Why doesn't the "actual" path matter for line integrals?

Question

We have the following definition given in our textbook:

Let $U \subseteq \mathbb{R}^n$ be open and $F: U \rightarrow \mathbb{R}$ be continuously partial differentiable. If $a, b \in U$ and $\gamma$ is a piecewise differentiable path from a to b, that lies completely in $U$ ($[a,b]\in U$), then: $$\int_\gamma (\operatorname{grad} F) \cdot dx = F(b)-F(a)$$

This is obviously super useful for solving line integrals $$\int_\gamma f\,dx$$ where we can find $F$ such that $\operatorname{grad} F = f$. My question is:

why doesn't the path matter in these cases? If I have two paths $\gamma$ and $\gamma^*$ with the same origin/destination but with completely different paths, this tells me the line integral is the same. Why does this make sense?

Answer 1

First, note that this formula only applies to (in fact only makes sense for) line integrals of certain vector fields $\bf F$, namely conservative ones, that is, those of the form $${\bf X} = \operatorname{grad} F$$ for some function $F : U \to \Bbb R$ and (when $n > 1$) this is a very strong restriction: In a sense that can be made precise, most vector fields are not conservative.

So, procedurally, the formula (called the "Fundamental Theorem of Calculus for line integrals") works because we restrict our attention to line integrals of vector fields for which the formula will work. And it's not surprising that the formula works for conservative vector fields: After all, $\operatorname{grad} f \cdot d{\bf s}$ is the infinitesimal change of $f$ along the path $\gamma : [a, b] \to U$ with length element $d{\bf s}$, so by definition $\int_\gamma \operatorname{grad} F \cdot d{\bf s}$ is just the total change of $F$ along the path from $a$ to $b$. But we already know $F$ and the values $F(a), F(b)$ at its endpoints, so the total change along the path is $$\int_\gamma \operatorname{grad} F \cdot d{\bf s} = F(b) - F(a) ,$$ which doesn't depend on the path $\gamma$ connecting those endpoints.

Answer 2

So the physical intuition is as below

As you may know $\nabla F$ is a vector which represents how much your function $F$ is changing and its direction tells you in what direction the function change is maximum.

Now in the integral you have the quantity $\nabla F .d \vec{l} $ which tells you the infinitesimal change of your function $F$ along the direction $d\vec{l}$ which is actually the tangent to the path on which you evaluate your integral on.

So basically your integral is summing over the infinitesimal variations in your function $F$ as you travel along the path.

Now recall the First Fundamental Theorem of Calculus

\begin{equation*} \int_a^b F’(x) \, dx =F(b)-F(a) \end{equation*}

The equation we have is a multivariable analog of the First Fundamental Theorem of Calculus and so the same reasoning applies in both cases. Here $F’(x)$ represents the infinitesimal variation in our function $F$. The physical intuition behind the Fundamental Theorem of Calculus is that if you sum up the infinitesimal variations (which is given by the left side of the above equation), then that sum must be equal to the total variation of the function (given by the right side).

Now lets get back to our case and label our end points as $a$ and $b$. Now the total variation in our function $F$ from $a$ to $b$ is given by $F(b)-F(a)$. Now lets take 2 paths $\gamma$ and $\gamma’$ going from $a$ to $b$. Since the integral along any of these 2 paths calculates the sum of infinitesimal variations along the path, the total value must be equal to the total variation of the function going from $a$ to $b$ (this is what First Fundamental theorem of Calculus tells us). Hence no matter what path we take we must always get $F(b)-F(a)$.

Note: This holds true in an open set $U\subset \mathbb{R}^n$ but not for domains which have holes in them.

Answer 3

This is true for more or less the same reason that the fundamental theorem of calculus (for integrals over subsets of the real axis) is true. Thus, you shouldn't expect one to be more intuitively clearer than the other. It now follows that if you understand why the latter holds, then you should not be far from seeing why it's true in general.

Note that this theorem can be intuitively understood as claiming that if you're integrating a continuous function over some compact domain in the real axis, then you may think of this function as the derivative of some function. It then follows that you're integrating something of the form $$g'(x)\mathrm d x,$$ from $a$ to $b,$ say. This means you're integrating an infinitesimal change in $g(x)$ (since you're multiplying a rate of change by an infinitesimal change in $x$) over $[a,b],$ which is the total value of $g(x)$ accumulated on this interval. Now since $g(x)$ is continuous throughout the interval, it follows that this value is given by the difference in the amounts of $g$ at the initial and final points, namely $g(b)-g(a).$

Now since any continuously differentiable path can be given in terms of a parameter varying over an interval $[a,b],$ and we're assuming our field is conservative (this term is very apt; can you see why?), namely that we can think of it as a gradient; then eventually the computation reduces to a sum of two integrals of a real-valued function of a real variable which are derivatives -- and this we know depends only on the end values.