Double application of Stokes' theorem to find the area of a simplex

[norg012pp]

I'm having trouble using Stokes' theorem in order to find a simple formula for the area found within a two dimensional simplex. I know the formula, but I'm interested in the derivation. For simplicity, I've been working with a unit triangle with vertices at the coordinates (0,0), (0,1), and (1,0). In this way, the area enclosed by these points should be 1/2.

The standard integration by parts formula tells us that

$$\int_{\Omega} \nabla u \cdot \mathbf{v} = \int_{\partial\Omega} (u \mathbf{v})\cdot n-\int_\Omega u \nabla\cdot\mathbf{v}$$

where $$n$$ denotes the outer unit normal, $$\Omega$$ denotes the domain, and $$\partial\Omega$$ denotes the boundary of the domain. In order to use this formula, we let $$\Omega$$ be equal to the unit triangle, $$\mathbf{v}=(1,1)$$, and $$u=(x+y)/2$$. In this way, we have that

$$\int_{\Omega} 1 = \int_{\partial\Omega} ((x,y)\in\Re^2\mapsto (x+y)/2) ((1,1)\cdot n)$$

Since the unit triangle has nice straight lines, the unit norm is constant along each face. Therefore, we can break up the above integral into three pieces where $$\partial\Omega_1=\{(x,y)\in\Re^2 : x=0, 0\leq y\leq 1\}$$, $$\partial\Omega_2=\{(x,y)\in\Re^2:y=1-x,0\leq x\leq 1\}$$, $$\partial\Omega_3=\{(x,y)\in\Re^2:y=0,0\leq x\leq 1\}$$. This allows us to express the integral as

$$\left((1,1)\cdot(-1,0)\int_{\partial\Omega_1} (x,y)\in\Re^2\mapsto (x+y)/2 \right)+\left((1,1)\cdot(1,1)\int_{\partial\Omega_2} (x,y)\in\Re^2\mapsto (x+y)/2\right)+\left((0,-1)\cdot(1,1)\int_{\partial\Omega_3} (x,y)\in\Re^2\mapsto (x+y)/2\right)$$

Simplifying, we have that the area is equal to

$$-1/4 + 2(1/2) - 1/4$$

which gives the correct area as $$1/2$$. Now, I would like to use this trick a second time so that the area of the triangle is expressed purely by its vertices. In order to do this, we let $$u=xy/2$$. In this way, we have that

$$\sum\limits_{i=1}^3 ((1,1)\cdot n_i) \int_{\partial\Omega_i} (x,y)\in\Re^2\mapsto (x+y)/2=\sum\limits_{i=1}^3 ((1,1)\cdot n_i) \int_{\partial\partial\Omega_i} ((x,y)\in\Re^2\mapsto xy/2)(\tilde{n}\cdot (1,1))$$

where $$\tilde{n}$$ denotes the outer unit normal to $$\partial\partial\Omega_i$$. Herein lies the trouble. The function $$(x,y)\in\Re^2\mapsto xy/2$$ evaluates to zero at every vertex of the unit triangle. Therefore, the above equivalence is not correct as the second integral evaluates to zero.

Using differential forms leads to a similar problem. In differential forms, Stokes' theorem has the form

$$\int_\Omega du = \int_{\partial\Omega} u$$

In order to evaluate the area of a triangle by only considering some function at the vertices, we would need the relations

$$\int_\Omega d(du) = \int_{\partial\Omega} du = \int_{\partial\partial\Omega} u$$

where $$d(du)=1$$. However, this is impossible since $$d(du)=0$$ for any smooth function $$u$$.

Is it possible to apply Stokes' theorem (or integration by parts) twice in order to determine the area of some simplex?

 

[norg012pp]

This question was cross posted on the mathhelpforum and resolved there:

http://www.mathhelpforum.com/math-help/f57/double-application-stokes-theorem-find-area-simplex-167156.html

Of course, if someone has a different explanation, it's welcome.

 

[xaos]

the boundary of a boundary should be empty, so no...