infinitesimal volume element - solve for function V if dV = dxdydz?

Question

I have always struggle with infinitesimal equations and obtaining it from a proper equation.

But I recently came across this answer Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio? which helped me alot. Now I can calculate $df$ if $f$ is function and its definition is known. So

if $\overrightarrow{r}:\left(x,y,z\right)\longmapsto\left\langle x,y,z\right\rangle$ then $d\overrightarrow{r}=\left\langle dx,dy,dz\right\rangle$

if $\overrightarrow{r}:\left(r,\theta,\phi\right)\longmapsto\left\langle rsin\theta cos\phi,rsin\theta sin\phi,rcos\theta\right\rangle$ then $d\overrightarrow{r}=dr\hat{r}+rd\theta\hat{\theta}+rsin\theta d\phi\hat{\phi}$

I cant seem to obtain $dV=dxdydz$ with same reasoning. I have tried $V:\left(x,y,z\right)\longmapsto xyz$ .

How do I obtain it ?

Answer 1

Let $e_1, e_2, e_3$ be a set of Cartesian basis vectors. Let $s = r e_1 + \theta e_2 + \phi e_3$ be a vector in some 3d vector space. If you can imagine, this is like having a stretched and deformed coordinate system where the spherical coordinates are instead rectangular.

This is clearly not the vector space we're used to, but there is a diffeomorphism--an invertible, differentiable map--between this vector space and the usual one. Let that map be denoted $f$:

$$f(s) = s' = r \sin \theta \cos \phi e_1 + r \sin \theta \sin \phi e_2 + r \cos \theta e_3 = x e_1 + y e_2 + z e_3$$

The Jacobian $\underline f$ of this map (which I will not calculate) gives the coordinate basis vectors of Euclidean space with respect to spherical coordinates.

$$\begin{align*} \underline f(e_1) &= e_r = \hat r \\ \underline f(e_2) &= e_\theta = r \hat \theta \\ \underline f(e_3) &= e_\phi =( r \sin \theta) \hat \phi \end{align*}$$

When you write something like $d\ell$ in an integral, you're writing it in terms of these spherical basis vectors $e_r, e_\theta, e_\phi$. You are not writing them in terms of unit vectors. Rather, you have

$$d\ell = e_r \, dr + e_\theta \, d\theta + e_\phi \, d\phi$$

If $\ell(t)$ is some curve, then what you're saying is that

$$\frac{d\ell}{dt} = e_r \frac{dr}{dt} + e_\theta \frac{d\theta}{dt} + e_\phi \frac{d\phi}{dt}$$

The quantity $d\ell/dt$ is a "tangent vector". It is literally the vector tangent to the curve at a given point. This is the key to understand about integration in multiple variables: we're always using some curve or surface or whatever and integrating by considering a small piece that is tangent to the curve or surface or whatever and summing over many such small pieces.

Volume integrals are no different. At each point, we take an infinitesimal volume. We build up these volumes by using three curves through a point, so their tangent vectors form a basis for the 3d space. The most convenient thing to do is to use the coordinate lines themselves.

Finally, we build up a representation of a tangent volume using what's called a "wedge product". That is, we define

$$dV \equiv (e_r \, dr) \wedge (e_\theta \, d\theta) \wedge (e_\phi \, d\phi) = (e_r \wedge e_\theta \wedge e_\phi) dr \, d\theta \, d\phi$$

(Note: in differential forms, the quantity $e_r \, dr$ is sometimes confusingly called $dr$ by itself. This, to me, is massively misleading.)

Wedge products are made to build up generalizations of vectors called $k$-vectors. The 3-vector $e_r \wedge e_\theta \wedge e_\phi$ can be written in terms of the unit vectors instead:

$$e_r \wedge e_\theta \wedge e_\phi = r^2 \sin \theta (\hat r \wedge \hat \theta \wedge \hat \phi) = \epsilon r^2 \sin \theta$$

$\epsilon$ is seen as the unit right-handed 3-vector. This choice of symbol is intentional; the components of $\epsilon$ are those of the Levi-Civita tensor. Any three unit vectors that are orthogonal form through wedges a 3-vector that is either $\pm \epsilon$. In particular, if you used Cartesian coordinates instead, you get

$$dV = (e_x \, dx ) \wedge (e_y \, dy) \wedge (e_z \, dz) = (\hat x \wedge \hat y \wedge \hat z) \, dx \, dy \, dz = \epsilon (dx \, dy \, dz)$$

Clearly, then, we have

$$dV = \epsilon (dx \, dy \, dz) = \epsilon r^2 \sin \theta \, (dr \, d\theta \, d\phi)$$

In multivariable calculus, you're not even prepared to deal with an object like $\epsilon$, so it is often silently thrown away, and all you're offered in explanation is that the volume $dV$ transforms according to determinants of Jacobians and such and such. All very true, but by throwing away $\epsilon$, the geometric picture of what is being done--how a tangent volume is constructed using the coordinate basis vectors--is utterly lost.

Forming volumes with the coordinate basis vectors is the very origin of how volume elements transform in new coordinate systems. The wedge product is key to doing this; vector calculus merely imitates this using the scalar triple product.

Answer 2

I'll treat polar coordinates, just so that I can deal with fewer variables.

The important thing is that we have scalar fields $x, y, r, \theta$ satisfying the equations $x = r \cos \theta$ and $y = r \sin \theta$. At this point, it doesn't matter so much on what they are fields, just that they are, and have this relationship.

From this, we can compute equations that their differentials satisfy:

$$ dx = \cos \theta \, dr - r \sin \theta \, d\theta $$ $$ dy = \sin \theta \, dr + r \cos \theta \, d\theta $$

These are also called "1-forms". While you can multiply a scalar field (which can also be called 0-forms) by a 1-form in the "normal" way, you can't do the same with two 1-forms. Instead, there is something called the wedge product, and you can take the wedge product of an $m$-form with an $n$-form to get an $(m+n)$-form.

In particular, $dx \wedge dy$... or just $dx \, dy$ for short... is a $2$-form.

The wedge product behaves similarly to the ordinary product, with one key difference: if $\omega$ and $\eta$ are both $1$-forms, then $\omega \eta = -\eta \omega$. So you have to pay attention to the order in which things are multiplied. More generally, if $\omega$ is an $m$-form and $\eta$ is an $n$-form, then $\omega \eta = (-1)^{mn} \eta \omega$.

Note that this means $(dx)^2 = 0$, and similarly for any $1$-form.

In particular, you can expand products in the normal way. So we can write

$$ \begin{align}dx \, dy &= (\cos \theta \, dr - r \sin \theta \, d\theta) (\sin \theta \, dr + r \cos \theta \, d\theta ) \\&= (\cos \theta \, dr) (\sin \theta \, dr) - (r \sin \theta \, d\theta) (\sin \theta \, dr) + \ldots \\&= \cos \theta \sin \theta \, (dr)^2 - r \sin^2 \theta \, d\theta \, dr + \ldots \\&= (r \sin^2 \theta + r \cos^2 \theta) \, dr \, d\theta \\ &= r \, dr \, d\theta \end{align}$$

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Now, sometimes we might consider two different spaces. We might consider $E$, the ordinary Euclidean plane, with coordinate functions (i.e. scalar fields) $x$ and $y$, with the property that any point satisfies

$$ P = (x_P, y_P) $$

where $x_P$ is notation for the value of $x$ at the point $P$: i.e. for $x(P)$ if we view $x$ as a function. So we might just say that a generic point is $(x,y)$.

We might also have the space $X$ which is again a Euclidean plane, but this one has the coordinate functions $r$ and $\theta$.

And, as you wanted to do, we have a function $X \to E$ defined by

$$ f(r, \theta) = (r \cos \theta, r \sin \theta) $$

Or more precisely,

$$ f(P) = f(r_P, \theta_P) = (r_P \cos \theta_P, r_P \sin \theta_P)$$

so that if $Q = f(P)$, then

$$ (x_Q, y_Q) = (r_P \cos \theta_P, r_P \sin \theta_P)$$

We can actually use this to define new scalar fields $\bar{x}$ and $\bar{y}$ on $X$, by the equations

$$\bar{x}_P = x_{f(P)} \qquad \qquad \bar{y}_P = y_{f(P)} $$

Now, $\bar{x}, \bar{y}, r, \theta$ are all scalar fields on the same space $X$. Using $f$ in this way to take scalars on $E$ and turning them into scalars on $X$ is called a "pull back". This also extends to differential forms: in particular, $dx$ pulls back to $d\bar{x}$, and similarly.

So the area form $dx \, dy$ on the Euclidean plane $E$ pulls back to the $2$-form $d\bar{x} \, d\bar{y}$ on $X$. If $U,V$ are subsets of $X$ and $E$ respectively with the property that $f$ is a bijection from $U$ to $V$, then we can actually compute the integrals of a $2$-form (e.g. $dx \, dy$) over the region $V$ by instead computing the integral of the pullback (e.g. $d\bar{x} \, d\bar{y}$) over $U$.

This amounts to one means of doing a rigorous treatment of "change of variables".

By using the calculation above, this tells us

$$ \begin{align} \iint_V f(x,y) \, dx \, dy &= \iint_U f(\bar{x}, \bar{y}) \, d\bar{x} \, d\bar{y} \\ &= \iint_U f(r \cos \theta, r \sin \theta) r\, dr \, d\theta \end{align}$$

This shows how we can use the framework of differential forms to rigorously treat change of variables.

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A dual thing we can do is a push forward. If, rather than differentials, we work with "tangent vectors" -- e.g. a vector field $e_r$ on $X$ that, at every point, is a "unit" vector that points in the direction in which $r$ grows if you hold $\theta$ constant, and similarly for $e_\theta$.

then, you can push these forward, and get new vector fields $\bar{e}_r$ and $\bar{e}_\theta$ on $E$. These are more often written as $f_*(e_r)$, or maybe $df \cdot e_r$ (we can give $df$ a precise meaning as a vector of differential forms, if we wanted to).

This is a very useful tool for many things... but if you're trying to understand differentials so that you can integrate with them, I think it makes things much more complicated to try and express things in terms of this.