- #1
norg012pp
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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?
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?