Computing $\int_{\mathbb{S}^2}(xdy\wedge dz+ydz\wedge dx+zdx\wedge dy)$

Question

Let $M$ be the $3-$ball in $\Bbb{R}^3$ $\{(x,y,z):x^2+y^2+z^2 \leq 1\}$. Let $\omega$ be the $2-$form given by

$$\omega = x \, dy \wedge dz + y \, \wedge dz \wedge dx + z \,dx \wedge dy.$$

Verify stokes theorem by verifying both RHS and LHS of the theorems statement.

So for the LHS, i computed $d \omega$ to be

$$d \omega = 3 \, dx \wedge dy \wedge dx.$$

So can I just say since

$$\int_M dx \wedge dy \wedge dz$$

is the volume of the $3-$ball that the final value is $3$ times the volume of a $3-$ball in $3$ space which is $4\pi$. Then for the RHS, I know the boundary of the $3-$ball is the unit $2$ ball $S^2$, which is all $(x,y,z) \in \Bbb{R}^3$ such that $x^2+y^2+z^2=1$. Then I need to compute

$$\int_{S^2} x \, dy \wedge dz + y \, dz \wedge dx + z\, dx \wedge dy$$

Which I still dont know how to compute integrals with wedge products, so is this a sum of $3$ triple integrals? Or could i swap to spherical coodrinates by setting

$$x = r \sin \theta \cos \phi, y = r \sin \theta \sin \phi, z = r \cos \theta$$

Where my $r=1$ Then my integral becomes

$$\int_0^{2 \pi} \int_0^{2 \pi} \int_0^1 f(r,\theta,\phi) r^2 \sin \theta d r d \theta d \phi$$

Where $f(r,\theta,\phi)$ is $\omega$ with spherical coordinates. But that would get rather messy, no? So what’s the fastest method for computing the LHS and is my RHS ok as far as justification goes? With the polar swap I get

\begin{align} \omega&= (\sin \theta \cos \phi)d(\sin \theta \sin \phi) \wedge d(\cos \theta) + (\sin \theta \sin \phi ) d(\cos \theta) \wedge d(\sin \theta \cos \phi)+(\cos \theta) d(\sin \theta \cos \phi) \wedge d( \sin \theta \sin \phi)\\ &=\sin^3 \theta \cos \phi +\sin^4 \theta \sin \phi \cos^2 \phi d \theta \wedge d \phi \wedge \cos \theta \cos \phi + \cos^2 \theta \cos \phi d \theta - (2 \sin \theta \sin \phi) d \phi \end{align}

Answer 1

By symmetry the answer will be $6$ times the integral of $z \,dx \wedge dy$ over the upper half of the sphere. Since $z = \sqrt{1 - x^2 - y^2}$, you can just do this in coordinates as the integral $$\int_{x^2 + y^2 < 1} \sqrt{1 - x^2 - y^2} \,dA$$ The above integral is easily determined to be ${\displaystyle {2\pi \over 3}}$ after a routine change to polar coordinates. Multiplying this by $6$ gives the desired answer of $4\pi$.

What Stokes theorem is basically doing here is turning the integral of the constant function $3$ over the interior of the sphere into the sum of three integrals of the constant function $1$, and then integrating each of these first in one variable and then in the remaining two variables. The above calculation is one of the two integrals you have left when you integrate in $z$ first.

Answer 2

The problem amounts to computing: $$\color{blue}{\int_{\partial M}\omega := \int_{\mathbb{S}^2}(xdy\wedge dz + ydz\wedge dx + zdx\wedge dy)}$$

We will be using the "Spherical Coordinates" parameterization (which covers $\mathbb{S}^2$ with one map), together with the definition the integral and using the pullback formula for differential forms. For reference, use the following from Lee's "Introduction to Smooth Manifolds" (p.408):

Prop 16.8: (Integration Over Parameterizations): Let $\mathcal{N}$ be an oriented smooth $n$-manifold and let $\omega$ be a compactly supported $n$-form on $\mathcal{N}$. Suppose $D_1,...,D_k$ are open domains of integration in $\mathbb{R}^n$, and for $i=1,...,k$, we are given smooth maps $F_i:\overline{D}_i\to \mathcal{N}$ satisfying:

(i) $F_i$ restricts to an orientation-preserving diffeomorphism from $D_i$ onto an open subset $W_i\subseteq \mathcal{N}$;
(ii) $W_i\cap W_j = \emptyset$ when $i\neq j$;
(iii) $supp(\omega)\subseteq \overline{W}_1\cup ... \cup \overline{W}_k$.

Then $$\int_{\mathcal{N}}\omega = \sum\limits_{i=1}^k\int_{D_i}F^*_i(\omega).$$

Accordingly, take $\mathcal{N}:= \mathbb{S}^2$, $\color{blue}{D:= (0,2\pi)\times(0,\pi)}\subseteq \mathbb{R}^2,$ and $F:\overline{D}\to \mathcal{N}$ by: $$\color{blue}{F(\theta,\phi) := \big(cos(\theta) sin(\phi),\text{ }sin(\theta) sin(\phi),\text{ }cos(\phi)\big)}.$$ Checking (i)-(iii) holds above, we have: $$\color{maroon}{\int_{\partial M}\omega = \int_DF^*(\omega)}.$$

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The formula for the pullback of differential forms by a smooth map (in a smooth chart) is: $$\color{maroon}{F^*(\omega_Idx^I) = (\omega_I\circ F)\cdot d(x^{i_1}\circ F)\wedge ... \wedge d(x^{i_k}\circ F)},$$ where $I = (i_1...i_k)$ is an ordered multi-index, the summation convention is applied, and $d$ is the total differential (or exterior derivative operator). Reference Lemma 14.16 (c) (p.361)

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Rewriting: $$\omega = xdy\wedge dz - ydx\wedge dz + zdx\wedge dy$$ Thus in our case we have the 3 ordered multi-indices, respectively: $I = (23)$, $(13)$, and $(12)$, with their component functions $\omega_I = x$, $-y$, and $z$.

Next, compute:

$d(x\circ F) = d(cos(\theta)sin(\phi)) = -sin(\theta)sin(\phi)d\theta+cos(\theta)cos(\phi)d\phi,$

$d(y\circ F) = d(sin(\theta)sin(\phi)) = cos(\theta)sin(\phi)d\theta+sin(\theta)cos(\phi)d\phi,\text{ and }$

$d(z\circ F) = d(cos(\phi)) = -sin(\phi)d\phi.$

Finally, plug in the results to the formula:

$F^*(\omega) = (cos(\theta)sin(\phi))\cdot \big(cos(\theta)sin(\phi)d\theta+sin(\theta)cos(\phi)d\phi\big)\wedge \big(-sin(\phi)d\phi\big)$ $$+ (-1)*(sin(\theta)sin(\phi))\cdot \big(-sin(\theta)sin(\phi)d\theta+cos(\theta)cos(\phi)d\phi\big)\wedge \big(-sin(\phi)d\phi\big)$$ $$+(cos(\phi))\cdot \big(-sin(\theta)sin(\phi)d\theta+cos(\theta)cos(\phi)d\phi\big)\wedge \big(cos(\theta)sin(\phi)d\theta+sin(\theta)cos(\phi)d\phi\big)$$ Simplifying:

$\small = (-cos^2(\theta)sin^3(\phi))d\theta\wedge d\phi+(-sin^2(\theta)sin^3(\phi))d\theta\wedge d\phi+(-sin^2(\theta)sin(\phi)cos^2(\phi)-cos^2(\theta)sin(\phi)cos^2(\phi))d\theta\wedge d\phi$ $$= \big(-cos^2(\theta)sin^3(\phi)-sin^2(\theta)sin^3(\phi)-sin^2(\theta)sin(\phi)cos^2(\phi)-cos^2(\theta)sin(\phi)cos^2(\phi)\big)d\theta\wedge d\phi$$ $$=(-sin^3(\phi)-sin(\phi)cos^2(\phi))d\theta\wedge d\phi$$ $$ = -sin(\phi)d\theta\wedge d\phi$$

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The result: $$\int_DF^*(\omega) = \int_0^{2\pi}\int_0^{\pi} sin(\phi)d\phi d\theta = 4\pi.$$

Final note: We skipped checking (i)-(iii). In the process of showing the map is orientation-preserving, one would have chosen orientations on the domain and codomain of $F$ such that this was the case. The orientations that make this work are $[\partial_{\phi},\partial_{\theta}]$ and the (outward) normal on the sphere. We need the orientation on the domain to list the nested integrals in the correct order.