# What does $\iint_S (\hat{\mathbf{n}} \times \nabla \psi) \; dS$ mean?

In the Wikipedia article on vector calculus identities, we have the following

$$\oint_{\partial S} \psi \; d\mathbf{\ell} = \iint_S (\hat{\mathbf{n}} \times \nabla \psi) \; dS$$

The right hand side is an integral of a vector field over a surface integral against $dS$, and not $\cdot dS$ or $\cdot \mathbf{\hat{n}} \; dS$. So if it's not a flux integral, I don't know what this means.

I don't think the right hand side is the flux integral of $\mathbf{\hat{n}} \times \nabla \psi$. Otherwise, by Stokes' theorem, this would equal $$-\int_{\partial S} \psi \mathbf{\hat{n}} \; \cdot d\ell = 0$$ since the normal vector is normal to the tangent vector on the boundary. Then the identity is false.

So what is this called? The only way I know how to integrate a vector field along a surface it by its flux, but this clearly is not what this is. I haven't seen this notation anywhere else, does anyone know where I can find a precise, rigorous definition of what it's supposed to mean?

EDIT: Here is my guess. These are vector valued integrals. The left hand side is $$\oint_{\partial S} \psi \; \mathbf{d\ell}$$ What this means is the vector $$\left(\int_{\partial S} (\psi, 0, 0) \cdot d\ell, \int_{\partial S} (0, \psi, 0) \cdot d\ell, \int_{\partial S} (0, 0, \psi) \cdot d\ell \right)$$ where the integrals are the usual line integrals of vector fields along $\partial S$.

The right hand side is $$\iint_S (\hat{\mathbf{n}} \times \nabla \psi) \; dS$$ Observe that $\hat{\mathbf{n}} \times \nabla \psi$ gives a vector that can be written as $$(\mathbf{G}_1 \cdot \mathbf{\hat{n}}, \mathbf{G}_2 \cdot \mathbf{\hat{n}}, \mathbf{G}_3 \cdot \mathbf{\hat{n}})$$ for some vector fields $\mathbf{G}_i$, $1 \leq i \leq 3$. The right hand side is thus the vector $$\left(\int_S \mathbf{G}_1 \cdot \mathbf{\hat{n}} \; dS, \int_S \mathbf{G}_2 \cdot \mathbf{\hat{n}} \; dS, \int_S \mathbf{G}_3 \cdot \mathbf{\hat{n}} \; dS\right)$$ where the components are the flux integrals of the $\mathbf{G}_i$.

Still not sure what the $dS$ is supposed to be, since it seems a bit useless?

EDIT 2:

I think I understand now, the integral of a vector valued function is just the integral of the component functions with respect to the volume form of what we are integrating over ($dS$ or $d\ell$). So there are two ways to integrate a vector field over a surface. One, is to take the usual flux integral, which yields a scalar. The second, which is presented here, is the integrals of the component functions over the surface, which yields a vector. Funnily enough, I've never seen this in any of my math books...perhaps it's more common in physics?

• So, these aren't flux integrals. What kind of answer are you expecting, then? – Muphrid Jun 15 '14 at 22:59
• A precise, rigorous definition of this expression. – user908123 Jun 15 '14 at 23:00
• I asked a very similar question recently. math.stackexchange.com/questions/833261/… – David H Jun 15 '14 at 23:00
• Ah, thanks. Nice to see I'm not the only one confused by this notation... – user908123 Jun 15 '14 at 23:01
• @user908123 I do not see how a rigorous definition would be any different than a "flux" integral; all that is different is that you're taking a cross product with the normal vector instead of a dot product. Why is that confusing? – Muphrid Jun 15 '14 at 23:01

Your guess is a bit off base: you need to actually calculate the unit tangent vector $\mathbf{\hat t}$ and multiply it by $\psi$ for the line integral. Here, you've chosen a non-unit vector of $(1,1,1)$, which has no relationship whatsoever to the geometry of the problem.

As far as the surface integral, I think you're too concerned with trying to shoehorn the expression into a flux integral; surface integrals are more flexible than you're giving them credit. When evaluating a surface integral that is vector-valued, the common approach is to break the integral into components of some fixed basis. Let $\mathbf F = \nabla \psi$, and

$$\left( \int_S \hat n_2 F_3 - \hat n_3 F_2 \, dS, \ldots, \ldots \right)$$

I would not posit that there exists a vector field such that this whole expression can be written as a dot product with the normal vector and said vector field (at the least, if you do posit this, you should be prepared to prove it). It is not necessary to construct said vector field, so why worry about it?

The surface element $dS$ stems from the coordinates used. For instance, in spherical coordinates on a spherical surface, it would commonly be said that $\hat{\mathbf n} = \hat{\mathbf r}$ and that $dS = r^2 \sin \theta \, d\theta \, d\phi$. At first, this seems very arbitrary, for $dS$ would have very different descriptions in other coordinates, and you would end up needing a whole table of these elements to get them all organized and figured out.

One remedy to this logistical mess is to instead write the integrals in terms of non-unit normal vectors. Then we get

$$\mathbf n = r^2 \hat{\mathbf r} \sin \theta = \mathbf{e_\theta} \times \mathbf{e_\phi}$$

And we just write $\mathbf n \, d\theta \, d\phi$ for the (vectorial) surface element $d\mathbf S$. This is a general and safe approach that easily extends itself to arbitrary coordinate systems, and it keeps you from having to do algebraic gymnatics to calculate both the normal vector and $dS$ separately (which often involves canceling factors from $\mathbf n$ to make it unit only to put them back into $dS$ to compensate; the above approach is much more direct).

Edit: incidentally, it's not strictly necessary to pick a unit vector $\hat{\mathbf t}$ either, for similar arguments above. It all depends on the choice of parameterization for the curve.

• "I would not posit that there exists a vector field such that this whole expression can be written as a dot product with the normal vector and said vector field (at the least, if you do posit this, you should be prepared to prove it)." Can't we just factor? So the above example $F_3 n_2 - F_2 n_3$ is $(0, F_3, -F_2) \cdot n$. Then just take the integral of this. – user908123 Jun 15 '14 at 23:50
• You'll note I wrote out only one component of the vector-valued surface integral. The others would be quite incompatible with your suggestion. – Muphrid Jun 16 '14 at 3:25

Hopefully, this offers some intuition.

We can rotate a small patch of the surface so that the normal is out of the image

$\hspace{5cm}$

I've broken the vector integral into two parts: vertical pieces and horizontal pieces. In the top image, we sum the vertical contributions to the vector integral. The difference between the values on the right edge and the values on the left edge is $\partial_x\psi\,\mathrm{d}x$. Incorporating the length and direction we get $$\boldsymbol{j}\,\partial_x\psi\,\mathrm{d}x\,\mathrm{d}y$$ In the bottom image, we sum the horizontal contributions to the vector integral. The difference between the values on the upper edge and the values on the lower edge is $\partial_y\psi\,\mathrm{d}y$. Incorporating the length and direction we get $$-\boldsymbol{i}\,\partial_y\psi\,\mathrm{d}y\,\mathrm{d}x$$ Adding the two contributions yields $$\boldsymbol{k}\times\nabla\psi\,\mathrm{d}x\,\mathrm{d}y$$ Rotating this to the direction of the normal and summing over all the patches with edges cancelling, shows that the vector integral of $\psi$ along boundary of the surface is the surface integral of the cross product of the normal and the gradient of $\psi$ $$\oint \psi\,\mathrm{d}\boldsymbol{\ell}=\int\boldsymbol{n}\times\nabla\psi\,\mathrm{d}S$$