Let $a > b > 0$. Consider the torus $\mathbb{T}^2$ obtained by rotating the circle of radius $b$ centered at $(a, 0, 0)$ (lying in the $xz$ plane) around the $z$ axis. It can be parametrized by local charts involving the restriction of the map $(θ, ϕ) → ((a + b \cos ϕ) \cos θ,(a + b \cos ϕ) \sin θ, b \sin ϕ)$ to squares $I × J$ with $|I|, |J| < 2π$. This is same as $z^2 = 4b^2 − (\sqrt{x^2 + y^2}−(a − b))^2$
Compute the Riemannian metric induced on $\mathbb{T^2}$ from $\mathbb{R}^3$ in the local coordinates $(θ, ϕ)$.
The only thing that I know is Riemannian metric on a manifold $M$ is an inner product on $T_pM$ such that for each chart $(U,x)$ on $M$ the functions $g_{ij}=\langle\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\rangle$ are differentiable on $U$. This definition is really abstract and it does make any sense to me to apply it here. Could somebody please tell me how to find that inner product here?
So you have $x_1=\theta$ and $x_2=\phi.$ You have a parametrization $$f(\theta,\phi) = \big((a+b\cos\phi)\cos\theta,(a+b\cos\phi)\sin\theta,b\sin\phi\big).$$ The vector field $\dfrac{\partial}{\partial x_i}$ is nothing fancier than $\dfrac{\partial f}{\partial x_i}$, so compute the vectors $\dfrac{\partial f}{\partial\theta}$ and $\dfrac{\partial f}{\partial\phi}$ and calculate the dot products!