Can someone help me in understanding how to apply stokes theorem here?

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Quick remark, the question couldn't be answered at this point so I'm happy go read more answers

I have some problems in understanding how to apply stokes theorem. So we had the following version:

Let $U\subset \Bbb{R}^n$ be open and $M\subset U$ an orientable k-dimensional submanifold and $\omega$ a continuous differentiable $k-1$ form on $M$. Then for each compact $A\subset M$ with smooth boundary $$\int_A d\omega=\int_{\partial A} \omega $$where $\partial A$ has the induced orientation.

Now I have some general problems in applying this theorem since I don't really know what I all need to do especially with the orientation of the boundary. So if I remember correclty in the lecture we always took first a parametrisazion of the whole $A$ such that somehow the manifold is determined by the first coordinate beeing negative and then looked if it is orientation faithful (sorry I don't know the exact word in english). Then from this parametrization we have deduced a parametrization for the boundary.

Somehow it makes sense but when I consider the following example I struggle a bit finding a parametrization for the whole $A$ such that it is determined by the first coordinate being negative.

Take $A=\{(x,y,z)\in \Bbb{R}^3: x+y+z=1, x,y,z\geq 0\}$. This is clearly the triangle with edges $(0,0,1),(0,1,0),(1,0,0)$. then I thought that I could take the parametrization $\phi(x,y)=(x,y,1-x-y)$ but is my set $A$ then determined by the first coordinate being negative?

Example of the lecture

Let $A=\left\{x^2+y^2+z^2\leq 1: |z|\leq \frac{\sqrt{2}}{2}\right\}$. Let $\omega=x~dy+z~dx$. We want to use stoke to compute $\int_A d\omega$. Since this is a ball let us consider the general parametrisazion $$\phi(\theta, \rho)=(\cos\theta\sin\rho,\sin\theta\sin\rho,\cos\rho)$$ where $\theta \in [0,2\pi]$ and $\rho \in [\frac{\pi}{4},\frac{3\pi}{4}]$. Now remark that $$\partial A=\partial A_1 \cup \partial A_2$$ where in $\partial A_1$ we take $z=\frac{\sqrt{2}}{2}$ and in $\partial A_2$ we take $z=\frac{-\sqrt{2}}{2}$. Now wee need a parametrisation such that $$A\leftrightarrow \{x_1\leq 0\}$$ We take the follwoing parametrisation $$\phi_1(x_1,x_2)=\left(\cos(x_1)\sin(\frac{\pi}{4}-x_1), \sin(x_2)\sin(\frac{\pi}{4}-x_1), \cos(\frac{\pi}{4}-x_1)\right)$$. Then $\det(D\phi)=\det\left(D\phi_1\circ \begin{pmatrix} 0 & 1\\-1 & 0 \end{pmatrix}\right)>0$ Therefore it is oriantation faithful. Now we can built from this a parametrisation for $\partial A_1$. $$\psi_1(x_2)=\phi_1(0,x_2)=\left(\cos(x_2)\frac{\sqrt{2}}{2}, \sin(x_2) \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$

Then with this parametrization we computed the pullpack and then $\int_{\partial_A} \omega$ but this is clear. Then we did the same for $\partial A_2$

I hope you see what I mean, so the same procedure I wanted to apply to the exercise with the triangle.

Thanks for your help