Proving that $\int_{M}1d\sigma$ is equal $\int_{U}\sqrt{1+\left\Vert \nabla f\right\Vert ^{2}}du_{1}\ldots du_{n-1}$ when M is a graph of f

Question

I am trying to prove that if M is a manifold of n-1 dimension, that can be described as a graph of function $f:U\rightarrow\mathbb{R}$ than it's surface area is $\int_{U}\sqrt{1+\left\Vert \nabla f\right\Vert ^{2}}du_{1}\ldots du_{n-1}$.

So first of all, I know that if we have $$r\left(x_{1},\ldots,x_{n-1}\right)=\left(x_{1},x_{2},\ldots,x_{n-1},f\left(x_{1},\ldots,x_{n-1}\right)\right)$$ and $$G=\begin{pmatrix}\left\Vert \partial_{1}r\right\Vert ^{2} & \left\langle \partial_{1}r,\partial_{2}r\right\rangle & \ldots & \left\langle \partial_{1}r,\partial_{n-1}r\right\rangle \\ \left\langle \partial_{1}r,\partial_{2}r\right\rangle & \left\Vert \partial_{2}r\right\Vert ^{2}\\ \vdots & & \ddots\\ \left\langle \partial_{1}r,\partial_{n-1}r\right\rangle & & & \left\Vert \partial_{n-1}r\right\Vert ^{2} \end{pmatrix}$$ we have $\intop_{M}1d\sigma=\intop_{U}\sqrt{\text{det}G}$ so basically i want to prove $\text{det}G=1+\left\Vert \nabla f\right\Vert ^{2}$ I also realized that $$G_{i,j}=\begin{cases} \partial_{i}f\partial_{j}f & i\neq j\\ 1+\left(\partial_{i}f\right)^{2} & i=j \end{cases}$$ so I tried looking at G-I and proving that it has eigenvalues $\partial_{1}f,\ldots,\partial_{n-1}f$ But i am stuck here, I got that for every vector u i get: $$\left(G-I\right)\left(v\right)=\begin{pmatrix}\partial_{1}f\left(\partial_{1}fv_{1}+\ldots+\partial_{n-1}fv_{n-1}\right)\\ \partial_{2}f\left(\partial_{1}fv_{1}+\ldots+\partial_{n-1}fv_{n-1}\right)\\ \vdots\\ \partial_{n-1}f\left(\partial_{1}fv_{1}+\ldots+\partial_{n-1}fv_{n-1}\right) \end{pmatrix}$$ but i didn't manage to find any eigenvalues.

Any help is welcomed, thank you :)

Answer 1

You are very close to the answer. From your computations, you have shown that $$G=I_{n-1} +\nabla f \otimes \nabla f $$ where $I_{n-1}$ is the $(n-1)$-identity matrix, $\nabla$ is the gradient in $\mathbb R^{n-1}$, and $u\otimes v$ is the outer product of $u$ and $v$. Then it follows from the matrix determinant lemma that $$ \det G = 1 + \nabla f \cdot \nabla f = 1 + \vert \nabla f \vert^2 $$ which completes the proof.