Is the Equation for Integrating an Exact Differential Correct?

[Cathr]

Let's say we have $df=2xy^3dx + 3x^2y^2dy$ - this is an exact differential.
In integrating, to find f, can we write $f = \int 2xy^3 \, dx + \int 3x^2y^2 \, dy = 2x^2y^3 + C$
Or am I getting it wrong?

 

[BvU]

We can write it (as you demonstrate)

Only thing that's wrong is the 2 in $2x^2y^3$

 

[Cathr]

We can write it (as you demonstrate)

Only thing that's wrong is the 2 in $2x^2y^3$

Why? This result is exactly the sum of what we get from the integrals, we get $x^2y^3$ two times... don't we?

 

[dRic2]

Why? This result is exactly the sum of what we get from the integrals, we get x2y3x2y3x^2y^3 two times... don't we?

Because if you take the differential of $2x^2y^3$ you don't get your starting differential. In fact $df = \frac {\partial f}{\partial x}dx + \frac {\partial f}{\partial y}dy$ and if you put $2x^2y^3$ inside the equation you would get $df = 4xy^3dx + 6x^2y^2dy$ instead of $df = 2xy^3dx + 3x^2y^2dy$.

I think (personal opinion) this

can we write f=∫2xy3dx+∫3x2y2dy

is a misleading way to write it. You may integrate in the wrong way the function.

 

[Cathr]

Because if you take the differential of $2x^2y^3$ you don't get your starting differential. In fact $df = \frac {\partial f}{\partial x}dx + \frac {\partial f}{\partial y}dy$ and if you put $2x^2y^3$ inside the equation you would get $df = 4xy^3dx + 6x^2y^2dy$ instead of $df = 2xy^3dx + 3x^2y^2dy$.

I think (personal opinion) this

is a misleading way to write it. You may integrate in the wrong way the function.

I see. I think the problem is I don't really understand what we do with the $\frac {\partial f}{\partial y}dy$ part when we integrate with respect to x. Even if it is a dy differential, it still depends of x, but we say it's a constant.
I guess the right way to do it is to ignore the dy part when we integrate with respect to x, and add a constant depending on y, then integrate the dy part and add a C(x). Then to say $x^2y^3 + C(x) = x^2y^3 + C(y)$ Then it works, but to me it still seems a bit confusing.

 

[PeroK]

I see. I think the problem is I don't really understand what we do with the $\frac {\partial f}{\partial y}dy$ part when we integrate with respect to x. Even if it is a dy differential, it still depends of x, but we say it's a constant.
I guess the right way to do it is to ignore the dy part when we integrate with respect to x, and add a constant depending on y, then integrate the dy part and add a C(x). Then to say $x^2y^3 + C(x) = x^2y^3 + C(y)$ Then it works, but to me it still seems a bit confusing.

You need to take a step back and look at what is actually happening here. The differentials themselves are something of a nice shortcut, but you can get into trouble if you try to deconstruct things from them. So:

We have a function $f(x, y)$ of these two variables. We know that:

$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = 2xy^3 dx + 3x^2y^2 dy$

This gives us a couple of differential equations for $f$:

$\frac{\partial f}{\partial x} = 2xy^3$

$\frac{\partial f}{\partial y} = 3x^2y^2$

Now, we can integrate each of these to give:

$f(x, y) = x^2y^3 + C_1(y)$

$f(x, y) = x^2y^3 + C_2(x)$

Where $C_1, C_2$ are any functions of a single variable.

From this we see that $C_1(y) = C_2(x)$. There is now a simple but powerful argument (can you see it?) that:

$C_1(y) = C_2(x) = C$

For some constant $C$. And therefore:

$f(x, y) = x^2y^3 + C$

 

[Cathr]

You need to take a step back and look at what is actually happening here. The differentials themselves are something of a nice shortcut, but you can get into trouble if you try to deconstruct things from them. So:

We have a function $f(x, y)$ of these two variables. We know that:

$df = \frac{\partial f}{\partial x}dx + \frac{\partial f}{\partial y}dy = 2xy^3 dx + 3x^2y^2 dy$

This gives us a couple of differential equations for $f$:

$\frac{\partial f}{\partial x} = 2xy^3$

$\frac{\partial f}{\partial y} = 3x^2y^2$

Now, we can integrate each of these to give:

$f(x, y) = x^2y^3 + C_1(y)$

$f(x, y) = x^2y^3 + C_2(x)$

Where $C_1, C_2$ are any functions of a single variable.

From this we see that $C_1(x) = C_2(y)$. There is now a simple but powerful argument (can you see it?) that:

$C_1(x) = C_2(y) = C$

For some constant $C$. And therefore:

$f(x, y) = x^2y^3 + C$

Thank you! Now it's crystal clear :)

 

[Delta2]

if we do definite integration over a curve $C$ of the xy plane then the equation holds as
$\int_C df=\int_Cf_xdx+\int_C f_ydy$ or equivalently

$f(x_1,y_1)-f(x_0,y_0)=\int_C f_x dx +\int_C f_y dy$ (1)

where $(x_0,y_0)$ an $(x_1,y_1)$ are the end points of the curve C and $f_x=\frac{\partial f}{\partial x}, f_y=\frac{\partial f}{\partial y}$. Notice that the two integrals at the RHS of (1) are integrals over the curve $C$ and not necessarily over the straight line from $x_0$ to $x_1$ (over the x-axis, which is the "default curve" when we integrate over x) or the line from $y_0$ to $y_1$ (over the y-axis which is the default curve when we integrate over y). Ofcourse we can choose the curve C to be anything we want, and if we choose it to be the straight line over the x-axis from $x_0$ to $x_1$ then the second term of (1) will be zero because y doesn't vary over this curve C. Similarly if we choose the curve C to be the straight line over the y-axis from $y_0$ to $y_1$ then the first term of (1) will be zero because x doesn't vary over that curve C.

If we do indefinite integration then the equation just doesn't hold otherwise we ll get the absurd result that $f=2f$. To be more precise when we do indefinite integration $\int df=\int f_x dx+\int f_y dy$ we have to (silently) choose the variable of integration which if we choose to be x, then one integral (the $\int f_x dx$ ) will give f and the other will give zero (because when we choose the variable of integration to be x, it means that only x varies and that y doesn't vary at all so it is like integrating y from $y_0$ to $y_0$ that is taking the integral $\int_{y_0}^{y_0}f_ydy$ which is by definition zero), so the equation will be $\int df=\int f_x dx+0 \Rightarrow f=f$.