Integrate the 2-form $\omega=\frac1x dy\wedge dz-\frac1y dx\wedge dz$ over the following surface:
The top half of the unit sphere using the following parametrization.
$(r,\theta)\to (r\cos\theta, r\sin\theta,\sqrt{1-r^2})$, where $0\leq \theta\leq 2\pi$ and $0\leq r\leq 1$.
I am currently working through "A Geometric Approach to Differential Forms" by David Bachman, which gives a rather informal introduction to the topic, and this is a task, which I am unable to solve.
The reason I struggle here is that I have the following formula:
$\int_M \omega =\int_R \omega_{\phi(x,y)}\left(\frac{\partial\phi}{\partial x}(x,y),\frac{\partial\phi}{\partial y}(x,y)\right) dx\wedge dy$.
And I now want to integrate the 2-form above, which is part $dy\wedge dz$ and $dx\wedge dz$.
So I am confused on how to apply this formula in this case, where we have "$dx\wedge dz-dy\wedge dz$" and not just "$dx\wedge dy$". Can you help me out?
Thanks in advance.
Best Answer
I now solved this as follows.
I calculate the integrals seperatly. So first I calculate the integral for $\frac{1}{x} dy\wedge dz$
Then $\int_M \frac{1}{x}dy\wedge dz=\int_R \frac{1}{r\cos(\theta)} \left(\left\langle \cos(\theta),\sin(\theta),\frac{-r}{\sqrt{1-r^2}}\right\rangle,\left\langle-r\sin(\theta),r\cos(\theta),0\right\rangle\right)dr\wedge d\theta$
Where $\langle\cdot\rangle$ just notes a vector.
Now to respect $dy\wedge dz$ it seems I have to calculate the determinant given by the y- and z-coordinates. Hence
$\int_R \frac{1}{r\cos(\theta)}\begin{vmatrix}\sin(\theta)& r\cos(\theta)\\ \frac{-r}{\sqrt{1-r^2}}&0\end{vmatrix}dr\wedge d\theta=\int_R \frac{r}{\sqrt{1-r^2}}dr\wedge d\theta=\int_0^{2\pi}\int_0^1 \frac{r}{\sqrt{1-r^2}}dr\,d\theta=2\pi$
And with a similar calculation we achiev
$\int_M -\frac{1}{y}dx\wedge dz=2\pi$.
So the result is $2\pi+2\pi=4\pi$, which is correct according to the solutions in the named book.
One question remains: How does the integral "remember" the coordinates we started with after employing the formula for the parametrization. In the first integral I respected y- and z-coordinate.
In the second integral I take the determinant with regards to the x- and z-coordinate, which leads to
$\begin{vmatrix}\cos(\theta)&-r\sin(\theta)\\ \frac{-r}{\sqrt{1-r^2}}&0\end{vmatrix}$.
Since the solution is correct, I assume that this is the right way to do it, but I do not know why we have to take these specific determinants, as the formula is the same in both cases. So how is $dx\wedge dz$ respectively $dy\wedge dz$ respected in the formula?
Thanks in advance.