Lets consider the following coordinates on $\Bbb R^3$: $$x_1=r\cos(\xi),x_2=r\sin(\xi) w_2,x_3=r\sin(\xi) w_3$$ with $w_2^2+w_3^2=1,r=||x||$ and $0\leq \xi\leq \pi$. Further let $f$ be a function on $\Bbb R^3$ depending only on $r$ and $\xi$. I want to calculate the Laplacian of $f$ in the coordinates above.
In the paper I am reading this Laplacian is given but I dont get how they do it. It only says that $$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2d\xi^2+r^2\sin(\xi)^2(dw_2^2+dw_3^2)$$ and therefore, since $\Delta f=\frac{1}{\sqrt g}\sum_k \partial_k(\sum_i g^{ik}\sqrt g\partial_if)$, $$\Delta f=\frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial}{\partial r}f+\frac{1}{r^2}\frac{1}{\sin(\xi)}\frac{\partial}{\partial \xi}\sin(\xi)\frac{\partial}{\partial \xi}f.$$ Could someone explain in some more detail what is happening here? What do I have to calculate? Thanks in advance for any help.
Edit: I am able to obtain the first equation now since $$dx_1=\cos(\xi)dr-r\sin(\xi)d\xi\\ dx_2=\sin(\xi)w_2dr+r\cos(\xi)w_2d\xi+r\sin(\xi)dw_2\\ dx_3=\sin(\xi)w_3dr+r\cos(\xi)w_3d\xi+r\sin(\xi)dw_3$$ so that $$dx_1^2+dx_2^2+dx_3^2=dr^2+r^2d\xi^2+r^2\sin(\xi)^2(dw_2^2+dw_3^2)$$ follows be squaring each row and using $w_2^2+w_3^2=\cos(\xi)^2+\sin(\xi)^2=1$. Now if I use the matrix $$\begin{pmatrix} 1&0&0\\ 0&r^2&0\\ 0&0&r^2\sin(\xi)^2 \end{pmatrix}$$ as matrix tensor $g=g_{ij}dx_i\otimes dx_j$ in the new coordinates I obtain the desired result by the formula for the Laplacian above.
However I am not sure how rigorous this is since $dr,d\xi,dw_2,dw_3$ don't form a basis of the cotangent bundle since we have the restriction $w_2^2+w_3^2=1$ (note that I also only added one $r^2\sin(\xi)^2$ term on the diagonal of $g$ despite I have it for $w_2$ and $w_3$). Do I have to write $w_2,w_3$ as $\cos(t),\sin(t)$ at first, so that I have the same number of variables or is there another way?