Confusion with the 2-form: $z \, dx\wedge dy$ I'm a bit new to forms and orientations of manifolds, and I'm having a bit of trouble understanding the following simple question.
The integral of the two form $w=z \, dx\wedge dy$ over the surface of a sphere of radius $R$ should give the volume of a sphere, $\frac{4}{3}\pi R^3$.
The question is why doesn't oddity of $w$ in $z$ kill the integral and give zero, as one would expect in conventional integrals? 
I think there is something going on with the sphere being an oriented manifold (I barely know what that means), and thus the notion of $dx\wedge dy$ must also flip sign when crossing the $xy$-plane, but this is a guess and I do not understand it. Is it simply that using $dx\wedge dy$ in the integral is an abuse of notation, and one should perhaps write $w=z \, dA$, so as to not presume a parameterization of the sphere? Or is writing $dx\wedge dy$ actually precise, and I am missing something?
This is my first post here, so please feel free to alert me if my formatting or something is undesirable.
Thank You.
EDIT: As @James S. Cook has kindly shown, we see that the calculation works if we change to the usual polar angles. But shouldn't the question be answerable without any appeal to a different parameterization? Can we make some statement about why the upper and lower half contributions do not cancel that is entirely self contained to the cartesian coordinates, $x,y,z$? Are they intrinsically bad coordinates for this calculation?
Edit2: As @James S. Cook has pointed out, we have a relation $$x^2+y^2+z^2=R^2.$$ Taking the $d$ of this and wedging on the right by $dy$ we can write that $$dx\wedge dy=\frac{z}{x}dy \wedge dz.$$ This expression is odd in $z$. Therefore, we can conclude that the integrand is actually not odd in $z$, as we may have naively thought... I am no mathematician, can anyone comment on the validity of such reasoning?
 A: Interesting question, I like your suggestion that this integral should be trivial on the face of $\int_S z \, dx \wedge dy$ since $z$ is odd and symmetrically so over the sphere. Seems reasonable to suppose this is zero.
Of course if $w = z \, dx \wedge dy$ then $dw = dz \wedge dx \wedge dy = dx \wedge dy \wedge dz$ therefore, as $S = \partial B$ where $B$ is the solid ball of radius $R$ we have:
$$ \int_{\partial B} w = \int_B dw = \int_B dx \wedge dy \wedge dz = vol(B) = \frac{4}{3}\pi R^3.$$
Clearly not zero. But, why? 
To calculate on the sphere we use
$$ x = R \cos \theta \sin \phi, \ \ y = R\sin \theta \sin \phi, \ \ z = R \cos \phi. $$
We calculate,
$$ dx \wedge dy = (-R \sin \theta \sin \phi d \theta + R \cos \theta \cos \phi d \phi) \wedge (R \cos \theta \sin \phi d \theta + R \sin \theta \cos \phi d \phi)$$
But, as $d\phi \wedge d\phi = 0$ and $d\theta \wedge d\theta=0$ and $d\phi \wedge d\theta = -d\phi \wedge d\theta$ the above simplifies nicely to
$$ dx \wedge dy = R^2 \cos \phi \sin \phi \, d\phi \wedge d\theta $$
hence, using a common abuse of notation, the pullback of $w$ is,
$$ w = z dx \wedge dy = R^3 \cos^2 \phi \sin \phi d\phi \wedge d\theta $$
The integral of a form on a manifold is defined in terms of pulling back the form to the parameter space and integrating that form by dropping the wedges and doing the usual multivariate integrals. The orientation is tied to the ordering of the parameters. In the usual case, ordering $\phi, \theta$ generates an outwards pointing normal. I do the same here.
$$ \int_S w dx \wedge dy = \int_0^{2\pi} \int_0^{\pi} R^3 \cos^2\phi \sin \phi \, d\phi \, d\theta = \frac{4}{3}\pi R^3. $$
In this formalism, the reason the $z>0$ (corresponds to $0 < \phi < \pi/2$) and $z<0$ (corresponds to $\pi/2 < \phi < \pi$) do not cancel is the $\cos\phi$ from $dx \wedge dy$ is matched by another $\cos \phi$ from $z$.
Your idea that $dx \wedge dy$ flips sign from above to below the $xy$-plane is in some sense correct... since $x^2+y^2+z^2=R^2$ we can also think formally that $xdx + ydy + zdz = 0$. The $dx, dy, dz$ are not independent here. They're on the sphere. 
A: 
...the integrand is not odd, as one might have naively thought

Likewise the integral of the 1-form $w=z \, dx$ (restricted to the circle) over the circle of radius $R$ (where $z=z(x)$) should give the area of the disk, $2\pi R^2$. However the integration splits into integration over 2 semicircles. On the lower semi-circle not only you now have $z=-\sqrt{R^2-x^2}$ but also the integration limits of the x-parameter are now from $R/2$ to $-R/2$.
