Computing volume element in spherical coordinates

Question

Suppose $y = (r, \theta^1, \theta^2)$ are spherical coordinates in $(\mathbb{R}^3,g)$. What is the $d\text{vol}$ in these coordinates?

I solved it but I don't know if it's right.

My solution:

We want $d\text{vol} = \sqrt {\det (g^y)} dr \, d\theta^1 \, d\theta^2$

Let $x = (x^1,x^2,x^3)$ be cartesian coordinates in $\mathbb{R}^3$. Then

$$x^1 = r\cos\theta^1\sin\theta^2; \quad x^2 = r\sin\theta^1\sin\theta^2; \quad x^3 = r\cos\theta^2$$

I know that $\det (g^y) = \det (g^x) (\det J)^2$, where $\displaystyle J_{ij} = \frac{\partial x^j}{\partial y^i}$. It follows that

$$g^x_{ij} = \left\langle \frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j} \right\rangle_{\mathbb{R}^3} = \delta_{ij} \\ \implies g^x = \text{Id} \\ \implies \det (g^y) = (\det J)^2 \\ \implies \sqrt{\det (g^y)} = |\det J| $$

Then $d\text{vol} = |\det J| \, dr \, d\theta^1 \, d\theta^2$ where $\det J = -r^2 \sin(\theta^2) (\sin(\theta^1) \sin(\theta^2)+\cos^2(\theta^2))$

Is it correct?

Answer 1

For the record, another solution: put $$(x,y,z) = \Phi(r,\theta,\varphi) \doteq (r\cos\theta\cos\varphi,r\cos\theta\sin \varphi,r \sin \theta),$$for $(r,\theta,\varphi) \in \Bbb R_{>0} \times \left]-\pi,\pi\right[ \times \left]0,2\pi\right[$. Then $$\begin{align}\Phi_r(r,\theta,\varphi) &=(\cos\theta\cos\varphi,\cos\theta\sin \varphi, \sin \theta) \\ \Phi_\theta(r,\theta,\phi) &= (-r\sin\theta\cos\varphi,-r\sin\theta\sin \varphi,r \cos \theta) \\ \Phi_\varphi(r,\theta,\varphi)&= (-r\cos\theta\sin\varphi,r\cos\theta\cos \varphi,0).\end{align}$$So identifying $r \leftrightarrow 1$, $\theta \leftrightarrow 2$ and $\varphi \leftrightarrow 3$, we have: $$(g_{ij})_{i,j=1}^3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \cos^2\theta\end{pmatrix}.$$This tells us that $$\Phi^*g = {\rm d}r^2 + r^2\,{\rm d}\theta^2 + r^2\cos^2\theta\,{\rm d}\varphi^2 \quad \mbox{and}\quad {\rm d}({\rm vol}) = r^2 \cos \theta\;{\rm d}r \wedge {\rm d}\theta \wedge {\rm d}\varphi.$$

Answer 2

If this is in $\mathbb{R}^3$ then use the vector triple product.

Given a volume with point location vector ${\bf r}(\alpha,\beta,\gamma)$, a volume element is described by

$$ {\rm d} V = \left( \frac{{\rm d}{\bf r}}{{\rm d}\alpha} \cdot \left( \frac{{\rm d}{\bf r}}{{\rm d}\beta} \times \frac{{\rm d}{\bf r}}{{\rm d}\gamma} \right) \right) {\rm d} \alpha\,{\rm d} \beta \, {\rm d} \gamma $$

Example:

$${\rm r}(t,\varphi,\theta) = \pmatrix{t \cos \varphi \cos\theta \\ t \sin \varphi \\ t \cos\varphi \sin \theta} $$

yields

$${\rm d}V = (t^2 \cos\varphi) {\rm d}t\,{\rm d}\varphi \, {\rm d}\theta $$