Let $(M,g)$ be a Riemannian manifold, $(U,\varphi), (V,\psi)$ be two charts and $\phi:U\rightarrow V$ be some isometry. I would like to prove the formula $$\int_Vf dV_g=\int_U(f\circ\phi)dVg$$
I tried to prove this formula through the definition of the integral as follows: \begin{align*} \int_Vf dV_g&=\int_{\psi(V)}f(\psi^{-1}(x))\sqrt{\det g(\partial_i \psi,\partial_j\psi)_{\vert\psi^{-1}(x)}} dx\\ &= \int_{\varphi(U)}(f\circ\phi)(\varphi^{-1}(x))\cdot \sqrt{\det(d(\tilde{\phi})_x)}\sqrt{\det g(\partial_i \psi,\partial_j\psi)_{\vert\psi^{-1}(\tilde{\phi}x)}} dx, \end{align*} where I have used for the last equality the common substitution rule for euclidean space with respect to the diffeomorphism $\tilde{\phi}:=\psi\circ\phi\circ \varphi^{-1}.$
What remains to show is the very complicated looking identity:
$$\sqrt{\det(d(\tilde{\phi})_x)}\sqrt{\det g(\partial_i \psi,\partial_j\psi)_{\vert\psi^{-1}(\tilde{\phi}x)}}=\sqrt{\det g(\partial_i \varphi,\partial_j\varphi)_{\vert\varphi^{-1}(x)}}$$
I could not prove this equality yet. Can someone give me an idea how to attack this problem?
Best wishes
$\phi$ is an isometry so $|det(d\phi)|=1$ We have $$\int_V fdV_g=\int_U (f\circ\phi) |det(\phi)| dV_g,$$