Let $M = \{(x,y,z) \in \mathbb{R}^3 | x^2+y^2+z^2=1,z>0\}$, and let $\omega = xdy$. I would like to verify that $\int_M d\omega = \int_C \omega$, where $C = \partial M$ given by $C = \{(x,y,z) \in \mathbb{R}^3 | x^2+y^2 = 1, z=0\}$.
I first parametrize $M$ as the image of the set $A = \{(\theta, \phi) \in \mathbb{R}^2 | -\frac{\pi}{2} < \theta < \frac{\pi}{2}, 0 < \phi < 2\pi\}$ under the spherical transformation $\alpha(\theta, \phi) = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$.
So we have that, $$\int_M d\omega = \int_M dx \wedge dy = \int_A \alpha^* (dx \wedge dy) = \int_A d\alpha_1 \wedge d\alpha_2.$$ Since $$d\alpha_1 = \cos\theta\cos\phi \ d\theta - \sin\theta\sin\phi \ d\phi,$$ and $$d\alpha_2 = \cos\theta\sin\phi \ d\theta + \sin\theta\cos\phi \ d\phi,$$ $d\alpha_1 \wedge d\alpha_2 = \sin\theta\cos^2\phi \ d\theta \wedge d\phi - \sin\theta\cos\theta\sin^2 \phi \ d\phi \wedge d\theta = \sin\theta\cos\theta \ d\theta \wedge d\phi$, so the integral becomes $$\int_A \sin\theta \cos\theta \ d\theta \wedge d\phi = \int_0^{2\pi} \int_{-\pi / 2}^{\pi / 2} \sin\theta \cos \theta \ d\theta d\phi = 0.$$
On the other hand, we can parametrize $C$ as the image of $U = \{\theta \in \mathbb{R} | 0 < \theta < 2\pi \}$ under $\gamma(\theta) = (\cos\theta, \sin\theta)$. Then $$\int_C \omega = \int_C xdy = \int_U \gamma^*(xdy) = \int_U \cos\theta \ d\gamma_2 = \int_U \cos^2\theta \ d\theta = \int_0^{2\pi} \frac{1+\cos2\theta}{2} \ d\theta = \pi.$$
Why do my answers not agree?
Your mistake is in the parametrization $\alpha$ you are using, or better in the definition of $A$. Your $\theta$ should only run from $0$ to $\frac{\pi}{2}$.