]

if you put the differential dx, dy, and dz in the front of the integrand how do you order it so that it matches the order of which variable you want to integrate first?

 

[HallsofIvy]

That depends on the "convention" you use. Most people write the differentials after the integrand but even if you write them in front the same convention is used: The differential that appears first (yes, I see your point: you want to use the one nearest the integrand, the last one, don't you) that you integrate with respect to:
\int \int \int dxdydz f(x,y,z)
implies that you integrate with respect to x first.

 

[Pere Callahan]

Another notation which I like very well is to always keep the integral sign and the dx together, sort of an operator notation:

<br /> \int dz \int dy \int dx \,f(x,y,z)<br />

The integrations are done from right to left.

 

[HallsofIvy]

In general, unless you are given specific limits of integration, some of which involve one or two of the variables, it doesn't matter what order you use. If the limits of integration involve variables, you can deduce the correct order of integration by: the limits of integration on the final integral (the one done last) must be constant and the limits of integration on the second integral cannot involve the variable with respect to which is being integrated nor the variable used in the first integral. Of course, to use that you have to be able to decide what variable goes with what limits of integration. That is why, I expect, Pere Callahan prefers the form where each variable is next to its limits of integration.

I personally prefer to put that information on the integral itself. For example
\int_{x= yz}^z\int_{z= 0}^1\int_{y= z}^{z^2} f(x,y,z)dxdydz[/itex]<br /> I can tell, no matter how &quot;dxdydz&quot; is written, that the &quot;outer integral&quot; is with respect to z since its limits of integration are constants, that the &quot;middle integral&quot; is with respect to y since its limits of integration depend only on z and that the &quot;inner integral&quot; is with respect to x since its limits of integration depend on both y and z.