Definition of surface measure and integration on submanifolds?

Question

I'm reading some lecture notes of Terence Tao and at one point he says

Of course we give $S^{n-1}$ the usual surface measure $d\sigma$.

though I couldn't find a definition of "the surface measure", just a Wikipedia article about surface area.

However, I remember that in our Calc III course we introduced integration on submanifolds of $\mathbb R^n$ using the Gramian matrix and I wonder whether I can obtain the surface measure from that framework? In particular, I'm wondering whether the surface measure for some $k$-dimensional submanifold $M$ of $\mathbb R^n$ with parameterization $\varphi\colon U\subseteq\mathbb R^k\to M$ is given by $$\sigma\colon\mathcal B^n\cap M\to[0,\infty],A\mapsto\int_AdS:=\int_{\varphi^{-1}(A)}\sqrt{\det D\varphi(x)^TD\varphi(x)}d\lambda^k(x),$$ where $\lambda^k$ is the Lebesgue-Borel measure on $\mathbb R^k$. References regarding this topic are also welcome.

Comment to Boris Bilich's answer: (I'm appending this to my question instead of commenting under Boris Bilich's answer, because this comment is rather heavy on formulas).

Even though your answer was actually helpful for me in that it helped me find a decent derivation of the surface measure, I think your statement about the integral being dependent on the choice of the parameterization is wrong, because I remember proving the contrary in Calc III.

Consider the map $\psi(x):=\varphi(\lambda x),\lambda\neq0$ from your answer, where $\varphi$ is the parameterization from my question above. For $A=\varphi(B)=\psi(\frac1\lambda B)$ we have $\varphi^{-1}(A)=B=\lambda\psi^{-1}(A)$ and $$ D\psi(x) =D(\varphi\circ\lambda)(x) =D\varphi(\lambda x)\circ D\lambda(x) =\lambda D\varphi(\lambda x). $$ I think you already see, where this is going. Using the transformation theorem the $\lambda$ in the argument of the derivative will cancel the $\lambda$ outside the derivative: $$ \begin{align*} &\int_{\psi^{-1}(A)}\sqrt{\det D\psi(x)^TD\psi(x)}d\lambda^k(x) \\ &=\int_{\psi^{-1}(A)}\sqrt{\det\lambda^2D\varphi(\lambda x)^TD\varphi(\lambda x)}d\lambda^k(x) \\ &=\int_{\psi^{-1}(A)}\sqrt{\det D\varphi(\lambda x)^TD\varphi(\lambda x)}\cdot\underbrace{|\det D\lambda(x)|}_{=|\lambda|^k}d\lambda^k(x) \\ &=\int_{\lambda\psi^{-1}(A)}\sqrt{\det D\varphi(x)^TD\varphi(x)}d\lambda^k(x) \\ &=\int_{\varphi^{-1}(A)}\sqrt{\det D\varphi(x)^TD\varphi(x)}d\lambda^k(x) \end{align*} $$

Answer 1

According to your definition measure depends strongly on parametrization i.e. if you choose another parametrization $\phi(x)=\varphi(\lambda x)$ for some real $\lambda$ the measure would be multiplied by $\lambda$ too.

Let $\iota \colon M \rightarrow \mathbb{R}^n$ be inclusion map. It induces differentials on tangent spaces, which can be used to define Riemann metric on M: if $v,w\in T_p M,~g(v,w)=(d\iota~ v, d\iota~ w)$. Then Riemann metric can be used to define volume form on M.

For details I recommend the book "Smooth Manifolds" by John Lee.