I'm a beginner at vector calculus so I'm not really good at this yet. Trying to learn from the book I stated below:
The fundamental theorem for gradients from the Vector Analysis section of Griffith's Introduction to Electrodynamics textbook:
$$\int_a^b(\nabla T) \bullet d \vec I = T(b) - T(a)$$
In the example question, it gives a scalar $T = x y^2$ and gave two points and the question was to check for the fundamental theorem of gradients.
He did so by taking 2 paths and comparing the line integrals, which is 2.
However, isn't there a chance that the line integrals of that 2 specific paths are coincidentally the same? If that's the case, it's doesn't really prove the theorem.
So, I decided to try to prove by definitions instead.By definition, $$(\nabla T) \bullet d \vec I = (\frac{\partial T}{\partial x} \hat x + \frac{\partial T}{\partial y} \hat y + \frac{\partial T}{\partial z} \hat z) \bullet (dx \hat x + dy \hat y + dz \hat z)$$ $$= \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z} dz$$ which, by a theorem of partial derivatives, $= dT$
So,I'm left with$$\int_a^b dT$$ Here, if we do the integral regularly, it's $= b-a$. Clearly not the case.
So, here's what I tried because the book doesn't talk about it, I was thinking that T is actually in terms of $x,y,z$ and $a,b$ should also be $(x,y,z)$ itself (a cartesian coordinate). I was thinking that it's really something like this:$$\int_{r=a}^{r=b} dT(r)$$ which would yield $T(b) - T(a)$.
I just chose r arbitrarily such that $r = (x,y,z)$. Is my logic correct? Have I somehow broken any vector rules on the way but still got the end answer through coincidence? If I'm right, what better notations or ways that I can use to instead of $r = b, r = a$ part
I was also thinking its something like this:$$\int_{z_a}^{z_b} \int_{y_a}^{y_b} \int_{x_a}^{x_b} dT(x,y,z) dx dy dz$$ Is this right? Never seen triple integrals for a line integral before. Probably no reason to do so but I just want to understand it better.
In the first part your intuition is correct, if $T$ is smooth enough then
$$ {\rm d}T = \nabla T \cdot {\rm d}{\bf l} $$
so if the path $c$ along which the integral is being carried out is parametrized as $c = \{{\bf x} | ~{\bf x} = {\bf f}(r), ~~ r_a<r<r_b \}$, then
$$ \int_c {\rm d}T = \int_{\bf a}^{\bf b} \nabla T \cdot {\rm d}{\bf l} = \int_{r_a}^{r_b}\nabla T\cdot\frac{{\rm d}{\bf l}}{{\rm d}r}{\rm d}r= T({\bf b}) - T({\bf a}) $$
where ${\bf a} = {\bf f}(a)$ and ${\bf b} = {\bf f}(b)$.
As for the second part, you can think of the integral you wrote as a sum of volumes weighted by the quantity ${\rm d}T$, which definitely is not the same as the line integral you calculated in the first part.