Find $dr\wedge d\alpha \wedge d\beta$ in terms of $dx \wedge dy \wedge dz$.

Question

\Let $$x=r\cos(\alpha)\sin(\beta)$$ $$y=r\sin(\alpha)\sin(\beta)$$ $$z=r\cos(\beta)$$ be the spherical coordinates. Find $dr\wedge d\alpha \wedge d\beta$ in terms of $dx \wedge dy \wedge dz$.

I have never seen a question like this so I'm a bit confused. Do I need to differentiate $x,y,z$ respectively to $r,\alpha,\beta$ ?

My first approach was $xdx=r\cos(\alpha)\sin(\beta)dx$ , $ydy=r\sin(\alpha)\cos(\beta)dy$ and $zdz=r\cos(\beta)dz$.

$(xyz) dx\wedge dy\wedge dz=(r^3\cos(\alpha)\cos^2(\beta)\sin(\alpha)\sin(\beta))dx\wedge dy\wedge dz$.

Is this a right path to follow or am I missing something? Thanks!

Answer 1

I will compute the expression in opposite direction. i.e. attempt to express $dx \wedge dy \wedge dz$ in terms of $dr\wedge d\alpha \wedge d\beta$ first.

$$\begin{align} & dx \wedge dy \wedge dz\\ = & \frac1z dx \wedge dy \wedge zdz\\ = & \frac1z dx \wedge dy \wedge (xdx + ydy + zdz)\\ = &\frac{1}{2z} dx\wedge dy \wedge d(x^2+y^2+z^2)\\ = & \frac{1}{2z} dx\wedge dy \wedge dr^2\\ = & \frac{r}{z} d(x \wedge dy \wedge dr\\ = & \frac{1}{\cos\beta}d(r\cos\alpha\sin\beta) \wedge d(r\sin\alpha\sin\beta) \wedge dr\\ = &\frac{r^2}{\cos\beta}d(\cos\alpha\sin\beta)\wedge d(\sin\alpha\sin\beta) \wedge dr \end{align}$$ Notice $$\begin{align} & d(\cos\alpha\sin\beta)\wedge d(\sin\alpha\sin\beta)\\ = & (-\sin\alpha\sin\beta d\alpha + \cos\alpha\cos\beta d\beta) \wedge (\cos\alpha\sin\beta d\alpha + \sin\alpha\cos\beta d\beta)\\ = & \sin\beta\cos\beta( -\sin^2\alpha^2 - \cos^2\alpha) d\alpha\wedge d\beta)\\ = & -\sin\beta\cos\beta d\alpha\wedge d\beta\end{align}$$

We obtain

$$dx \wedge dy \wedge dz = -\frac{r^2}{\cos\beta}\sin\beta\cos\beta d\alpha\wedge d\beta\wedge dr = - r^2 \sin\beta dr\wedge d\alpha \wedge d\beta$$

From this, it is easy to see $$dr \wedge d\alpha \wedge d\beta = -\frac{dx \wedge dy \wedge dz}{\sqrt{(x^2+y^2+z^2)(x^2+y^2)}}$$

Answer 2

Hint: the functions $x,y,z$ depend on $r,\alpha,\beta$, so their differentials are \begin{align*} dx &= \cos \alpha \sin \beta dr-r\sin\alpha\sin \beta d\alpha+ r\cos \alpha \cos \beta d\beta, \\ dy &= \sin \alpha \sin \beta dr+r\cos \alpha \sin \beta d\alpha + r\sin \alpha \cos \beta d\beta, \\ dz &= \cos \beta dr - r\sin \beta d\beta. \end{align*} You can write this as a system: $$ \left( \begin{matrix} dx \\ dy \\ dz \end{matrix} \right) = \left( \begin{matrix} \cos \alpha \sin \beta & -r \sin \alpha \sin \beta & r\cos \alpha \cos \beta \\ \sin \alpha \sin \beta & r\cos \alpha \sin \beta & r\sin \alpha \cos \beta \\ \cos \beta & 0 & -r\sin \beta \end{matrix} \right) \left( \begin{matrix} dr \\ d\alpha \\ d\beta \end{matrix} \right). $$ Now you need to invert this system. This way you obtain explicit expressions of $dr,d\alpha$, and $d\beta$. Assume you find $dr = adx+bdy+cdz, d\alpha = a'dx+b'dy+c'dz$, then the rule to compute $dr \wedge d\alpha$ is $$dr \wedge d\alpha = (ab'-a'b)dx\wedge dy+(ac'-a'c)dx \wedge dz + (bc'-b'c)dy \wedge dz.$$ From here you can compute $dr \wedge d\alpha \wedge d\beta$.

Answer 3

The first step is to compute the differentials $dy^i = \sum_j {\partial y^i\over\partial x^j}\,dx^j$, that is, $$dx = \cos\alpha\sin\beta\,dr - r\sin\alpha\sin\beta\,d\alpha + r\cos\alpha\cos\beta\,d\beta$$ and so on. You could then grind through computing all of those wedge products term by term, but a simpler approach is to relate the wedge product to a determinant, as in this question. In this case, we have $$dr\wedge d\alpha\wedge d\beta = {\partial(r,\alpha,\beta)\over\partial(x,y,z)}\,dx\wedge dy\wedge dz,$$ so the problem reduces to computing the Jacobian determinant of the coordinate transformation. This is the analog of the change-of-variables formula for integrals that you probably learned in a previous calculus course. Instead of inverting the coordinate transformation you’ve been given, it’s easier to compute its Jacobian determinant and invert that, then rewrite it in terms of $x$, $y$ and $z$ if needed.

In fact, what you’re being asked to do is to compute the pullback of the volume form on $\mathbb R^3$ under the map $f:(r,\alpha,\beta)\mapsto(r\cos\alpha\sin\beta,r\sin\alpha\sin\beta,r\cos\beta)$. There’s a somewhat complex explicit general formula for the pullback of a $k$-form, but for an $n$-form in $\mathbb R^n$ it reduces to a single Jacobian as above.