An Example (part 3)
Now we can take our differential form and our singular cube and put them together. That is, we can integrate the 
First we write down the definition:
![\int\limits_{c_a}\omega=\int\limits_{[0,1]}{c_a}^*\omega](http://s0.wp.com/latex.php?latex=%5Cint%5Climits_%7Bc_a%7D%5Comega%3D%5Cint%5Climits_%7B%5B0%2C1%5D%7D%7Bc_a%7D%5E%2A%5Comega&bg=e6e6e6&fg=333333&s=0 "\int\limits_{c_a}\omega=\int\limits_{[0,1]}{c_a}^*\omega")
to find this pullback of ![[0,1]](http://s0.wp.com/latex.php?latex=%5B0%2C1%5D&bg=e6e6e6&fg=333333&s=0 "[0,1]")
This actually isn’t that hard; there’s only the one basis vector 
We also have to calculate the composition
This lets us calculate
![\displaystyle\begin{aligned}\int\limits_{c_a}\omega&=\int\limits_{[0,1]}{c_a}^*\omega\\&=\int\limits_{[0,1]}\left[\omega(c_a(t))\right]\left({c_a}_{*t}\left(\frac{d}{dt}\right)\right)\,dt\\&=\int\limits_{[0,1]}\left[\frac{1}{a}\left(-\sin(2\pi t)dx+\cos(2\pi t)dy\right)\right]\left(2\pi a\left(-\sin(2\pi t)\frac{\partial}{\partial x}+\cos(2\pi t)\frac{\partial}{\partial y}\right)\right)\,dt\\&=2\pi\int\limits_0^1\sin(2\pi t)^2+\cos(2\pi t)^2\,dt=2\pi\int\limits_0^1\,dt=2\pi\end{aligned}](http://s0.wp.com/latex.php?latex=%5Cdisplaystyle%5Cbegin%7Baligned%7D%5Cint%5Climits_%7Bc_a%7D%5Comega%26%3D%5Cint%5Climits_%7B%5B0%2C1%5D%7D%7Bc_a%7D%5E%2A%5Comega%5C%5C%26%3D%5Cint%5Climits_%7B%5B0%2C1%5D%7D%5Cleft%5B%5Comega%28c_a%28t%29%29%5Cright%5D%5Cleft%28%7Bc_a%7D_%7B%2At%7D%5Cleft%28%5Cfrac%7Bd%7D%7Bdt%7D%5Cright%29%5Cright%29%5C%2Cdt%5C%5C%26%3D%5Cint%5Climits_%7B%5B0%2C1%5D%7D%5Cleft%5B%5Cfrac%7B1%7D%7Ba%7D%5Cleft%28-%5Csin%282%5Cpi+t%29dx%2B%5Ccos%282%5Cpi+t%29dy%5Cright%29%5Cright%5D%5Cleft%282%5Cpi+a%5Cleft%28-%5Csin%282%5Cpi+t%29%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+x%7D%2B%5Ccos%282%5Cpi+t%29%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial+y%7D%5Cright%29%5Cright%29%5C%2Cdt%5C%5C%26%3D2%5Cpi%5Cint%5Climits_0%5E1%5Csin%282%5Cpi+t%29%5E2%2B%5Ccos%282%5Cpi+t%29%5E2%5C%2Cdt%3D2%5Cpi%5Cint%5Climits_0%5E1%5C%2Cdt%3D2%5Cpi%5Cend%7Baligned%7D&bg=e6e6e6&fg=333333&s=0 "\displaystyle\begin{aligned}\int\limits_{c_a}\omega&=\int\limits_{[0,1]}{c_a}^*\omega\\&=\int\limits_{[0,1]}\left[\omega(c_a(t))\right]\left({c_a}_{*t}\left(\frac{d}{dt}\right)\right)\,dt\\&=\int\limits_{[0,1]}\left[\frac{1}{a}\left(-\sin(2\pi t)dx+\cos(2\pi t)dy\right)\right]\left(2\pi a\left(-\sin(2\pi t)\frac{\partial}{\partial x}+\cos(2\pi t)\frac{\partial}{\partial y}\right)\right)\,dt\\&=2\pi\int\limits_0^1\sin(2\pi t)^2+\cos(2\pi t)^2\,dt=2\pi\int\limits_0^1\,dt=2\pi\end{aligned}")
So, what conclusions can we draw from this? Well, Stokes’ theorem now tells us that the 
which we now know not to be the case. Similarly, 
It turns out that there’s a deep connection between the two halves of this example. Further, in a sense every failure of a closed 
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