Let $\alpha= f\,dx_1 \wedge \cdots\wedge dx_n$; where $f$ is continuous on $A$. Show that $\int_ \Phi \alpha =\int_ \Phi f$

Question

Let $A \subset \mathbb{R}_k$ be a rectangle (or box), and let $\Phi:A\to\mathbb{R}_k$, be the identity mapping.

Let $\alpha= f \, dx_1 \wedge \cdots \wedge dx_n$; where $f$ is continuous on $A$.

Show that

$$\int_ \Phi \alpha =\int_ \Phi f$$

I'm not sure what the formal way to proceed is, if we just go back to the formal definitions of $\alpha(\Phi)$ and $f(\Phi)$ I know what to do in the case of $\alpha$ but I'm not sure what to do for $f$ because it's a $0$-form.

So by looking at the definition, there is no jacobian to calculate for $f$ and for $\alpha$ the jacobian has to be one because we have an identity mapping? I'm not sure how to write this properly.

Thank you for your help as always!

Answer 1

According to Baby Rudin, $$\int_\Phi\alpha=\int_A (f\circ\Phi)\cdot\det D\Phi.$$

In this case ($\Phi=Id$), $$\int_\Phi\alpha=\int_A f.$$