Is the warped product: $dr^{2} + r^{2} \gamma$ always flat in 3 dimensions? Suppose you have a Riemannian 2-sphere $(S^{2} , \gamma)$. Define a metric $g$ on $M = S^{2} \times [1,\infty)$ in this way:
If $(x_{1}, x_2)$ is a coordinate chart on $S^{2}$ then 
$$g = dr^{2} + r^{2}\gamma_{ij}dx^i dx^j$$
Where $i,j = 1,2$. Notice that $\gamma_{ij}$ does not depend on $r$ and so this is a warped product. My question is: is this metric flat? And is this metric the Euclidean metric on $M$?
Let us fix a coordinate chart on $S^{2}$ call it $(\theta, \phi) \in U$ where $U$ is some open set of $\mathbb{R}^2$. Then we are looking at $(U\times [1,\infty) , g)$ where $g = dr^2 + \tilde{\gamma}$ where $\tilde{\gamma} = r^2 \gamma$.
I think this means that the hypersurfaces $\{ r = r_{0} \}$ are all totally umbilical since the second fundamental form $K$ satisfies:
$$K_{ij} = -\frac12 \frac{d}{dr} \tilde{\gamma} = -\frac1r \tilde{\gamma} $$ 
So the principle curvatures are $-\frac1r $ and so the Gaussian curvature is $\frac{1}{r^2} $ . Using this, I computed the Rici curvature components of M and found them to be $0$ (But I don’t trust my computations at all). This implies that the Rieman curvature tensor vanishes on $M$ and so $M$ is flat (since the Weyl part of Rieman curvature tensor vanishes for 3-dim manifolds, the vanishing of the Ricci curvature implies the vanishing of the Rieman Curvature tensor). Is there any mistake in what I am doing?
 A: There is an error in the computation of the Ricci curvature, but without more details it's not possible to say where.
From OP's comment under the question it appears that $\gamma$ is an arbitrary metric on $S^2$. The coordinates are local, so this question is more or less just asking about the behavior of the cone metric $$\hat \gamma := dr^2 + r^2 \pi^* \gamma$$ over a Riemannian surface $(\Sigma, \gamma)$ on $\hat\Sigma := \Bbb R_+ \times \Sigma$. Here, $\pi : \hat\Sigma = \Bbb R_+ \times \Sigma \to \Sigma$ is just the canonical projection onto the second factor. Computing directly$^*$ gives that the respective Ricci curvatures $\operatorname{Ric}, \widehat{\operatorname{Ric}}$ of $\gamma, \hat\gamma$ are related by
$$\widehat{\operatorname{Ric}} = \pi^*(\operatorname{Ric} - \gamma) ,$$
and then forming traces gives that the respective scalar curvatures $R, \hat R$ are related by
$$\hat R = r^{-2} \pi^* (R - 2) .$$
These two identities immediately give that $$\widehat{\operatorname{Ric}} = 0 \Leftrightarrow \operatorname{Ric} = \gamma \Leftrightarrow \hat R = 0 \Leftrightarrow R = 2 ,$$ which in particular recovers Yu Ding's observation from the comments.

$^*$One can compute these identities efficiently using isothermal coordinates: There are some local coordinates $(x_1, x_2)$ on $\Sigma$ for which $\gamma = e^{2 \Upsilon} (dx_1^2 + dx_2^2)$ for some smooth function $\Upsilon(x_1, x_2)$. Applying the Koszul Formula yields relations between $\hat\nabla$ and $\nabla$, e.g.,
$$\hat \nabla_{\partial_{x_i}} \partial_{x_j} = \nabla_{\partial_{x_i}} \partial_{x_j} - r \gamma(\partial_{x_i}, \partial_{x_j}) \partial_r ,$$
among simpler relations, and then compute the curvatures directly from these. (In this display equation, we've implicitly used the decompositions $T_{(r, p)} \hat\Sigma = T_r \Bbb R_+ \oplus T_p \Sigma$.)
