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This is Problem 23.5 from Tu's An Introduction to Manifolds.

Prove that the area form $\omega$ on $S^2$ in Example 23.11 is equal to the orientation form $$x dy \wedge dz - y dx \wedge dz + z dx \wedge dy$$ of $S^2$.

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The hint says to take the exterior derivative of $x^2 + y^2 + z^2 = 1$ to obtain a relation among the $1$-forms $dx, dy, dz$ on $S^2$. Then show for example that for $x \neq 0$, one has $dx \wedge dy = (z/x)dy \wedge dz$. However, I do not see how to solve this problem using this hint. I would appreciate any help.

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Following the path suggested, you get $x\,dx+y\,dy+z\,dz = 0$ on $S^2$. So, for example, when $z\ne 0$, we have $dz = -\dfrac xz\,dx - \dfrac yz\,dy$. Thus, \begin{align*} x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy &= (x\,dy-y\,dx)\wedge dz + z\,dx\wedge dy \\ &= (x\,dy-y\,dx)\wedge \big(-\frac xz\,dx+\frac yz\,dy\big) + z\,dx\wedge dy \\ &= \big(\frac{x^2}z +\frac{y^2}z\big)dx\wedge dy + z\,dx\wedge dy \\ &= \frac{x^2+y^2+z^2}z dx\wedge dy = \frac{dx\wedge dy}z, \end{align*} as required.

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