In $\mathbb{R}^3$, does $dS = \sqrt{dx^2dy^2+dy^2dz^2+dx^2dz^2}$ hold in surface integrals?

Question

I'm wondering whether the formula above is true or not. It looks like an extension of $dl = \sqrt{dx^2+dy^2+dz^2}$ used in line integrals. I don't see it in my textbook, so it is likely wrong since the formula looks too nice to omit. But if it is true, there are some proven implications.

1. Assume $dS = \sqrt{dx^2dy^2+dy^2dz^2+dx^2dz^2}$ is true. Prove that $$ dS = \sqrt{ \left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial z}{\partial u} \frac{\partial y}{\partial v} \right)^2 + \left( \frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial x}{\partial u} \frac{\partial z}{\partial v} \right)^2 + \left( \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v} \right)^2} dudv\text{.} $$

(The latter formula appears in Gilbert Strang's Calculus (p. 658). I found the book on a MIT's website.)

Let $x = x(u,v)$, $y = y(u,v)$ and $z = z(u,v)$. Then \begin{align*} dS &= \sqrt{dx^2dy^2+dy^2dz^2+dx^2dz^2} \\ &= dudv\sqrt{\frac{dx^2dy^2+dy^2dz^2+dx^2dz^2}{du^2dv^2}} \\ &= dudv\sqrt{\frac{dx^2dy^2}{du^2dv^2} + \frac{dy^2dz^2}{du^2dv^2} + \frac{dx^2dz^2}{du^2dv^2}} \\ &= dudv\sqrt{\left(\frac{dxdy}{dudv}\right)^2 + \left(\frac{dydz}{dudv}\right)^2 + \left(\frac{dxdz}{dudv}\right)^2} \\ \end{align*}

It's known that

$$\frac{dxdy}{dudv} = \left| \frac{D(x,y)}{D(u,v)} \right| = \left| \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} \right| = \left| \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \right|$$

And

$$\frac{dydz}{dudv} = \left| \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial y}{\partial v} \frac{\partial z}{\partial u} \right|$$

$$\frac{dxdz}{dudv} = \left| \frac{\partial x}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial z}{\partial u} \right|$$

Substitute these into the equation of (dS) \begin{align*} dS = dudv\sqrt{ \left( \left| \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \right| \right)^2 + \left( \left| \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial y}{\partial v} \frac{\partial z}{\partial u} \right| \right)^2 + \left( \left| \frac{\partial x}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial z}{\partial u} \right| \right)^2} \end{align*}

Drop the absolute value signs and rearrange. I get $$ dS = \sqrt{ \left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial z}{\partial u} \frac{\partial y}{\partial v} \right)^2 + \left( \frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial x}{\partial u} \frac{\partial z}{\partial v} \right)^2 + \left( \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v} \right)^2} dudv \text{ (Q.E.D)} $$

2. Assume $dS = \sqrt{dx^2dy^2+dy^2dz^2+dx^2dz^2}$ is true, prove that $dS = dxdy$ in $\mathbb{R}^2$.

In $\mathbb{R}^2$, for any pairs ((x,y)), $z = 0$ and $dz = 0$.

Therefore \begin{align*} dS &= \sqrt{dx^2dy^2+dy^2dz^2+dx^2dz^2} \\ &= \sqrt{dx^2dy^2+dy^20^2+dx^20^2} \\ &= \sqrt{dx^2dy^2} \\ &= dxdy \text{ (Q.E.D).} \end{align*}

There are more examples of implications, but I think two are enough to convey my thoughts.

If the formula is incorrect, can you prove it or provide a counterexample? Or can you point out where I went wrong? If it is correct, how can I prove it? Also, since it doesn't look as obvious as $dl = \sqrt{dx^2+dy^2+dz^2}$, how can I visualize its geometric meaning?

Answer 1

$\newcommand{\abs}[1]{\left|{#1}\right|}$ $\newcommand{\scalprod}[2]{\left\langle{#1},\,{#2}\right\rangle}$ $\newcommand{\norm}[1]{\left\Vert{#1}\right\Vert}$ $\newcommand{\transp}[1]{\vphantom{#1}^{\mathbf{t}}{#1}}$

The "actual" formulas for the "length element" $ds$ and for the "surface element" $dS$ are: $$ ds =\sqrt{\det G(dP)},\qquad dS =\sqrt{\det G(d_1^{}P,\,d_2^{}P)}, $$ where $G(\boldsymbol{v}_1,\,\dotsc,\,\boldsymbol{v}_m)$ is the Gram matrix of the family of vectors $(\boldsymbol{v}_1,\,\dotsc,\,\boldsymbol{v}_m)$, $dP$ is a tangent vector to the given curve at a point $P$, and $d_1^{}P$ and $d_2^{}P$ are tangent vectors to the given surface at a point $P$.^[1]

Thus, $$ ds =\sqrt{\scalprod{dP}{dP}} =\norm{dP} $$ and $$ dS =\sqrt{\scalprod{d_1^{}P}{d_1^{}P}\scalprod{d_2^{}P}{d_2^{}P} -\scalprod{d_1^{}P}{d_2^{}P}^2}. $$

Let now $$ P = (x, y, z),\qquad d_1^{}P = (d_1^{}x,\,d_1^{}y,\,d_1^{}z),\qquad d_2^{}P = (d_2^{}x,\,d_2^{}y,\,d_2^{}z), $$ $$ Q = (u, v),\qquad d_1^{}Q = (d_1^{}u,\,d_1^{}v),\qquad d_2^{}Q = (d_2^{}u,\,d_2^{}v), $$ where $P$ is the image of $Q$, and the tangent vectors $d_1^{}P$ are $d_2^{}P$ are the images (under the corresponding tangent/total derivative map) of the vectors $d_1^{}Q$ and $d_2^{}Q$. Let $J$ be the corresponding Jacobian matrix: $$ J =\begin{pmatrix} \dfrac{\partial x}{\partial u} &\dfrac{\partial x}{\partial v}\\[1ex] \dfrac{\partial y}{\partial u} &\dfrac{\partial y}{\partial v}\\[1ex] \dfrac{\partial z}{\partial u} &\dfrac{\partial z}{\partial v} \end{pmatrix}. $$

Then \begin{align*} G(d_1^{}P,\,d_2^{}P) &=\begin{pmatrix} d_1^{}x & d_1^{}y & d_1^{}z \\[1ex] d_2^{}x & d_2^{}y & d_2^{}z \end{pmatrix} \begin{pmatrix} d_1^{}x & d_2^{}x\\[1ex] d_1^{}y & d_2^{}y\\[1ex] d_1^{}z & d_2^{}z \end{pmatrix}\\[2ex] &=\transp{\begin{pmatrix} d_1^{}x & d_2^{}x\\[1ex] d_1^{}y & d_2^{}y\\[1ex] d_1^{}z & d_2^{}z \end{pmatrix}} \begin{pmatrix} d_1^{}x & d_2^{}x\\[1ex] d_1^{}y & d_2^{}y\\[1ex] d_1^{}z & d_2^{}z \end{pmatrix} \end{align*} and $$ \begin{pmatrix} d_1^{}x & d_2^{}x\\[1ex] d_1^{}y & d_2^{}y\\[1ex] d_1^{}z & d_2^{}z \end{pmatrix} = J\begin{pmatrix} d_1^{}u & d_2^{}u\\[1ex] d_1^{}v & d_2^{}v \end{pmatrix}, $$ and therefore, $$ G(d_1^{}P,\,d_2^{}P) =\transp{\begin{pmatrix} d_1^{}u & d_2^{}u\\[1ex] d_1^{}v & d_2^{}v \end{pmatrix}} \transp{J}J \begin{pmatrix} d_1^{}u & d_2^{}u\\[1ex] d_1^{}v & d_2^{}v \end{pmatrix}. $$ Using properties of determinants, this gives: $$ \det G(d_1^{}P,\,d_2^{}P) =\det(\transp{J}J) \begin{vmatrix} d_1^{}u & d_2^{}u\\[1ex] d_1^{}v & d_2^{}v \end{vmatrix}^2 =\det(\transp{J}J) \Bigl((d_1^{}u)(d_2^{}v) - (d_1^{}v)(d_2^{}u)\Bigr)^2. $$ Thus, $$ dS =\sqrt{\det(\transp{J}J)}\,\abs{(d_1^{}u)(d_2^{}v) - (d_1^{}v)(d_2^{}u)}. $$

Now, assume, as usually, that $$ d_1^{}u = du > 0,\quad d_1^{}v = 0,\quad d_2^{}u = 0,\quad d_2^{}v = dv > 0. $$ Then \begin{align*} d_1^{}P &=\frac{\partial P}{\partial u}d_1^{}u +\frac{\partial P}{\partial v}d_1^{}v =\frac{\partial P}{\partial u}du =\left(\frac{\partial x}{\partial u},\,\frac{\partial y}{\partial u},\,\frac{\partial z}{\partial u}\right)du,\\[2ex] d_2^{}P &=\frac{\partial P}{\partial u}d_2^{}u +\frac{\partial P}{\partial v}d_2^{}v =\frac{\partial P}{\partial v}dv =\left(\frac{\partial x}{\partial v},\,\frac{\partial y}{\partial v},\,\frac{\partial z}{\partial v}\right)dv, \end{align*} and \begin{gather*} \det G(d_1^{}P,\,d_2^{}P) =\det(\transp{J}J)(du\,dv)^2,\\[2ex] dS =\sqrt{\det(\transp{J}J)}\,du\,dv. \end{gather*}

Since $$ \det(\transp{J}J) =\left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial z}{\partial u} \frac{\partial y}{\partial v} \right)^2 +\left( \frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial x}{\partial u} \frac{\partial z}{\partial v} \right)^2 +\left( \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v} \right)^2, $$ this gives: $$ dS =\sqrt{ \left( \frac{\partial y}{\partial u} \frac{\partial z}{\partial v} - \frac{\partial z}{\partial u} \frac{\partial y}{\partial v} \right)^2 +\left( \frac{\partial z}{\partial u} \frac{\partial x}{\partial v} - \frac{\partial x}{\partial u} \frac{\partial z}{\partial v} \right)^2 +\left( \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial y}{\partial u} \frac{\partial x}{\partial v} \right)^2 }\,du\,dv. $$

As to the question of whether "$dS =\sqrt{dx^2dy^2+dy^2dz^2+dx^2dz^2}$," I do not see what is meant by "$dx$," "$dy$" and "$dz$," because, a priori, there are $d_1^{}x$, $d_1^{}y$, $d_1^{}z$ and $d_2^{}x$, $d_2^{}y$, $d_2^{}z$, but no any particular "$dx$," "$dy$," or "$dz$." (While "$du$" stands for $d_1^{}u$ and "$dv$" stands for $d_2^{}v$.)

However, if, abusing the Grassmann algebra exterior product notation, we write "$dx\wedge dy$" to mean $(d_1^{}x)(d_2^{}y) - (d_1^{}y)(d_2^{}x)$, and so on: \begin{align*} dx\wedge dy &\stackrel{\text{def}}{=}(d_1^{}x)(d_2^{}y) - (d_1^{}y)(d_2^{}x) =\begin{vmatrix} d_1^{}x & d_2^{}x\\[1ex] d_1^{}y & d_2^{}y \end{vmatrix},\\[2ex] dy\wedge dz &\stackrel{\text{def}}{=} (d_1^{}y)(d_2^{}z) - (d_1^{}z)(d_2^{}y) =\begin{vmatrix} d_1^{}y & d_2^{}y\\[1ex] d_1^{}z & d_2^{}z \end{vmatrix},\\[2ex] dz\wedge dx &\stackrel{\text{def}}{=} (d_1^{}z)(d_2^{}x) - (d_1^{}x)(d_2^{}z) =\begin{vmatrix} d_1^{}z & d_2^{}z\\[1ex] d_1^{}x & d_2^{}x \end{vmatrix}, \end{align*} then I think it is true that $$ dS =\sqrt{(dx\wedge dy)^2 + (dy\wedge dz)^2 + (dz\wedge dx)^2}. $$

The underlying reason is that each dimensional component of the Grassmann algebra of a Euclidian vector space is naturally also a Euclidian vector space. (Though, arguably, the scalar products in the dimensional components are not totally canonical and can be "rescaled" -- $1/m!$ for the $m$-dimensional component being a popular coefficient.) Using the Grassmann algebra with its standard scalar product in each dimension, I think we have the formula: $$ dS =\norm{d_1^{}P\wedge d_2^{}P}, $$ where ($\wedge$) is the actual exterior product.

The possible choices of "rescaling"/"normalisation" of the norm $\norm{d_1^{}P\wedge d_2^{}P}$ here depend on two other choices:

1. whether we want the surface element $dS$ represent the area of a parallelogram, or that of a triangle spanned by $d_1^{}P$ and $d_2^{}P$, and

2. what unit we want to use to measure the area: a square meter, or a triangular meter, or a round meter, etc.

I may extend the details when I have time.

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^[1] Traditionally, instead of "$d_1^{}$," "$d_2^{}$," $\dotsc$ as decorations of "differential variables" (variables that describe objects in the tangent space), when a single "$d$" was not enough, different symbols resembling "$d$" -- such as "$\delta$" -- were used. See, for example, page 387 in Hermann Weyl, Reine Infinitesimalgeometrie (1918), English translation here.

Answer 2

Not a rigorous proof, but an indication of the geometry behind the proposed area differential.

Think of a triangle in three-dimensional space having vertices at $(a,0,0),(0,b,0),(0,0,c)$. We work out the lengths of the sides using the Pythagorean Theorem and then plug into Heron's Formula for the area $S$ of this triangle. After cancelling out a lot of terms from the Heron's Formula radical we end up with

$S^2=\frac14(a^2b^2+a^2c^2+b^2c^2).$

Now project this triangle onto the coordinate planes. It is not difficult to show that the area of the $xy$ projection is $\frac12ab$ and similarly for projections onto the other two coordinate planes. Comparing these results with the full triangle area above, we find that we have basically promoted the Pythagorean Theorem from two dimensions to three!

This result also applies to infinitesimal triangular elements on a general surface, so adding the squares of the coordinate plane projections of the area differential will indeed give the squared area differential on the surface:

$dS^2=dx^2dy^2+dx^2dz^2+dy^2dz^2.$