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In $\mathbb{R}^3$ consider following 2-form given by $$\omega = xy \: dx \wedge dy + 2x \: dy \wedge dz + 2y \: dx \wedge dz$$ and $$A = \{(x,y,z) \in \mathbb{R}^3 | x^2+y^2+z^2=1, z\geq 0\}.$$ Show, that the Integral of $\omega$ over $A$ vanishes.

Using the map $$\phi\colon\mathbb{R}^2\rightarrow \mathbb{R}^3, \:\:(u,v)\mapsto\:(sin(u)cos(v), sin(u)sin(v), cos(u)).$$

I calculated the pullback $\phi^\ast\omega$:

$d\phi_1(u,v):=d(x\circ \phi)(u,v) = \frac{\partial\phi_1}{\partial u}(u,v)du+\frac{\partial\phi_2}{\partial v}(u,v)dv=cos(u)cos(v)du-sin(u)sin(v)dv\\ d\phi_2(u,v):=d(y\circ \phi)(u,v) = \frac{\partial\phi_1}{\partial u}(u,v)du+\frac{\partial\phi_2}{\partial v}(u,v)dv=cos(u)sin(v)du+cos(v)sin(u)dv \\ d\phi_3(u,v):=d(z\circ \phi)(u,v) = \frac{\partial\phi_1}{\partial u}(u,v)du+\frac{\partial\phi_2}{\partial v}(u,v)dv=-sin(u)du$

$\phi^\ast\omega=d\phi_1\wedge d\phi_2\wedge d\phi_3(u,v)=(cos(u)vos(v)du-sin(u)sin(v)dv)\wedge (cos(u)sin(v)du+cos(v)sin(u)dv) \wedge (-sin(u)du)=(cos(u)cos^2(v)sin(u)dudv + sin(u)sin^2(v)cos(u)dudv) \wedge (-sin(u)du)= (cos(u)sin(u)dudv) \wedge (-sin(u)du)$

But now it would follow that the pullback is zero. Can somebody tell me where the mistake is?

My next step shall be using that $\phi^{-1}=[-1,1]\times[-1,1]$ and calculating the integral $\int_{[-1,1]\times[-1,1]}\phi^\ast\omega$. Is this a valid way?

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  • $\begingroup$ Why is a pullback of a $2$-form a $3$-form? $\endgroup$ – Tunococ Feb 8 '16 at 9:47
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I hope this answer satisfies the sadistic requirement of the problem.

The correct pullback $\phi^*\omega$ is \begin{align} \phi^*\omega & = \left( xy \begin{vmatrix} \frac{\partial\phi_1}{\partial u} & \frac{\partial\phi_1}{\partial v}\\ \frac{\partial\phi_2}{\partial u} & \frac{\partial\phi_2}{\partial v} \end{vmatrix} + 2x \begin{vmatrix} \frac{\partial\phi_2}{\partial u} & \frac{\partial\phi_2}{\partial v}\\ \frac{\partial\phi_3}{\partial u} & \frac{\partial\phi_3}{\partial v} \end{vmatrix} + 2y \begin{vmatrix} \frac{\partial\phi_1}{\partial u} & \frac{\partial\phi_1}{\partial v}\\ \frac{\partial\phi_3}{\partial u} & \frac{\partial\phi_3}{\partial v} \end{vmatrix}\right) du \wedge dv \\ & = \left(xy\cos u \sin u + 2x \sin^2 u \cos v - 2y \sin^2 u \sin v \right) du \wedge dv \\ & = \left(\sin^3u \cos u \sin v \cos v + 2 \sin^3 u \cos^2 v - 2 \sin^3 u \sin^2 v \right) du \wedge dv. \end{align} Directly computing the integral on $R = [0, \pi) \times [0, 2\pi)$ yields \begin{align*} \int_R \phi^*\omega & = \int_0^\pi \int_0^{2\pi} \left(\sin^3u \cos u \sin v \cos v + 2 \sin^3 u \cos^2 v - 2 \sin^3 u \sin^2 v\right) dv du \\ & = 0 + \frac{8\pi}3 - \frac{8\pi}3 = 0. \end{align*}

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Notice that $\int_A\omega$ is equal to the integral of $\omega$ over the lower half of the sphere and so $\int_A\omega=\frac12\int_Bd\omega$, where $B=\partial A$ is the unit ball. Since $d\omega=0$ we get $\int_A\omega=0$.

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  • $\begingroup$ You are using Stokes' theorem which is not allowed in this exercise. $\endgroup$ – monoid Feb 8 '16 at 9:42
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    $\begingroup$ But is is easy to check that this form is exact : $\omega = 1/4 dx^2 \wedge dy^2+2 d(xy)\wedge dz= d( 1/4x^2 d y^2+ 2(xy)dz)$. So you dont need Stokes, just Green Riemann $\endgroup$ – Thomas Feb 8 '16 at 10:43
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    $\begingroup$ @Thomas Green's theorem is very particular case of Stokes' theorem (for forms). $\endgroup$ – John B Feb 8 '16 at 12:27

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