Torus coordinates are given by $$x=(R+r \cos\theta)\cos\phi,\space y=(R+r \cos\theta)\sin\phi,\space z=r\sin\theta,$$ with $R>0$.
A given torus $T_b$ is defined by $$T_b=\{(r,\theta,\phi):0\leq r\leq b<R, 0\leq\theta\leq 2\pi,0\leq\phi\leq 2\pi\}.$$ Is it correct that the volume is given by $$\int_0^b\int_0^{2\pi}\int_0^{2\pi}1 Id\phi d\theta dr$$ where $I$ is the integrationfactor of the torus coordinates? How do I find $I$?
P.S. Is there an easy way to sketch this torus in cartesian coordinates?
Let's parametrize the torus (parametrize the volume in space). We can parametrize a circumference in the $y=0$ plane, with radius $r$, centered in the point $(R,0,0)$ by $$\alpha(u,v) = (R + u\cos v, 0, R+ u \sin v), \quad 0 < u < r, \quad 0 < v < 2\pi,$$ and rotating around the $z$ axis we have: $${\bf x}(u,v,w) = ((R+u\cos v)\cos w, (R+u \cos v)\sin w, R+u\sin v),$$ with $0 < w < 2\pi$. The oriented of the torus is: $$V = \int_0^{2\pi} \int_0^{2\pi} \int_0^r |\det({\bf x}_u,{\bf x}_v,{\bf x}_w)|\,{\rm d}u\,{\rm d}v\,{\rm d}w.$$
I don't think there is an easy way to look at this using cartesian coordinates.