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The Question

Let $K = \{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 \geq 1\}$, where $K$ is oriented via the canonical volume form on $\mathbb{R}^3$: $dx \wedge dy \wedge dz$.

Let $\mathbb{S}^2$ be the unit sphere, considered as the boundary of $K$, with the orientation on $\partial K$ given by the induced orientation from $K$.

Define the canonical inclusion map $j: \mathbb{S}^2 \rightarrow \mathbb{R}^3$ and the 2-form on $\mathbb{R}^3$

$\omega := (2z -x^2 - xy) dx \wedge dy - dy \wedge dz +dz \wedge dx$

Calculate

\begin{align} \int\limits_{\mathbb{S}^2} j^* \omega \end{align}

where $j^*$ denotes the pullback of $j$ in the usual way.

Attempt at a solution

Using Stokes' theorem, we have

\begin{align} \int\limits_{\mathbb{S}^2} j^* \omega &= \int\limits_{\mathbb{R}^3 \backslash K} d\omega\\ &= 4 \pi \int\limits_0^1 dr\,2 r^2\\ &= \frac{8\pi}{3}\\ \end{align}

The difficulty

Is this correct or should the answer be $-8\pi/3$? In class we've said that for manifolds with odd dimension we will adopt the convention that the boundary will have the opposite orientation as the one induced by a volume form on the manifold. To me this means that we pick up a minus sign in this case, because by integrating $r$ from 0 to 1 we are adopting the orientation given by the volume form on $\mathbb{R}^3$. Is this true?

It also seems that, using basic calculus reasoning, the result of the integral should differ by a sign change depending on if we view $\mathbb{S}^2$ as the boundary of $K$ or as the boundary of $\mathbb{R}^3 \backslash K$ (because the definition of "outwards" would differ for the normal vectors). Is this true? If so, how is it imposed formally? What does this mean for my question in the preceding paragraph?

I'm very much interested in the "why" for a general case, not just the solution to this particular problem.

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