As a practice, I compute the scalar curvature of $T^2$ with induced metric of $R^3$. First,give a coordinate $$ u:(x,y)\rightarrow((3+\cos x)\cos y, (3+\cos x)\sin y, \sin x) $$ Then $$ \partial_xu=(-\sin x \cos y,-\sin x\sin y,\cos x) \\ \partial_y u=(-(3+\cos x)\sin y,(3+\cos x)\cos y ,0) $$ Then,metric is $$ g_{11}=1 \\ g_{22}= (3+\cos x)^2 \\ others =0 $$ Then the Christoffel is $$ \Gamma^x_{xx}=-\frac{\sin x}{3+\cos x}~~~~~~others=0 $$
Then ,curvature Tensor is zero, Ricci tensor is zero, and Scalar curvature is zero. But by Gauss-Bonnet theorem, there is not a metric of scalar curvature identity to zero on $T^2$.What wrong is there ?
As Anthony, I have done some wrong in my compute, there is a result by software