# Two form surface integral over sphere

I'm trying to compute $$\int_M \omega$$ with \begin{align*} \omega &= x^4 dy \wedge dz + y^4 dz \wedge dx + z^4 dx \wedge dy, \\ M &: x^2 + y^2 + z^2 = R^2. \end{align*} I have done this in two ways: Stokes' theorem and direct computation of the wedge product via a spherical coordinate transformation. Using $$\phi$$ as azimuthal and $$\theta$$ as polar angle and applying Stokes I obtain \begin{align*} 4 \int_0^{2 \pi}\int_0^{\pi} (x^3 + y^3 + z^3) R^2 \sin \theta d\theta d\phi = 0, \end{align*} where I used the standard \begin{align*} x &= R \sin \theta \cos \phi \\ y &= R \sin \theta \sin \phi \\ z &= R \cos \theta. \end{align*} By directly computing the wedge products and plugging them in I obtain \begin{align*} dy \wedge dz &= R^2 \sin^2 \theta \cos \phi d\theta \wedge d\phi \\ dz \wedge dx &= R^2 \sin^2 \theta \sin \phi d\theta \wedge d\phi \\ dx \wedge dy &= R^2 \cos \theta \sin \theta d\theta \wedge d\phi \\ \int_M x^4 dy &\wedge dz + y^4 dz \wedge dx + z^4 dx \wedge dy = 0. \end{align*} The solution given by my professor was $$\frac{12}{5} \pi^2 R^5$$, which neither method agrees with. Then I repeat this procedure but instead of taking a sphere, I take a hemisphere. Changing the limits in the $$\theta$$ integrals to $$[0, \frac{\pi}{2}]$$, the Stokes' method yields $$2 \pi R^5$$ and the direct wedge computation yields $$\frac{\pi R^6}{3}$$. The solution given then is just $$\frac{6}{5} \pi^2 R^5$$, which also does not agree with my solutions. Any help with my mistakes would be greatly appreciated.

• Wolfram mathematica says that the integral, using Stokes, is zero. Maybe there is a typo in the exercise or the order in the differential forms Apr 21 at 18:30
• @MasacrosoYeah that is exactly the problem I had, but since the hemisphere's don't give the same solution for both methods I was unsure whether the approach was wrong Apr 21 at 18:37
• note that for a hemisphere we need to add a circular surface to complete the boundary, did you take that into account when using Stoke's theorem? Apr 21 at 19:07
• $\frac{\pi R^6}{3}$ is the correct answer and both methods yield the same result. You may have made a mistake to get $2\pi R^5$. Please see my working. Apr 21 at 19:34

In case of sphere, the surface integral is indeed zero.

Please note in below working of mine, I use $$\theta$$ as azimuthal angle and $$\phi$$ as polar. I think you have the other way round. So want to call out to avoid confusion.

In case of hemisphere above $$z = 0$$,

Parametrization of sphere is $$r(\phi, \theta) = (R \cos \theta \sin \phi, R \sin\theta \sin\phi, R \cos\phi)$$

For outward normal vector, $$\displaystyle r'_{\phi} \times r'_\theta = R^2 \sin\phi(\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi)$$

The integral becomes,

$$\displaystyle \small \int_0^{2\pi}\int_0^{\pi/2} R^2 \sin\phi \ (R^4 \cos^4\theta\sin^4\phi, R^4 \sin^4\theta\sin^4\phi, R^4 \cos^4\phi) \cdot (\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi) \ d\phi \ d\theta$$

$$= \displaystyle \frac{\pi R^6}{3} \$$, which is same as your working.

In case of applying divergence theorem, we close the surface with a disk at $$z = 0$$.

$$\small \nabla \cdot (x^4, y^4, z^4) = 4(x^3+y^3+z^3) = 4r^3(\cos^3\theta \sin^3\phi+ \sin^3\theta \sin^3\phi + \cos^3\phi)$$

where $$0 \leq r \leq R$$.

So volume integral is,

$$\displaystyle \small \int_0^{2\pi} \int_0^{\pi/2} \int_0^R 4r^5 \sin\phi (\cos^3\theta \sin^3\phi+ \sin^3\theta \sin^3\phi + \cos^3\phi) \ dr \ d\phi \ d\theta = \frac{\pi R^6}{3}$$.

Now note the flux through the bottom disk is zero so flux through spherical surface is $$\frac{\pi R^6}{3}$$, the same answer we received through direct surface integral.

• Ah yes thanks makes sense now. Guess the provided solutions were wrong. Apr 21 at 19:47
• You are welcome. Yes it is wrong or they mistyped the vector field / region. Apr 21 at 19:50