Geometric interpretation of Stokes' Theorem

Question

First of all could someone please explain my misunderstanding of the following example of Stokes' Theorem:

Example (Lee, SM, p. 414): Let $M$ be a smooth manifold and suppose $\gamma: [a,b]\to M$ is a smooth embedding, so that $S=\gamma([a,b])$ is an embedded 1-submanifold with boundary in $M$. If we give $S$ the orientation such that $\gamma$ is orientation-preserving, then for any smooth function $f\in C^\infty(M)$, Stokes’ theorem says that $$\int_\gamma df=\int_{[a,b]}\gamma^*{df}=\int_{S}df=\int_{\partial S}f=f(\gamma(b))-f(\gamma(a)).$$

Q1: What are the difference between $S$ and $\gamma$ in action? Isn't $\gamma$ means $\gamma([a,b])$ in the first integral?

Q2: What is the geometric interpretation of Stokes' Theorem?

By second question I don't mean the calculus and physics type interpretation. I don't mean explaining by De Rham cohomology (exact and closed forms) also. I just want to know how it is possible that integrating over whole manifold can be computed just by integrating over its boundary.

Perhaps the answer of Q2 lies in the following question:

Q3: What does $\int_\gamma df$ mean? Is it a kind of measurement the length of $\gamma$? If so, why its value does not depend on any curve with same endpoints of $\gamma$?

Sorry if my questions related to the proof of the theorem somehow.

Answer 1

Q1: As stated in the excerpt from Lee's book, $S$ is the image of $\gamma$. And yes, in the first integral, the notation $\int_\gamma df$ means the same thing as $\int_S df$.

Q2: This question is pretty vague, so I'm not sure what kind of answer you're looking for. How can there be an interpretation that's not a "calculus" interpretation? Are you comfortable with the "Fundamental Theorem of Calculus", that $\int_a^b f'(t) ~ dt = f(b) - f(a)$? Or are you really asking why that's true?

• If you're ok with the Fundamental Theorem, then maybe my answer to Q3 below will satisfy you, because the reason it only depends on the endpoints (and not the path) is essentially because this 1-dimensional example is equivalent to the Fundamental Theorem of Calculus.
• If my answer below for Q3 is not satisfying, then maybe what you're really asking is why $\int_a^b f'(t) ~ dt = f(b) - f(a)$ is even true in the first place.

Q3: As in the quote you give from Lee: $\int_\gamma df = \int_{[a,b]} \gamma^* df$. Taking this a step further, and unwrapping the definition of $\gamma^* df$, this is the following: $$ \int_\gamma df = \int_{[a,b]} \gamma^* df = \int_a^b \frac{d}{dt} \Big( f(\gamma(t)) \Big) ~ dt $$ So by the usual Fundamental Theorem of Calculus, the result is the difference of the function values of $f(\gamma(t))$ at $t=a$ and $t=b$.