# Integrating a 2-from

I'm working through McDuff and Salamon's `Introduction to Symplectic Topology' (highly recommend). In example 3.1.2, we are given the example of a symplectic manifold: the 2-sphere with its standard area form. Spcifically, $$S^2 = \{(x_1,x_2,x_3)\in\mathbb{R}^{3}\:|\: x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=1\}$$ with induced area form $$\omega_{x}(\xi,\eta) = \langle x,\xi\times\eta\rangle$$ for $$\xi,\eta\in T_{x}S^{2}$$. The authors then comment that the total area is $$4\pi$$.

So my question is: what is the easiest way to see that the total area is $$4\pi$$? I would like to know how I can integrate this 2-form $$\int_{S^{2}}\omega$$ possibly using Stoke's theorem. I would also have a preference for coordinate free solutions!

The easiest coordinates in this case are cylindrical ones, you can check that writing the form in those terms gives $$dz\wedge d\theta$$, now the integral is easy to take.
• Thankyou @AnonymousMemer. How would I explicitly compute this using cylindrical coordinates? Here is my try: $S^{2} = f^{-1}(0)$ where $f(r,\theta,z) = r^{2}+z^{2}-1$. Thus $\nabla f = (2r,0,2z)$ and thus $\xi$ and $\eta$ would be in the span of $(0,1,0)$ and $(z,0,-r)$. I then compute $\langle x, \xi\times\eta\rangle$ to be $-r^{2}-z^{2} = -1$. So $\int_{S^{2}}\omega = \int_{S^{2}}-1 d\theta\wedge dz = \int_{S^{2}} dz\wedge d\theta = 4\pi$. I'm not happy with my working though. Commented Nov 11, 2019 at 3:58