Integration of a differential form along a curve Given the differential form $\alpha = x dy - \frac{1}{2}(s^2+y^2) dt$, I'd like to evaluate $\int_\gamma \alpha$ where $\gamma(s)=(\cos s,\sin s, s)$ and $0\leq s\leq \frac{\pi}{4}$. When attempting to evaluate this in a way similar to a line integral, first I need to find the derivative of each component of $\gamma$ and then take the square root of the sum of their squares:
$$
dt = \sqrt{x'(s)^2+y'(s)^2+t'(s)^2} = \sqrt{2} ds
$$
But when I'm trying to evaluate 
$$
\int_\gamma x dy- \frac{1}{2}(s^2+y^2) dt
$$
I'm not sure how to handle the different differentials. The first statement I made about $dt$ doesn't seem to make sense in this case because I don't know how $dy$ and $dt$ compare. I honestly have no idea how to approach this and would appreciate a step-by-step guide to evaluating integrals of this type.
 A: Differential forms exist to eat vector fields.  The only vector field in sight is the velocity field on $\gamma$, so we may as well feed it to $\alpha$:
\begin{eqnarray*}\alpha(\gamma'(s)) &= \cos(s)dy(-\sin(s),\cos(s),1) - \frac{1}{2}(s^2 + \sin^2(s))dt(-\sin(s),\cos(s),1) \\
&= \cos^2(s) - \frac{1}{2}(s^2 + \sin^2(s)).\end{eqnarray*}
This gives us a nice function on $[0,\frac{\pi}{4}]$ which we can integrate.  That's the integral of $\alpha$ on $\gamma$:
$$\int_\gamma\alpha = \int_0^\frac{\pi}{4} \alpha(\gamma'(s))ds = \int_0^1 \cos^2(s) - \frac{1}{2}(s^2 + \sin^2(s))\ ds.$$
I trust you can evaluate this.

As a general definition, to integrate a differential $k$-form $\omega$ on some oriented smoothly embedded $k$-dimensional submanifold $\Phi: U\hookrightarrow \mathbb{R}^n$, pull the differential form back via $\Phi$ and integrate over $U$:
$$\int_U\Phi^*\omega.$$
The pullback form $\Phi^*\omega$ is defined by $\Phi^*\omega(V_1,V_2,\ldots,V_k) = \omega(d\Phi V_1,d\Phi V_2,\ldots, d\Phi V_k)$.  (You'll notice that's exactly what we did with $\gamma$ and $\alpha$, since $\gamma'(s) = d\gamma_s(\frac{\partial}{\partial s})$.)  Since the pullback is a $k$-form on a $k$-dimensional domain, it's a multiple of the volume form and so it can be integrated as an ordinary function.
In practice, for one-forms, you'll compute the velocity field, feed it to the one-form to get a function, and then integrate that function with respect to the curve parameter.

Here's an exercise to tie integrating one-forms with more familiar line integrals of vector fields.  If $\gamma$ is a smooth curve on (say) $[0,1]$ into $\mathbb{R}^3$ and $V(p) = (V_x(p),V_y(p),V_z(p))$ is a smooth vector field on $\mathbb{R}^3$, you have defined
$$\int_\gamma V = \int_0^1 \langle V(\gamma(s)), \gamma'(s)\rangle ds.$$
On $\mathbb{R}^3$, to any vector vield $V$, we associate the one-form $\vartheta$ defined by $\vartheta(X) = \langle V,X\rangle$.  In terms of differentials, $\vartheta = V_xdx + V_ydy + V_zdz$.  Now look at the integrand of $\int_\gamma V$: taking the inner product of $V$ and $\gamma'$ is the same as feeding $\gamma'(s)$ to $\vartheta$.  The integral of $V$ over $\gamma$ is nothing but
$$\int_\gamma \vartheta.$$
