# Relevance of Differential Forms

I recently started reading about differential forms, and I am trying to figure out their purpose. Lets say $\omega=y\,dx+x\,dy$, and we want to evaluate $\int_C \omega$ over the curve parametrized by $\phi(t)=(t^2,t^3)$ from 0 to 1. So we have $\int_C \omega=\int_C y\,dx+x\,dy$...now I am trying to figure out what the purpose in defining differential forms as functions that send points to $T^*_p C$...when I parametrize the integral, am I suppose to evaluate $\omega$ at $\omega_{\phi(t)}(\phi(t))\,dt$, so that $\int_C \omega=\int_0^1 \omega_{\phi(t)}(\phi(t))\,dt$, where $\omega_{\phi(t)}=t^3\,dx_\phi+t^2\,dy_\phi$, and so $\omega_{\phi(t)}(\phi(t))=\omega_{\phi(t)}(t^2\partial_x,t^3\partial_y)=2t^5$, where I used that $dx\partial_x=1, dx\partial_y=0$. I know that this is wrong though.

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You're asking two separate questions here (what the point of differential forms is, and what you did wrong in this specific computation). You'd be better off separating them. –  Qiaochu Yuan Jun 25 '12 at 4:28
The thing is, I know how to do the substitution from calculus, but I'm trying to figure it out from the point of view of functionals, if there is another way to look at it. –  JLA Jun 25 '12 at 4:34
Though I edited it. –  JLA Jun 25 '12 at 4:43
Your definition of differential form as "functions that send points in $T_pC$ to $T^*_pC$" is incorrect. You want something like "functions that send points in $M$ or $\mathbb{R}^n$ to elements of $T^*_pC$". So a 1-form assigns to each point a function which eats a vector and spits out a number. –  Adam Saltz Jun 25 '12 at 5:12
It's $$\int_C \omega=\int_0^1 \omega_{\phi(t)}(\phi'(t))\,dt\ .$$ –  Christian Blatter Jul 28 '12 at 8:21

The problem is that you need to look at $\omega_{\phi(t)}(\phi'(t))$, and not $\omega_{\phi(t)}(\phi(t))$ as you were doing.
Remember, $\omega_{\phi(t)}$ is an element of $T_{\phi(t)}^*C$, so $\omega_{\phi(t)}\colon \ T_{\phi(t)}C \to \mathbb{R}.$ This means that $\omega_{\phi(t)}$ has to take tangent vectors as inputs.
Since $\omega_{\phi(t)}(\phi'(t)) = (t^3\,dx_\phi + t^2\,dy_\phi)(2t\,\partial_x + 3t^2\partial_y) = 2t^4 + 3t^4 = 5t^4,$ we have $$\int_C \omega = \int_{[0,1]}\phi^*\omega = \int_0^1 \omega_{\phi(t)}(\phi'(t))\,dt = \int_0^1 5t^4\,dt = 1,$$ where $\phi^*\omega$ is the pullback of $\omega$.