Nowhere $0$ form on the sphere? Consider the differential form on $\mathbb R^3$ given by $ x dy \wedge dz + y dz \wedge dx + z dx \wedge dy$. I converted this to spherical coordinates using a laborious calculation, and when I'm done, by some miracle (which would be cool if someone could explain exactly how that works), I get something really compact and elegant: $\rho^3 \sin \phi d\phi \wedge d\theta$. 
If we call this form $\Omega$, my task is to compute $i^*\Omega,$ where $i$ represents the inclusion from $\mathbb S^2 \rightarrow \mathbb R^3$, which seems to be pretty easy also - on the unit sphere we just have $\rho = 1$.
Now why is this nowhere $0$ on the sphere? It looks like it's $0$ along the half plane where $\phi = 0$!

First of all, there's just a point on the unit sphere where $\phi=0$ — the north pole (but there's also the south pole, where $\phi=\pi$, to worry about). But remember that spherical coordinates actually fail to give a coordinate system at these points (and we can debate what happens when $\theta = 0$ or $2\pi$).
In the original cartesian coordinates, you see that at the poles the $2$-form is given by $z\,dx\wedge dy = \pm dx\wedge dy$, and, since the tangent plane to the sphere is the $xy$-plane, this $2$-form is definitely non-zero on the sphere at those points.
