Integrating a form and using Gauss' theorem. 
Given the 2-form
  $$
\varphi = \frac{1}{(x^2+y^2+z^2)^{3/2}}\left( x\,dy\wedge dz +y\,dz\wedge dx + z\,dx\wedge dy\right) \ .
$$
  (a) Compute the exterior derivative $\textbf{d}\varphi$ of $\varphi$.
(b) Compute the integral of $\varphi$ over the unit sphere oriented by the outward normal.
(c) Compute the integral of $\varphi$ over the boundary of the cube of side 4, centered at the origin, and oriented by the outward normal.
(d) Can $\varphi$ be written $\textbf{d}\psi$ for some 1-form $\psi$ on $\mathbb{R}^3-\{0\}$?

This is problem 6.23 from Hubbard's Vector Calculus text. This is not homework, I am just studying for my final.
The only part I am having any trouble with is part (c). For part (a), note that $\varphi = \Phi_{\vec{F}}$ where
$$
\vec{F} = \frac{1}{(x^2+y^2+z^2)^{3/2}}\begin{bmatrix} x\\y\\z \end{bmatrix} \ .
$$
Then $\textbf{d}\varphi = M_{\nabla\cdot\vec{F}}=0.$ For part (b), we pick an orientation-preserving parametrization of the sphere, call it $\partial S$, and use it to evaluate the integral. Namely, we use
$$
\gamma : \begin{pmatrix} \alpha \\ \beta \end{pmatrix} \mapsto \begin{pmatrix} \sin\alpha\cos\beta \\ \sin\alpha\sin\beta \\ \cos\alpha \end{pmatrix}
$$
where $\alpha\in[0,\pi]$ and $\beta \in [0,2\pi)$. Then the pullback is
$$
\gamma^*\varphi = (\sin^3\alpha+\sin\alpha\cos^2\alpha)\,d\alpha\,d\beta = \sin\alpha\,d\alpha\,d\beta
$$
so that
$$
\int\limits_{\partial S}\varphi = \int\limits_0^{2\pi}\int\limits_0^{\pi} \sin\alpha\,d\alpha\,d\beta = 4\pi.
$$
Part (c) is where I am a bit stuck. Consider the region $R$ bounded between the described cube, say $C$ with boundary $\partial C$, and the unit ball at the origin, say $S$. We can use Guass' theorem since $\varphi$ is well defined there. We have
$$
\int\limits_{\partial C} \varphi = \int\limits_C \textbf{d}\varphi = \int\limits_R \textbf{d}\varphi \, + \int\limits_S \textbf{d}\varphi = \int\limits_S \textbf{d}\varphi \stackrel{?}{=} 4\pi
$$
by part (a). How do I go about obtaining the last equality since $\varphi$ is not defined at the origin? I'm not sure how to justify appealing to my result in part (b).
 A: I would rather write
$$0=\int_Rd\varphi=\int_{\partial R}\varphi=\int_{\partial C}\varphi-\int_{\partial S}\varphi,$$
i.e. both surface integrals coincide, and the origin is not involved at any moment.
A: There are a couple approaches to this.  You could integrate $\varphi$ directly on the cube; this should be doable with a table of integrals.
You can instead appeal to the divergence theorem and argue that the volume integral has a constant value of $4\pi$ so long as it contains the origin, no matter what the size or shape of the volume.  Since $\mathrm d\varphi$ is zero everywhere except the origin, this should help justify such an argument.
Problems like these, involving forms like $\varphi$, are fairly common: $\varphi$ can be seen as a Green's function for $\mathrm d$--that is, $\mathrm d\varphi = \star \delta$, where $\delta$ is the Dirac delta function.  This should convince you why the integral of $\mathrm d\varphi$ depends only on whether the integration region contains the origin.
A: Why not try dig out a small sphere of radius $\epsilon$, centered at the origin, say $B(0,\epsilon)$. Let $S_{\epsilon} = \partial B(0,\epsilon)$, then
$$
 \int_{C\backslash B(0,\epsilon)} \mathbf{d}\varphi = -\int_{S_{\epsilon}} \varphi + \int_{\partial C} \varphi.
$$
