Let $M$ be a submanifold on $\mathbb{R}^4$: $M=\{(\omega, x, y,z)\in\mathbb{R}^4\mid \omega^2+x^2=1, y^2+z^2=1\}$. I want to compute $\int_M i^\ast\omega$ where $i:M\rightarrow \mathbb{R}^4$ is the inclusion map and $\omega=xyzd\omega\wedge dy.$
First thing to do is probably parametrizing. Can I do $g(\theta_1, \theta_2)=(\cos\theta_1,\sin\theta_2,\cos\theta_2,\sin\theta_2)$ where $\theta_1,\theta_2\in(0,2\pi)$. Then I should apply stokes theorem but I am stuck at this point. Any help is greatly appreciated!
Just reparametrizing is fine, I think: write $w+ix=e^{i\alpha}$, $y+iz=e^{i\beta}$, then $i^*\omega=\sin^2{\alpha}\,\sin^2{\beta}\,\cos{\alpha}\,d\alpha \wedge d\beta$, so that $\int_M{i^*\omega}=\int_{(0,2\pi)^2}{\sin^2{x}\,\sin^2{y}\,\cos{y}\,dxdy}=\int_0^{2\pi}{\sin^2{x}\,dx}\times \int_0^{2\pi}{\cos{y}\,\sin^2{y}\,dy}=0$.
Here’s a trick though: note that $ydy+zdz=0$ on $M$, so that $i^*\omega=-xy^2dw \wedge dy=\frac{-1}{3}xdw \wedge d(y^3)$. So $-3i^*\omega=\pm d(xy^3dw)\pm y^3d(xdw)$. Now $d(xdw)=dx \wedge dw$ and thus $xd(xdw)=xdx \wedge dw=-w dw \wedge dw=0$, so that $-3i^*\omega=\pm d(xy^3dw)$, hence the integral is zero by Stokes.