# 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).

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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.

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That is much nicer! I really appreciate it. –  5space Jun 11 '13 at 22:09
very nice +1... –  B. S. Jun 12 '13 at 13:36

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.

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$\delta$ seems to come up a lot for problems like this. Sadly, we have not covered it. –  5space Jun 11 '13 at 22:08

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 $$\int_{C\backslash B(0,\epsilon)} \mathbf{d}\varphi = -\int_{S_{\epsilon}} \varphi + \int_{\partial C} \varphi.$$

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