How to understand the notation of area element vector

Question

I've already known that the area element of vector surface integral can be written as $dA = \frac{\partial \textbf{r}}{\partial s}\times\frac{\partial \textbf{r}}{\partial t}dsdt$, where r(s,t) is a parameterization of the surface. However, the area element can also be written as $dA=dxdy \hat k + dydz \hat i + dxdz \hat j$. So, do these two forms equal? How to derive(and understand) the latter form?

Answer 1

Intro

(This section was largely copied from my answer to the related question A doubt regarding change of variables in Double Integrals..)

There are two main types of ways to think about things like $\mathrm{d}x\mathrm{d}y$ in multivariable calculus, and we often switch between them depending on the context. (This clarification was inspired in part by Terry Tao's preprint on “differential forms and integration”.) $\mathrm{d}x$ can either act kind of like a positive number representing a small length, or act kind of like a vector with a direction/orientation baked in.

For the “number” interpretation, there are things like limit arguments in analysis or infinitesimals in nonstandard analysis to make things rigorous. For the “vector” interpretation, there are things like “differential forms” or “geometric calculus” to make things rigorous. But I'm going to gloss over those details because there are many ways to make things formal, and the exact choices don't affect the intuition here.

Orientation

If you care about the direction of the normal area vector $\mathrm d\mathbf A$ (e.g. because you would like to integrate $\mathbf F\cdot\mathrm d\mathbf A$ for some vector field $\mathbf F$), then we're in a situation where orientation matters.

Background

For instance, maybe $\dfrac{\partial\mathbf{r}}{\partial s}\times\dfrac{\partial\mathbf{r}}{\partial t}$ goes in the correct direction, but swapping $s$ and $t$ would give $\dfrac{\partial\mathbf{r}}{\partial t}\times\dfrac{\partial\mathbf{r}}{\partial s}$ which goes in the wrong direction. But this is a little weird, because an integral shouldn't care about how you labeled the parameters of the surface, as long as $\mathrm d\mathbf A$ points in the right direction in the end.

There are many ways to address this. One approach is to think about orientation also mattering in the parameter domain in the $st$-plane. For comparison, with regular $1$-d integrals, you can write things like $\int_4^3f(x)\,\mathrm dx$ to signify something like "integrate over $[3,4]$, but give a negative sign because we're doing it backwards". Similarly, if you think about the domain in the $st$-plane, it could be oriented positively (imagine something like $\mathrm ds\times\mathrm dt$) or negatively (imagine something like $-\mathrm ds\times\mathrm dt=\mathrm dt\times\mathrm ds$). And if $\mathrm ds\mathrm dt$ secretly meant something similar in spirit to $\mathrm ds\times\mathrm dt$ that swaps signs when the order swaps (unlike regular multiplication of numbers), then $\mathrm d\mathbf A=\dfrac{\partial\mathbf{r}}{\partial s}\times\dfrac{\partial\mathbf{r}}{\partial t}\,\mathrm ds\mathrm dt=\dfrac{\partial\mathbf{r}}{\partial t}\times\dfrac{\partial\mathbf{r}}{\partial s}\,\mathrm dt\mathrm ds$ since the two minus signs cancel out!

Across multiple different ways of formalizing things, a symbol used for this non-commutative multiplication is $\wedge$. Note that $\mathrm ds\wedge\mathrm ds=-\mathrm ds\wedge\mathrm ds$ by swapping the two factors, so that $\mathrm ds\wedge\mathrm ds=\boldsymbol 0$, and similarly for $\mathrm dt$.

Verification

Let's say $\mathbf r(s,t)=\left\langle x(s,t),y(s,t),z(s,t)\right\rangle$. With this noncommutative multiplication $\wedge$ and orientation mattering, we can justify a version of your formula by using formulas like you might have seen for linear approximations of multivariable functions: e.g. $\mathrm dx=\dfrac{\partial x}{\partial s}\mathrm{d}s+\dfrac{\partial x}{\partial t}\mathrm{d}t$. Then we have:

\begin{align} \mathrm d\mathbf A &=\dfrac{\partial\mathbf{r}}{\partial s}\times\dfrac{\partial\mathbf{r}}{\partial t}\,\mathrm ds\wedge\mathrm dt\\ &=\det\begin{bmatrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\boldsymbol{k}}\\ \dfrac{\partial x}{\partial s} & \dfrac{\partial y}{\partial s} & \dfrac{\partial z}{\partial s}\\ \dfrac{\partial x}{\partial t} & \dfrac{\partial y}{\partial t} & \dfrac{\partial z}{\partial t} \end{bmatrix}\,\mathrm{d}s\wedge\mathrm{d}t\\ &=\left(\left(\dfrac{\partial y}{\partial s}\dfrac{\partial z}{\partial t}-\dfrac{\partial z}{\partial s}\dfrac{\partial y}{\partial t}\right)\hat{\boldsymbol{\imath}}+\left(\dfrac{\partial z}{\partial s}\dfrac{\partial x}{\partial t}-\dfrac{\partial x}{\partial s}\dfrac{\partial z}{\partial t}\right)\hat{\boldsymbol{\jmath}}+\left(\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}-\dfrac{\partial y}{\partial s}\dfrac{\partial x}{\partial t}\right)\hat{\boldsymbol{k}}\right)\mathrm{d}s\wedge\mathrm{d}t \end{align}

And we also have (making sure to switch the order of the “$dxdz$” product into $\mathrm dz\wedge\mathrm dx$):

\begin{align} &\phantom{=}\mathrm{d}x\wedge\mathrm{d}y\hat{\boldsymbol{k}}+\mathrm{d}y\wedge\mathrm{d}z\hat{\boldsymbol{\imath}}+\mathrm{d}z\wedge\mathrm{d}x\hat{\boldsymbol{\jmath}}\\&=\left(\dfrac{\partial x}{\partial s}\mathrm{d}s+\dfrac{\partial x}{\partial t}\mathrm{d}t\right)\wedge\left(\dfrac{\partial y}{\partial s}\mathrm{d}s+\dfrac{\partial y}{\partial t}\mathrm{d}t\right)\hat{\boldsymbol{k}}+\cdots\\&=\left(\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial s}\mathrm{d}s\wedge\mathrm{d}s+\dfrac{\partial x}{\partial t}\dfrac{\partial y}{\partial s}\mathrm{d}t\wedge\mathrm{d}s+\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}\mathrm{d}s\wedge\mathrm{d}t+\dfrac{\partial x}{\partial t}\dfrac{\partial y}{\partial t}\mathrm{d}t\wedge\mathrm{d}t\right)\hat{\boldsymbol{k}}+\cdots\\&=\left(\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial s}*0+\dfrac{\partial x}{\partial t}\dfrac{\partial y}{\partial s}\mathrm{d}t\wedge\mathrm{d}s+\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}\mathrm{d}s\wedge\mathrm{d}t+\dfrac{\partial x}{\partial t}\dfrac{\partial y}{\partial t}*0\right)\hat{\boldsymbol{k}}+\cdots\\&=\left(\dfrac{\partial x}{\partial t}\dfrac{\partial y}{\partial s}\mathrm{d}t\wedge\mathrm{d}s+\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}\mathrm{d}s\wedge\mathrm{d}t\right)\hat{\boldsymbol{k}}+\cdots\\&=\left(\dfrac{\partial x}{\partial t}\dfrac{\partial y}{\partial s}\left(-\mathrm{d}s\wedge\mathrm{d}t\right)+\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}\mathrm{d}s\wedge\mathrm{d}t\right)\hat{\boldsymbol{k}}+\cdots\\&=\left(\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}-\dfrac{\partial y}{\partial s}\dfrac{\partial x}{\partial t}\right)\hat{\boldsymbol{k}}\,\mathrm{d}s\wedge\mathrm{d}t+\cdots=\mathrm{d}\mathbf{A}\checkmark \end{align} A version of this formula is mentioned in Spivak's Calculus on Manifolds and quoted in the question Calculating Surface Area using Differential Forms.

Numbers

If you don't care the direction of the normal area vector $\mathrm d\mathbf A$ (e.g. because you would like to integrate $\mathbf f\Vert\mathrm d\mathbf A\Vert$ for some scalar field $\mathbf f$), then orientation doesn't matter, and we really just care about the length of $\mathrm d\mathbf A$.

Background

If you just cared about volume under the graph of a scalar function, you might imagine adding up volumes of rectangular boxes, and you'd always write single integrals in an iterated integral with increasing bounds (like $\int_{0}^5\int_{-1}^1f(x,y)\,\mathrm dy\,\mathrm dx$ since \int_{5}^0 would obviously give the wrong sign). In this sort of context, something like $\mathrm dx$ would represent a small length, and $\mathrm dx\mathrm dy$ would be the area of small rectangle in (or parallel to) the $xy$-plane.

With this kind of setup, we have to give up on an idea like $\mathrm dx=\dfrac{\partial x}{\partial s}\mathrm{d}s+\dfrac{\partial x}{\partial t}\mathrm{d}t$ since the expression on the right couldn't represent a length if the partials were negative. And $\mathrm{d}x\mathrm{d}y\hat{\boldsymbol{k}}+\mathrm{d}y\mathrm{d}z\hat{\boldsymbol{\imath}}+\mathrm{d}x\mathrm{d}z\hat{\boldsymbol{\jmath}}$ would have three positive components, and so it would always point towards the first octant, even if that's not either direction of $\pm\mathrm d\mathbf A$. But it turns out that we can show that it has the right length under this interpretation anyway.

Verification

Let's say $\mathbf r(s,t)=\left\langle x(s,t),y(s,t),z(s,t)\right\rangle$.

First, let's consider the simple case when $z(s,t)$ is a constant $z_0$. Then $\mathrm dA=\left(\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}-\dfrac{\partial y}{\partial s}\dfrac{\partial x}{\partial t}\right)\hat{\boldsymbol{k}}$ with length $\left(\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}-\dfrac{\partial y}{\partial s}\dfrac{\partial x}{\partial t}\right)^2$.

Since $z(s,t)$ is constant, $\mathrm dz$ (a small change in the $z$ coordinate as we move along the surface) is $0$, so the candidate vector reduces to $\mathrm{d}x\mathrm{d}y\hat{\boldsymbol{k}}$. Now, if we had to integrate $f(x,y,z)$ along the surface, we could substitute in $z=z_0$ and worry about $\int f\left(x,y,z_0\right)\,\mathrm dx\mathrm dy$ for $(x,y)$ in the projection of the surface. By multivariable substitution/change of variables, we could use a Jacobian determinant to rewrite $$\int f\left(x,y,z_0\right)\,\mathrm dx\mathrm dy=\int f\left(x(s,t),y(s,t),z_0\right)\,\left|\det\begin{bmatrix}\dfrac{\partial x}{\partial s}&\dfrac{\partial x}{\partial t}\\\dfrac{\partial y}{\partial s}&\dfrac{\partial y}{\partial t}\end{bmatrix}\right|\mathrm ds\mathrm dt\text{.}$$ In other words, $$\mathrm dx\mathrm dy=\left|\det\begin{bmatrix}\dfrac{\partial x}{\partial s}&\dfrac{\partial x}{\partial t}\\\dfrac{\partial y}{\partial s}&\dfrac{\partial y}{\partial t}\end{bmatrix}\right|\mathrm ds\mathrm dt=\left|\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}-\dfrac{\partial y}{\partial s}\dfrac{\partial x}{\partial t}\right|\mathrm ds\mathrm dt\text{.}$$ Thus, $\mathrm{d}x\mathrm{d}y\hat{\boldsymbol{k}}=\pm\mathrm d\mathbf A$ and the length is still $\left(\dfrac{\partial x}{\partial s}\dfrac{\partial y}{\partial t}-\dfrac{\partial y}{\partial s}\dfrac{\partial x}{\partial t}\right)^2$.$\checkmark$

Now, consider the general case. Assuming $\mathbf r$ is nice and $\Delta s$ and $\Delta t$ are small, the four points $\mathbf r(s,t)$, $\mathbf r(s+\Delta s,t)$, $\mathbf r(s,t+\Delta t)$, and $\mathbf r(s+\Delta s,t+\Delta t)$ will approximate the vertices of a parallelogram.

It's a geometric fact that I won't reprove here that the square of the area of a parallelogram in 3D space is the sum of the squares of the areas of the projections onto the three coordinate planes. (A natural generalization is proven in Beyond Monge's Theorem: A Generalization of the Pythagorean Theorem. And this specific case is mentioned in Mikhail Katz's answer to Is it true that $d\textbf{S} = dy dz\textbf{ i }+ dx dz\textbf{ j }+ dx dy\textbf{ k }$. A closely related fact is shown in this answer to Area of projected parallelogram onto a plane.)

For each coordinate plane, an argument like the one I gave for the constant $z(s,t)$ case applies in discussing the area element of the projection onto that plane. So the three components of $\mathrm{d}x\mathrm{d}y\hat{\boldsymbol{k}}+\mathrm{d}y\mathrm{d}z\hat{\boldsymbol{\imath}}+\mathrm{d}x\mathrm{d}z\hat{\boldsymbol{\jmath}}$ have the right squares for the length to equal the length of $\mathrm d\mathbf A$, as desired.