Total differential integration help

[TromboneNerd]

Ok, so I'm having a little trouble with total differentiation. I know the total derivative is:

$$dz=\frac{}{}\partial z/\partial x dx+\frac{}{}\partial z/\partial y dy$$

but when i try to integrate it, the right side of the equation is equal to z times the number of dimensions you're dealing with.

$$\int dz=\int\frac{}{}\partial z/\partial x dx+\int\frac{}{}\partial z/\partial y dy$$

$$z=z+z$$

Is there something unique in total derivatives that I am missing that doesn't allow me to integrate like this? or am i making a horrible assumption with integrating partial derivatives in that they are the same as regular derivatives when integrating in respect to the same thing? My high school calculus teacher can't help me because he hasn't done this kind of math since college, which was quite some time ago. This isn't a homework problem (my actual homework is quite boring), just something i ran into on accident while studying ahead in my book that has me confused.

 

[TromboneNerd]

And i just figured out what i did wrong in the latex, so don't even mention it. this is what i meant, though i really doesn't change anything, it just looks better.

$$\int dz=\int\frac{\partial z}{\partial x}dx+\int\frac{\partial z}{\partial y}dy$$

 

[tiny-tim]

Welcome to PF!

Hi TromboneNerd! Welcome to PF!

(have a curly d: ∂ and an integral: ∫ )

… when i try to integrate it, the right side of the equation is equal to z times the number of dimensions you're dealing with.

If you write the limits to the integrals, you'll see why it doesn't work.

z is a function of both x and y, so in the first integral, what limits are you integrating between?

and what are your limits for the second and third integrals, in x and y respectively?

 

[Hurkyl]

An integral sign has no meaning on its own -- just affixing one to both sides should be somewhat suspect.


Normally, when you anti-differentiate, you are anti-differentiating a function respect to a variable, and you stick $\int \,\,\,\, dz$ around the function to indicate what is being anti-differentiated and with respect to what variable^*.

But you're not anti-differentiating a function with respect to a variable; you said you want to anti-differentiate a form. Are you sure that's really what you want? I don't think I've ever seen anti-derivatives of forms presented as if that was an integral-like operation.

Anyways, only exact forms have anti-derivatives... dz is exact, but $\partial z/\partial x \, dx$ is rarely so.




*: There are some awkward issues involved when you have more than one independent variable.

 

[TromboneNerd]

I should have put a constant of integration somewhere because I don't have any limits set up. is there a reason why indefinite integration won't work? It can't be in the constant, because that cancels if i do put limits on it. but about the z being a function of x and y, i think that is what i was overlooking. is this what it "should" look like then?

$$z(x,y)=z(x)+z(y)$$

does this make any sense at all? again, I'm still relatively fresh in multivariable functions, so i am making a lot of assumptions.

 

[TromboneNerd]

I'm not sure what I said made any mathematical sense, but here's basically what I meant and it appears to work.

lets say z=2x+y, so

$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$

$$dz=2dx+dy$$

$$\int dz=\int 2dx+\int dy$$

$$z=2x+y+C$$

So it was just my overlooking the fact that $$\int\frac{\partial z}{\partial x}dx$$ doesn't equal z, but just the x part of the function z.
question answered. Thanks guys

 

[Hurkyl]

Linear functions are extremely bad test cases when you're checking for odd behavior.

 

[TromboneNerd]

What kind of functions should I use then?

 

[Redbelly98]

lets say z=2x+y

I think the linear function used here provides a good, simple counterexample to show what was wrong with the original question.