Exact differentials (thermody)

[jenzao]

Homework Statement

This isn't exactly HW problem , but a "simple" worked problem that is supposed to illustrate exact differentials...

Consider a differential dZ =2xy⋅dx+x^2dy integrated
on two paths where Path I is x=y and Path II is x^2=y.

The integrations are from (x,y)=(0,0) to (x,y)=(1,1).

Then it works the problem, and says this:

delta Z sub-i = 2*integral from 1 to 0 xy*dx + integral from 1 to 0 x^2 dy

= 2 integral from 1 to 0 x^2dx + integral from 1 to 0 y^2dy = 1

Homework Equations

Why are they integrating twice, to arrive at x=1?

If this doesn't make any sense, I will upload the pdf file, I am have trouble writing out all the symbols using basic script.

The Attempt at a Solution

 

[nrqed]

Homework Statement

This isn't exactly HW problem , but a "simple" worked problem that is supposed to illustrate exact differentials...

It's hard to read what you wrote. Let me TeX it:

Consider a differential dZ =2xy dx+x^2 dy integrated
on two paths where Path I is x=y and Path II is x^2=y.

The integrations are from (x,y)=(0,0) to (x,y)=(1,1).

Then it works the problem, and says this:

delta Z sub-i =2*\int_1^0 xy~dx + \int_1^0 x^2 dy<br /> <br /> = 2 \int_1^0 x^2 dx + \int_1^0 y^2dy = 1

Homework Equations

Why are they integrating twice, to arrive at x=1?

If this doesn't make any sense, I will upload the pdf file, I am have trouble writing out all the symbols using basic script.

The Attempt at a Solution

is that what you meant to write? I don't understand why you put your integrations from 1 to 0 (instead of 0 to 1). The integrals you wrote down don't seem to make sense to me. Are you sure these are the correct ones? The first two ones are good, but the last two ones (using the line y=x^2) are wrong.

And why do you ask about integrating "twice"? What twice are you referring to? They simply want you to show that the integral around two different paths gives the same answer.

 

[nrqed]

It's hard to read what you wrote. Let me TeX it:


is that what you meant to write? I don't understand why you put your integrations from 1 to 0 (instead of 0 to 1). The integrals you wrote down don't seem to make sense to me. Are you sure these are the correct ones? The first two ones are good, but the last two ones (using the line y=x^2) are wrong.

And why do you ask about integrating "twice"? What twice are you referring to? They simply want you to show that the integral around two different paths gives the same answer.

For the first path: take from the origin to (x=1,y=0) and then from (x=1,y=0) to (x=1,y=1).

These are two segments. On the first segment (origin to (1,0), you have x varies from 0 to 1, y is constant at y=0 and dy=0. So set dy=0, y=1 and integrate x from 0 to 1.

For the second part of the segment (1,0) to (1,1), x is held constant at x=1 and y varies from 0 to 1. So set x=1, dx=0 and integrate y from 0 to 1.

Add the two results from the two segments to get your final answer.

Now, repeat for the second path which is along the parabola y=x^2. You may reexpress everything in terms of x. Replace y by x^2 and then replace dy by 2x dx. Integrate your expression over x from 0 to 1. Your final answer will give the same result as previously which will be equal to one.

 

[jenzao]

Thanks a lot for helping me with this.
your right, it should be from 0 to 1, my mistake.
The 2nd integrations are exactly as professor wrote them (I uploaded pdf file--its on pg 19).
This is why I am asking about integrating twice:
If we integrate xy with respect to x, we get x^2 right?
But then after we have integrated it (to get x^2) the integral sign should disappear shouldn't it?

But in order for the equation to = 1, you have to integrate a second time, specifically getting 2*(2/3) + (1/3) = 1

Isnt that integrating twice? Obviously I am stumbling on something but i don't know what it is.
Thanks for the help!

 

[jenzao]

sorry i meant to say 2*(1/3) + (1/3) = 1

 

[jenzao]

(here is pdf file that I pulling this from)

 

[nrqed]

Thanks a lot for helping me with this.
your right, it should be from 0 to 1, my mistake.
The 2nd integrations are exactly as professor wrote them (I uploaded pdf file--its on pg 19).

They don't make any sense to me! Are they supposed to be along the parabola y =x^2??

This is why I am asking about integrating twice:
If we integrate xy with respect to x, we get x^2 right?

It depends! It depends of the path you are following!
If you are going along the straight line (0,0) to (1,0) then y =0 so \int xy dx =0
If you are integrating along the straight line from (1,0) to (1,1), then x is held constant so dx=0 and the integral is again zero. If you integrate along the parabola y=x^2 from (0,0) to 91,1) then you have
\int_0^1 dx ~ x^3 = 1/4

So you must specify the path you are following! And you don't integrate twice.



You know, I just realized that the second expression would make sense if the integral was along the straight line y=x instead of the parabola y=x^2! Are you sure the prof did not say y=x??

 

[jenzao]

yes, he explains 2 paths. Path 1 is y=x.
But how can the above equation = 1 unless you integrate twice?

So first you go from xy to x^2
then (2nd time) you go from x^2 to (1/3)x^3

So, we're going from 0 to 1, so from the above equation you would have 2(1/3) + (1/3) = 1

But I arrived at 1, by integrating twice.
(??)

 

[jenzao]

OK I figured out why i though integrating twice.
The first part he inserts the equality y=x (for path 1). that's where i thought it was being integrated.