Intuitive Definition of Curl and Stokes' Theorem

I am asking this question on SE Physics because it's about an intuitive derivation, not a rigorously made proof of a theorem.

Reading Mathematical Methods for Physics and Engineering by Riley, et al. (actually the book does not matter) I found the following intuitive definition of the curl:

The notation might be a bit confusing - here $$\vec{a}$$ represents a vector field, $$d\vec{S}$$ is a surface differential with orientation.

Yes, I understand this. I can also do an intuitive proof on my own, reaching the conclusion with the following expression:

$$dxdydz \ (\nabla \times \vec{a}) = d\vec{S} \times \vec{a}$$

which is pretty much the same as the statement.

But another problem rises - the author states another intuitive definition of the curl:

I tried to derive this by applying the dot product with $$\vec{dA}=\hat{n} \ dA$$ to the above expression, where $$\hat{n}$$ is the normal vector to a specific point of the surface in three dimension.

It doesn't work. The problem is that I don't know how to deal with an arbitrary part of the surface, which obviously has an arbitrary orientation.

How do I show this? I noticed that it is, if integrated, equivalent to the Stokes' theorem in three dimension:

$$\int_{\partial D} \vec{a} \cdot d\vec{r} = \iint_D (\nabla \times \vec{a}) \cdot d\vec{S}$$

So the derivation of this formula should have greater importance than the above one.

• Find a pdf of Thomas' Calculus 13th edition. Observe figure 16.66 on page 1023. May 25, 2021 at 5:00

Your idea is correct: you have to exploit Stokes' formula for a decreasing sequence of surfaces $$\Sigma_n$$ tending to a point $$p$$. Here the surfaces of the sequence should be viewed as restrictions of an initially given surface $$\Sigma$$ containing $$p$$. When the surfaces tend to $$p$$, the averaged value with respect to the area of the line integral of $$\vec{a}$$ around the boundary of the surfaces, $$\frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r}$$ tends to the $$\vec{n}_p$$ component of the curl of $$\vec{a}$$ at $$p$$.
Indeed, taking advantage of Stokes' theorem, if $$\Sigma_n$$ is very concentrated around $$p$$, $$\frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r} = \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \nabla \times \vec{a}(q) \cdot \vec{n}_q \:dS(q) \sim \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \nabla \times \vec{a}(p) \cdot \vec{n}_p \:dS(q)$$ $$= \nabla \times \vec{a}(p) \cdot \vec{n}_p \frac{1}{S(\Sigma_n)}\int_{\Sigma_n}1 \:dS(q) = \nabla \times \vec{a}(p) \cdot \vec{n}_p$$

I write below a formal statement with a proof.

Theorem.

Let $$\Sigma$$ be a $$C^1$$ surface in $$\mathbb{R}^3$$ whose boundary $$\partial \Sigma$$ is a regular closed curve (without self-intersections). Consider a sequence of similar surfaces $$\Sigma_n\subset \Sigma$$, restrictions of $$\Sigma$$, such that $$\Sigma_n \to p \in \Sigma$$ for $$n\to +\infty$$

(in other words, $$\Sigma= \vec{x}(D_{r}(p_0))\subset \mathbb{R}^3$$ for some $$C^1$$ parametrization from the disk $$D_{r}(p_0) \subset \mathbb{R}^2$$ of radius $$r>0$$ such that $$\vec{x}(p_0) =p$$ and $$\Sigma_n= \vec{x}(D_{r_n}(p_0))$$ with $$r_n\to 0$$ as $$n\to +\infty$$. Obviously I am also assuming that $$\partial \Sigma_n = \vec{x}(\partial B_{r_n}(p_0))$$.)

If $$\vec{a}: O \to \mathbb{R}^3$$, where $$O\supset \Sigma$$ is an open set, is $$C^1$$ then $$\vec{n}_p \cdot \nabla \times\vec{a}(p) = \lim_{n\to +\infty} \frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r}$$
where $$\vec{n}_p$$ is the unit vector normal to $$\Sigma$$ at $$p$$ and the orientation of $$\partial \Sigma_n$$ is chosen positively with respect to $$\vec{n}$$, finally $$S(\Sigma_n) = \int_{\Sigma_n} 1 dS$$ is the area of $$\Sigma_n$$
.

PROOF. From Stokes' theorem, $$\frac{1}{S(\Sigma_n)}\oint_{+\partial \Sigma_n} \vec{a}\cdot d\vec{r}= \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \nabla \times \vec{a} \cdot \vec{n} \:dS = \nabla \times \vec{a}(p) \cdot \vec{n}_p + \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \left(\nabla \times \vec{a} \cdot \vec{n} - \nabla \times \vec{a}(p) \cdot \vec{n}_p\right)\:dS \:.$$ Now observe that $$\left| \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \left(\nabla \times \vec{a} \cdot \vec{n} - \nabla \times \vec{a}(p) \cdot \vec{n}_p\right)\:dS \right| \leq \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \left|\nabla \times \vec{a} \cdot \vec{n} - \nabla \times \vec{a}(p) \cdot \vec{n}_p\right|\:dS$$ $$\leq \frac{1}{S(\Sigma_n)}\int_{\Sigma_n} \sup_{q \in \Sigma_n}\left|\nabla \times \vec{a}(q) \cdot \vec{n}_q - \nabla \times \vec{a}(p) \cdot \vec{n}_p\right|\:dS = \sup_{q \in \Sigma_n}\left|\nabla \times \vec{a}(q) \cdot \vec{n}_q - \nabla \times \vec{a}(p) \cdot \vec{n}_p\right|\:.$$ Since $$\Sigma \ni q \mapsto \nabla \times \vec{a}(q) \cdot \vec{n}_q - \nabla \times \vec{a}(p) \cdot \vec{n}_p$$ is continuous (from the hypotheses) and $$\Sigma_n \to p$$, then $$\sup_{q \in \Sigma_n}\left|\nabla \times \vec{a}(q) \cdot \vec{n}_q - \nabla \times \vec{a}(p) \cdot \vec{n}_p\right| \to 0$$ as $$n\to +\infty$$ and the thesis follows. QED