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If you have $\omega=x_3 dx_1\wedge dx_2$, $\sigma(\theta,\phi) = (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi)$ defined on $[0,2\pi]\times[-\pi/2,\pi/2]$, calculate $\int_\sigma \omega$ directly.

What I did/thought:

So, I know that for a continuously differentiable mapping $\sigma':Q_2\to \mathbb{R^3}$, we should be able to write something like: $\int_{\sigma'} \omega = \int_{Q_2}\omega(D_1\sigma'(x),D_2\sigma'(x)) \quad dx$, where $Q_2$ is the 2-simplex. But, I'm not sure how to proceed, it's not immediate to me that our $\sigma$ is such a mapping from the $Q_2$ simplex and also, even if it's, I don't know how what to do beyond this:

$\int_{\sigma} \omega = \int_{Q_2}\omega(D_1\sigma(x),D_2\sigma(x)) dx = \int_{Q_2}\omega((-\sin\phi\cos\theta,\sin\phi\cos\theta,0),(-\sin\phi\cos\theta, -\sin\phi\sin\theta, cos\phi)) dx $

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$x_1=\sin\phi\cos\theta, x_2=\sin\phi\sin\theta,x_3=cos\phi$

$dx_1=-\sin\phi\sin\theta d\phi+\cos\phi\cos\theta d\theta$

$dx_2=\sin\phi\cos\theta d\phi+\cos\phi\sin\theta d\theta$

$$\omega=x_3dx_1\wedge dx_2=\cos\phi((-\sin\phi\sin\theta d\phi+\cos\phi\cos\theta d\theta)\wedge(\sin\phi\cos\theta d\phi+\cos\phi\sin\theta d\theta)=\cos\phi(-\cos^2\phi\sin^2\theta d\phi d\theta+\sin\phi\cos\phi\cos^2\theta d\theta d\phi)=\cos\phi(\sin\phi\cos\phi+\cos^2\phi) d\phi d\theta$$

$$\int_{\sigma([0,2\pi]\times[-\pi/2,\pi/2])}\omega=\int_{[0,2\pi]\times[-\pi/2,\pi/2]}\cos\phi(\sin\phi\cos\phi+\cos^2\phi) d\phi d\theta$$

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