I am trying to generalize from the following scenarios:
To find the length of a path $\gamma : I \rightarrow \mathbb{R}^n$, we can take $\int_0^1 ||\gamma'(t)|| dt$.
To find the volume of a surface parameterized as $\gamma : I \times I \rightarrow \mathbb{R}^3$, we can take $$\int_0^1 \int_0^1 \left| \left| \frac{\partial \gamma }{\partial x } \times \frac{\partial \gamma}{ \partial y} \right| \right| dx dy$$ Where $\times$ is cross product.
To find the n-volume of a continuous map $\gamma : I^n \rightarrow \mathbb{R}^n$, we can take
$$\int_0^1 \cdots \int_0^1 |\text{det}(\gamma'(t_1, ..., t_n))| dt_1 \cdots d t_n$$
So for the general situation, I have a continuous map $\gamma : I^k \rightarrow \mathbb{R}^n$ and I want to find the `size' of the image of $\gamma$. What is the correct/standard generalization here?
I am going to guess that the formula is. For a linear map $T : \mathbb{R}^n \rightarrow \mathbb{R}^m$, define $\Lambda^k T : \Lambda^k \mathbb{R}^n \rightarrow \Lambda^k \mathbb{R}^m$ sending $v_1 \wedge \cdots \wedge v_k$ to $T(v_1) \wedge \cdots \wedge T(v_k)$. Then my guess is that the desired formula is
$$\int_0^1 \cdots \int_0^1 || (\Lambda^n \gamma'(t_1, ..., t_n) )(e_1 \wedge \cdots \wedge e_n) || dt_1 \cdots dt_n$$ Where the norm inside the integral is taken inside $\Lambda^n \mathbb{R}^m$.
To motivate my guess, we might wonder the $n$-volume $\text{vol}(T(I^n))$ of $T(I^n)$ is for a linear map $T : \mathbb{R}^n \rightarrow \mathbb{R}^m$. If the volume is $$\text{vol}(T(I^n)) = ||\Lambda^n(e_1 \wedge \cdots \wedge e_n)||$$ then my guess is correct.