I have recently learnt about Green's Stokes' and the Divergence Theorems. I read here: http://www.supermath.info/CalculusIIIvectorcalculus2011.pdf.
On page 31, it describes a deformation you can make to show that one line integral is equal to another. I also saw a video describing the same idea for the divergence theorem. Here's how it works:
Suppose there is a field $\vec{F}$ defined on $R^2$ so that $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$, If you have an annulus bounded by C, then:
$$ \oint_C \vec{F} \cdot d\vec{r} = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = 0
$$
But:
$$ \oint_C \vec{F} \cdot d\vec{r} = \oint_{C_1} \vec{F} \cdot d\vec{r} - \oint_{C_2} \vec{F} \cdot d\vec{r} = 0
$$
Thus:
$$\oint_{C_1} \vec{F} \cdot d\vec{r} = \oint_{C_2} \vec{F} \cdot d\vec{r}
$$
The same idea applies to the surface boundary of a sphere like region with a hollowed hole in the middle: The flux of the 2 boundaries should be equal when the divergence is $0$ by the Divergence Theorem. The same goes for an annulus surface living in 3D because of Stokes' Theorem when the curl is $0$, since curl$\vec{F} \cdot \hat{n} = 0$.
Since I first read about it, I was always confused by this/these idea/s, I mean if the respected operation is $0$, then shouldn't all the boundary integrals also be $0$. For example, referring to the section above; shouldn't we be able to argue:
$$ \oint_{C_1} \vec{F} \cdot d\vec{r} = \iint_{R_1} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = 0 $$
($R_1$ is the region bounded by $C_1$) I apologize if this should be obvious, but I don't get why these theorems should at all be useful if all the terms are just $0$. Surely I am making a mistake somehow. Thank you for reading thus far, all answers are appreciated. :)
The thing you're overlooking in the usefulness of these theorems is the domain of definition of the vector field $F$. Let $U\subset\Bbb{R}^2$ be an open subset, and say $F=(P,Q):U\to\Bbb{R}^2$ is a $C^1$ vector field defined on $U$. If $C$ is a closed curve and $R$ is the region bounded by the curve, AND $R\subset U$ is completely contained in the domain of the given vector field $F$, then we have Stokes theorem: \begin{align} \int_C\vec{F}\cdot d\vec{r}&=\int_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA \end{align} If the curl of the vector field vanishes (i.e $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$ at every point of $U$), then sure, the RHS and hence the LHS are zero. To emphasize the importance of the domain $U$, and the significance of the theorem about deforming the contour/surface, let us consider two examples:
Example 1. Vector field Defined Everywhere
Consider the smooth vector field $F:\Bbb{R}^2\to\Bbb{R}^2$ defined as $F(x,y)=(e^xy, e^x)$. This clearly has vanishing curl. So, for any closed curve $C$ which bounds a region $R$, we have by Stokes theorem that \begin{align} \int_C\vec{F}\cdot d\vec{r}&=\int_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dA = 0 \end{align} Actually, this is still an interesting result, because $F$ is a vector field with such a special property that its integral over every closed curve which bounds a region vanishes.
Here, I gave a rather simple vector field $F$. I'm sure one can come up with VERY VERY VERY complicated vector fields, for which directly calculating the line integral by parametrizing the curve etc is completely unfeasible. However, if you know that the curl is $0$, then one can directly deduce that the integral over closed curves is $0$. You write
but what you're not realizing is that sometimes, showing a certain quantity equals zero is a very powerful statement. i.e knowing it is zero is valuable information in some contexts.
Example 2. Vector field not defined everywhere.
Here, consider the domain $U=\Bbb{R}^2\setminus\{(0,0)\}$, i.e the plane minus the origin. Consider the vector field $G:U\to\Bbb{R}^2$ defined as \begin{align} G(x,y)&=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right) \end{align} Note that we have to exclude the origin from the domain of definition because otherwise we'd be dividing by $0$, which is of course completely forbidden. You can once again verify directly that at every point of $U$, we have $\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}=0$. Now, suppose for example that $C$ and $R$ are given as in the figure below:
If in this case you claim "by Stokes theorem" \begin{align} \int_C\vec{G}\cdot d\vec{r}&=\int_R\left(\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\right)\,dA = 0 \end{align}
then you'd be completely wrong. This is wrong because one very important hypothesis has not been verified. In this case the bounded region $R$ contains the origin, however, the domain of $G$ is $U=\Bbb{R}^2\setminus\{(0,0)\}$, so Stokes theorem is just not applicable to this pair of $C$ and $R$. It is in this case that one uses Stokes theorem to "deform the contour". As I have drawn it, $C$ is rather complicated, so evaluating the line integral is next to impossible. What we can do is consider the following figure
Here the curve $C$ we consider has two pieces, $C_1$ which is the black curve the same as in the above picture, and $C_2$ which is the new red circle. The region bounded is $R$. Note that in this case, we have specifically ensured that $R$ does not contain the origin, i.e $R\subset U$. So, in this case, Stokes theorem is completely applicable: \begin{align} \int_C\vec{G}\cdot d\vec{r}&=\int_R\left(\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\right)\,dA = 0 \end{align} The LHS is $\int_{C_1}\vec{G}\cdot d\vec{r}-\int_{C_2}\vec{G}\cdot d\vec{r}$. Therefore, we get the conclusion \begin{align} \int_{C_1}\vec{G}\cdot d\vec{r}&= \int_{C_2}\vec{G}\cdot d\vec{r} \end{align} In other words, we have reduced the calculation of the complicated, horrendous, disgusting line integral $\int_{C_1}\vec{G}\cdot d\vec{r}$ over the complicated curve $C_1$ to the line integral over the simple circle $C_2$. In this case, by directly applying the definition of line integrals (i.e parametrizing the circle $C_2$ counterclockwise) one can show that \begin{align} \int_{C_1}\vec{G}\cdot d\vec{r}&= \int_{C_2}\vec{G}\cdot d\vec{r} = 2\pi. \end{align}
Therefore, using Stokes theorem to deform the contour often times simplifies computations. This is just one illustration of the utility of the Stokes theorem.