Line Integral - Independence of the way (Why these conditions)

Question

Given the following differential form $$ydx - x^2dy$$ It is desired to obtain the value of the line integral along the segment going from (0,0) to (1,1). In addition, it is stated that for any parameterization $x = y = f (t), 0 \leq t \leq 1, f'(t) > 0, f(0) = 0, f(1) = 1$ the value of the integral will be the same.

Well, the first part I already got, which is the value of the integral equal to 1/6.

However, in the second one, which is to verify the statements, I thought about using the independence of the path. For that, I checked if the field is conservative.

$$\frac{dP}{dy} = \frac{dQ}{dx}$$

But, equality is not satisfied. Implying that I cannot use the independence of the path to verify the claims. I'm wrong? What is the correct way to proceed. Furthermore why the need for the conditions mentioned ($x = y = f (t), 0 \leq t \leq 1, f'(t) > 0, f(0) = 0, f(1) = 1$)?

Thanks in advance!

Answer 1

"Not depending on parametrizations (of one path)" is different from "independent of paths".

The former says that the notion of the line integral using one particular parameterization is well-defined.

The latter is a property of the underlying vector field in the integral.

Answer 2

I show the reparametrizazion exercise using some notation on differential forms. Hope to not mess it up too much since I am rusty.

So the exercise suggests to take the generic parametrization of the path $\gamma$, defined through $f$ :

$p(t):[0,1]\rightarrow \mathbf{R}^2, p(t)=(f(t),f(t)), p(0)=(0,0),p(1)=(1,1)$

and than apply the definition of the line integral. Calling $z=ydx-x^2dy$ the 1-form on $\mathbf{R}^2$ this leads to:

$\int_{\gamma}z=\int_0^1 z(p(t))[p_*(\partial_x)]dt$

where $ z(f(t))$ is the one form evaluated at $f(t)$ and $p_*(\partial_x)$ is the pushforward of the tangent field in [0,1]. $z(x)[v]$ is the value of the differential form at $x$ applied to the tangent vector $v$.

Explicitely:

$z(p(t))=f(t)dx-f(t)^2dy$

$p_*(\partial_x)=f'(t)\partial_x+f'(t)\partial_y$

So putting together:

$\int_{\gamma}z=\int_0^1f(t)f'(t)-f(t)^2f'(t)dt$

And now we can change variables : $y=f'(t)$, so that:

$\int_{\gamma}z=\int_0^1 y-y^2 dy=1/6$

You may want to apply the same calculations using a different notation according to the one you are studying right now.