Working with differential forms

Question

The spherical to cartesian transformations are usually given in by the relation of coordinates, lets consider only the coordinate $x$ $$ \label{1} x = r \sin \theta \cos \phi . \tag{1} $$ Sometimes they are given by their differential, applying differential $d$ to equation \ref{1} we get $$ dx = -r \sin (\theta ) \sin (\phi ) d\phi+r \cos (\theta ) \cos (\phi )d\theta +\sin (\theta )\cos (\phi ) dr . $$ Now since by definition (using Einstein notation) $$ dx = \frac{\partial x}{\partial \xi^i}d\xi^i, $$ for $x = x(\xi^i)$,$\xi^i = (r,\theta,\phi)$ we should be able to recover the equation \ref{1} by $$ x=\int dx = \int \frac{\partial x}{\partial r}dr+ \int \frac{\partial x}{\partial \theta}d\theta + \int \frac{\partial x}{\partial \phi}d\phi. $$ But actually doing this I get $$ x = 3 r \sin (\theta ) \cos (\phi ). $$ Which differs from \ref{1} by a factor of $3$. Why is this?

Answer 1

As mentioned in a comment, the equation $$ x = \int \frac{\partial x}{\partial r} dr + \int \frac{\partial x}{d\theta} d\theta + \int \frac{\partial x}{d\phi} d\phi $$ doesn't make any sense. But even if we try to make sense of it, it's incorrect. You would expect each indefinite integral on the RHS to be equal to $x + C$ by the fundamental theorem of calculus, so we recover the equation $$ x = 3x + C, $$ which is the result you got. Note that you can replace $x$ here with any function $f : \mathbb{R}^3 \to \mathbb{R}$, and the same "reasoning" shows that $f = 3f + C$.

To correct this situation, note that you cannot take the indefinite integral of a differential form (in general, a differential form does not have an antiderivative), so we need to be explicit over what we're integrating over. By the fundamental theorem of line integrals, we have $$ \int_\gamma dx = x(\gamma(1)) - x(\gamma(0)), $$ where $\gamma : [0, 1] \to \mathbb{R}^3$ is any path. Let's consider a path $\gamma = (r, \theta, \phi)$ given in spherical coordinates. Then our computation becomes $$ \begin{aligned} \int_\gamma \frac{\partial x}{\partial r}dr &= \int_\gamma \sin(\theta)\cos(\theta)dr = \int_0^1 \sin(\theta(t))\cos(\theta(t))r'(t)dt \\ \int_\gamma \frac{\partial x}{\partial \theta}d\theta &= \int_\gamma r\cos(\theta)\cos(\phi)d\theta = \int_0^1 r(t)\cos(\theta(t))\cos(\phi(t))\theta'(t)dt \\ \int_\gamma \frac{\partial x}{\partial \phi}\,d\phi &= -\int_\gamma r\sin(\theta)\sin(\phi)d\phi = -\int_0^1 r(t)\sin(\theta(t))\cos(\phi(t))\phi'(t)dt, \end{aligned} $$ and we get $$ \int_\gamma \frac{\partial x}{\partial r} dr + \int_\gamma \frac{\partial x}{d\theta} d\theta + \int_\gamma \frac{\partial x}{d\phi} d\phi = \int_0^1 (x \circ \gamma)'(t)\,dt = x(\gamma(1)) - x(\gamma(0)) = \int_\gamma dx, $$ which is true.