\[\omega_p\]
be a differential p - form in \[\mathbb{R}^n\]
and let \[K\]
be a closed bounded rectangle in \[\mathbb{R}^p\]
Let
\[f : \mathbb{R}^p \rightarrow \mathbb{R}^n\]
be differentiable. In the diagram below, \[f(K)=S\]
If
\[\mathbf{u}=(u_1 , ..., u_p) \in K\]
then \[f\mathbf{(}\mathbf{u})=(f_1(\mathbf{u}),...,f_n (\mathbf{u})) \in S \]
We can apply
\[\omega^p\]
at a point \[\mathbf{f}(\mathbf{u})\]
to the set \[\frac{\partial \mathbf{f}}{\partial u_1} \mathbf{u}, ..., \frac{\partial \mathbf{f}}{\partial u_p} \mathbf{u} \]
t obtain\[\omega^p_{\mathbf{f}(\mathbf{u})}(\frac{\partial \mathbf{f}}{\partial u_1} \mathbf{u}, ..., \frac{\partial \mathbf{f}}{\partial u_p} \mathbf{u})\]
If
\[\Omega\]
is a grid cover of \[K\]
with vertices \[(\mathbf{u}_1 ,..., \mathbf{u}_\alpha )\]
and rectangles \[K_1,..., K_\alpha\]
we can form the sum\[\sum_{i=1}^\alpha \omega^p_{\mathbf{f}(\mathbf{u})}(\frac{\partial \mathbf{f}}{\partial u_1} \mathbf{u}, ..., \frac{\partial \mathbf{f}}{\partial u_p} \mathbf{u}) V(K_i) \]
where
\[V(K_i)\]
is the p - dimensional volume of \[K_i\]
The integral of
\[\omega^p\]
over \[S\]
is\[\int_S \omega^p = lim_{D(\Omega) \rightarrow 0} \sum_{i=1}^\alpha \omega^p_{\mathbf{f}(\mathbf{u})}(\frac{\partial \mathbf{f}}{\partial u_1} \mathbf{u}, ..., \frac{\partial \mathbf{f}}{\partial u_p} \mathbf{u}) V(K_i) \]
The sum on the right hand side is a Riemann sum, and if it exists then
\[\int_S \omega^p = \int_K \omega_{\mathbf{f} ( \mathbf{u})} (\frac{\partial \mathbf{f}}{\partial u_1} ,..., \frac{\partial \mathbf{f}}{\partial u_p} ) dV \]
(1)If p=1 then
\[K=[a,b]\]
and if \[t \in [a,b]\]
then \[\mathbf{f}(t) \in S \subset \mathbb{R}^n\]
.\[\mathbf{f}(t)\]
will be a curve in \[\mathbb{R}^n\]
If
\[\omega'{\mathbf{x}} =F_1 (\mathbf{x})dx_1 +...+F_n (\mathbf{x})dx_n \]
then (1) reduces to
\[\begin{equation} \begin{aligned} \int_S \omega'_{\mathbf{f}(t)} &= \int^b_a \omega'_{\mathbf{f}(t)} (\frac{\partial \mathbf{f}}{\partial t}) dt \\ &= \int^b_a (F_1 (\mathbf{x}) \frac{\partial f_1}{dt} +...+F_n (\mathbf{x}) \frac{\partial f_n}{dt}) dt \end{aligned} \end{equation}\]