Integrating a p - form

Let  
\[\omega_p\]
  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
\[\begin{equation} \begin{aligned} \int_S \omega'_{\mathbf{f}(t)} &= \int^b_a \omega'_{\mathbf{f}(t)} (\frac{\partial \mathbf{f}}{\partial t}) dt \\ &= \int^b_a (F_1 (\mathbf{x}) \frac{\partial f_1}{dt} +...+F_n (\mathbf{x}) \frac{\partial f_n}{dt}) dt \end{aligned} \end{equation}\]