# Integrating a 2-form over an ellipsoid

How can I calculate the integral of the 2-form $$w=xdx\wedge dy+xdy\wedge dz-dx\wedge dz$$ on the inferior hemisphere ($$z\leq 0$$) of the ellipsoid $$\frac{x^2}{9}+\frac{y^2}{9}+z^2=1$$ with the orientation determined by the normal vectors which have the third component negative on this part of the ellipsoid

I know first I should parametrize the ellipsoid: $$x = 3\cos\theta\sin\phi,\; y= 3\sin\theta\sin\phi,\; z=\cos\phi$$ Where $$(\theta, \phi) \in [0,2\pi]\times [\frac{\pi}{2},\frac{3\pi}{2}]$$ since ($$z\leq 0$$). And computing \begin{align*} \mathrm{d}x &= -3\sin\theta\sin\phi\mathrm{d}\theta + 3\cos\theta\cos\phi\mathrm{d}\phi\\ \mathrm{d}y &= 3\cos\theta\sin\phi\mathrm{d}\theta + 3\sin\theta\cos\phi\mathrm{d}\phi\\ \mathrm{d}z &=-\sin\theta\mathrm{d}\phi \end{align*}

I know I should use $$\int_M \omega = \int_U \varphi^*\omega$$

But I don't know how to compute $$\varphi^*\omega$$

I calculated everything and I have now to compute the integral $$\int_0^{2\pi}\int_{\pi/2}^{\frac{3\pi}{2}}-3\sin^2\phi(9\cos\theta \cos\phi+3\cos^2\theta \sin\phi + \sin\phi) d\theta d\phi = \int_0^{2\pi}\int_{\pi/2}^{\frac{3\pi}{2}}-9\cos^2\theta \sin^3\phi d\theta d\phi=0,$$

since the integrals $$\int_0^{2\pi}\cos(x)dx=\int_0^{2\pi}\sin(x)dx=\int_0^{2\pi}\cos(x)\sin(x)dx=0.$$

• What is $\omega$? You have only defined $w$. – user10354138 Jun 7 at 13:12

If $$\varphi(\theta,\phi)=(3\cos\theta\sin\phi,3\sin\theta\sin\phi,\cos\phi),$$ then you compute the pullback by substitution (general form given at end):

$$\varphi^*\omega=(3\cos\theta\sin\phi)d(3\cos\theta\sin\phi)\wedge d(3\sin\theta\sin\phi)+(3\cos\theta\sin\phi)d(3\sin\theta\sin\phi)\wedge d(\cos\phi)-d(3\cos\theta\sin\phi)\wedge d(\cos\phi).$$ Note that you've computed all of the $$d$$'s already in your answer, so you just need to substitute them in and then compute the wedge products. I'll leave that to you to try. Then, the integration is a standard multi-variable calculus computation, which you should be able to perform without issue.

In general, the pullback of a one-form $$\omega=\sum\limits_{j} a_jdx^j$$ in local coordinates under $$\varphi$$ has the form $$\varphi^*\omega=\sum\limits_{i,j} (a_i\circ \varphi)\frac{\partial \varphi^i}{\partial x^j}dx^j.$$

• I have edited my question with what I got. Could you please check if my result is correct? It is the first time I have done such calculus and I am unsure if what I did is ok. Thank you – codingnight Jun 7 at 16:38
• Sure, I'll read it over. – cmk Jun 7 at 17:32
• On a glance, it looks like in your integral, the last term in parenthesis should have a $\theta$ instead of a $\phi.$ The same is true, but in reverse, for your $dz$ expression. Just to be safe, I would encourage you to check it once more yourself, it is easy to make small mistakes on these types of things! Your process looks correct. – cmk Jun 7 at 17:54
• @codingnight to clarify, you $dz$ should've been $-\sin\phi d\phi$ (I realize that my phasing was confusing). – cmk Jun 7 at 19:19

The iterated integral should be $$\pm \int_{\pi/2}^\pi \int_0^{2 \pi} f(\phi, \theta) \, d\theta d\phi.$$ You have typos in $$dz$$ and in the integrand. To determine the sign, check the direction of the cross product of the tangent vectors given by your parametrization.

If $$\omega = F^1 \, dy \wedge dz + F^2 \, dz \wedge dx + F^3 \, dx \wedge dy$$, then $$d\omega = \nabla \cdot \mathbf F \, dx \wedge dy \wedge dz$$. Since the integral of $$x \, dx \wedge dy$$ over a disk in the $$x y$$-plane centered at the origin is zero, your integral can also be evaluated as $$\frac 1 2 \int_{x^2/9 + y^2/9 + z^2 \leq 1} dV.$$