2D surface with Ricci scalar R = 2

[Giammy85]

A two-dimensional Rienmannian manifold has a metric given by
ds^2=e^f dr^2 + r^2 dTHETA^2
where f=f(r) is a function of the coordinate r

Eventually I calculated that Ricci scalar is R=-1/r* d(e^-f)/dr

if e^-f=1-r^2 what this surface is?


In this case R comes to be equal to 2
I've read on wikipedia that Ricci scalar of a sphere with radius r is equal to 2/r^2

So, is this surface a sphere or radius r=1?

 

[Dick]

Yes, it is. If you want an explicit coordinate representation in 3-space, it's x=r*cos(theta), y=r*sin(theta), z=sqrt(1-r^2).