Imagine we have the following application: $$ f(\mathbf r): \mathbb{R}^n \rightarrow \mathbb{R}^m $$ We would like to find the tangent space to this application. Using the Jacobian matrix: $$ \mathbf{J} f (\mathbf r) = \begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} & \dots & \frac{\partial f_1}{\partial x_n}\\ \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} & \dots & \frac{\partial f_2}{\partial x_n}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \frac{\partial f_m}{\partial x_2} & \dots & \frac{\partial f_m}{\partial x_n} \end{pmatrix} \\ $$ Using the basis of the $\mathrm {Im}(\mathbf{J} f)$: $$ \mathcal B = \left \{ \begin{pmatrix} \frac{\partial f_1}{\partial x_1} \\ \frac{\partial f_1}{\partial x_2} \\ \vdots \\ \frac{\partial f_1}{\partial x_n} \end{pmatrix}, \begin{pmatrix} \frac{\partial f_1}{\partial x_1} \\ \frac{\partial f_1}{\partial x_2} \\ \vdots \\ \frac{\partial f_1}{\partial x_n} \end{pmatrix} \; \dots \; \begin{pmatrix} \frac{\partial f_m}{\partial x_1} \\ \frac{\partial f_m}{\partial x_2} \\ \vdots \\ \frac{\partial f_m}{\partial x_n} \end{pmatrix} \right \} = \left \{ \mathbf e_1, \mathbf e_2 \; \dots \; \mathbf e_k \right \} $$ It is well known, that the volume enclosed by this vectors (if $n = m$) is: $$ d^n\mathbf r = \det{ \left ( \mathbf{J} f \right)} dx_1 dx_2 \; \dots \;dx_n \ $$ Now, we want to see how much a vector $\mathbf e_i$ changes in every direction, given by the basis vectors of the space: $$ \lim_{t \to 0} \; \frac{\mathbf e_i (\mathbf r + t \mathbf e_j) - \mathbf e_i (\mathbf r)}{t} = \frac{\partial \mathbf e_i}{\partial x_j} $$ This derivative measures how much, the basis vector $\mathbf e_i$, changes along the direction of the basis vector $\mathbf e_j$.
We will define a matrix where the columns are the components of the vectors $\frac{\partial \mathbf e_i}{\partial x_j}$. $$ \frac{\partial \mathbf{J} f}{\partial x_j}: \mathbb{R}^n \rightarrow \mathbb{R}^m $$ $$ \frac{\partial \mathbf{J} f}{\partial x_j} = \begin{pmatrix} \frac{\partial \mathbf e_1}{\partial x_j} & \frac{\partial \mathbf e_2}{\partial x_j} \; \dots \; \frac{\partial \mathbf e_n}{\partial x_j} \end{pmatrix} \\ $$ Then the change of (hyper) volume in a certain direction is given by (if $n = m$): $$ d^n \mathbf r_j = \det \left ( \frac{\partial \mathbf{J} f}{\partial x_j} \right ) dx_1 dx_2 \; \dots \;dx_n \ $$ My question is, could this changes in volume, be related to curvature? If a surface isn't curved, like for example, on a hyper-plane, this volume wouldn't change. Is this reasonable? Did I made wrong assumptions?