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I have the scalar function $$f(x,y,z)=xy+y^2+x+z+1$$ defined over the tetrahedron $K$ with vertices $(x_1,y_1,z_1)=(1,2,3)$, $(x_2,y_2,z_2)=(1,2,4)$, $(x_3,y_3,z_3)=(1,3,5)$ and $(x_4,y_4,z_4)=(-1,0,1)$.

I need to calculate the integral of $f$ over the face $F$ (face of $K$) defined by the vertices $(x_1,y_1,z_1)$, $(x_3,y_3,z_3)$ and $(x_4,y_4,z_4)$.

To do this, I'm following the wikipedia formula with its notation.

  • It is easy so see than $$x-2y+z=0$$ is the formula of the plane over the face $F$.

  • The (natural) parametrization of this plane is $$\begin{cases}x(u,v)=u\\ y(u,v)=v\\ z(u,v)=2v-u\end{cases}$$

  • It follows that $$r_u\times r_v=\begin{vmatrix}i&j&k\\ \dfrac{\partial x}{\partial u}&\dfrac{\partial y}{\partial u}&\dfrac{\partial z}{\partial u}\\ \dfrac{\partial x}{\partial v}&\dfrac{\partial y}{\partial v}&\dfrac{\partial z}{\partial v}\end{vmatrix}=\begin{vmatrix}i&j&k\\ 1&0&-1\\ 0&1&2\end{vmatrix}=(1,-2,1)\Rightarrow \|r_u\times r_v\|=\sqrt{6}$$

With this: $$\int_F f(x,y,z)\, dS=\int_T f(x(u,v),y(u,v),z(u,v))\, \|r_u\times r_v\|\,dudv=\sqrt{6}\color{red}{\int_T (uv+v^2+2v-u+1)dudv}$$

and now I have two questions:

  • I'm on the right way?
  • How can I calculate the last integral? (the red one) I don't know wich are the restrictions that defines $T$ over the plane $uv$.
yemino
  • 814

1 Answers1

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You're pretty close except you forgot to add an $x=u$ to the integrand at the last step. Should be $uv+v^2+2v+1$. In the $uv$-plane the triangle is above $(1,2)$, $(1,3)$, and $(-1,0)$. The first two points share a common $u$, so it's probably easiest to take $u$ outer and $v$ inner. Between $(-1,0)$ and $(1,2)$ the two point formula gives $v=u+1$, while between $(-1,0)$ and $(1,3)$, $v=\frac32(u+1)$. Thus we get $$\sqrt6\int_{-1}^1\int_{u+1}^{\frac32(u+1)}(uv+v^2+2v+1)dv\,du$$ $$=\sqrt6\int_{-1}^1\left\{\frac12(u+1-1)\frac54(u+1)^2+\frac13\left(\frac{19}8\right)(u+1)^3+\frac54(u+1)^2+\frac12(u+1)\right\}du$$ $$=\sqrt6\left[\frac58\cdot\frac14(u+1)^4-\frac58\cdot\frac13(u+1)^3+\frac{19}{24}\cdot\frac14(u+1)^4+\frac54\cdot\frac13(u+1)^3+\frac12\cdot\frac12(u+1)^2\right]_{-1}^1$$$$=\frac{50}{\sqrt6}$$

user5713492
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