<s>[INST] Finding the centroid of a tetrahedron I have four points to form a tetahedron
$$A=(0,-\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
 \\B=(0,\frac{1}{2},-\frac{1}{4}\sqrt{\frac{1}{2}})
\\C=(-\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})
\\D=(\frac{1}{2},0,\frac{1}{4}\sqrt{\frac{1}{2}})$$
it looks like this :

I'm asked to find the centroid of the tetrahedron formed by these points.
I looked up in the internet and there is an easy way to find it:
$$G=(\frac{x_1+x_2+x_3+x_4}{4}+\frac{y_1+y_2+y_3+y_4}{4}+\frac{z_1+z_2+z_3+z_4}{4})$$
I tried to check it with the triple intregrals. In order to do it I find the plane at which A, D and B lie, and that would be my first integration limit and the plane formed by A,B and C.
The planes :
$$\overline{AB}=(0,1,0)\\
\overline{AD}=(\frac{1}{2},\frac{1}{2},\frac{1}{2}\sqrt{\frac{1}{2}})$$
And after takind the cross product of both, and then using $a(x-x_0)+b(y-y_0)+c(z-z_0)$ to obtain the plane I get $$\frac{1}{2}\sqrt{\frac{1}{2}}x-\frac{1}{2}z=\frac{1}{8}\sqrt{\frac{1}{2}}$$ and the other one containing A,B,C is $$ \frac{1}{2}\sqrt{\frac{1}{2}}x+\frac{1}{2}z=\frac{1}{8}\sqrt{\frac{1}{2}}$$
I divide it into this two planes, and then see how it looks like in the $x,y$ plane

And I get the integrals:
$$ \int_{0}^{\frac{1}{2}}\int_{-x+\frac{1}{2}}^{\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{-\frac{1}{4}\sqrt{\frac{1}{2}}+x\sqrt{\frac{1}{2}}} zdz dy dx dx+\int_{\frac{1}{2}}^{0}\int_{0}^{-x+\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{-\frac{1}{4}\sqrt{\frac{1}{2}}+x\sqrt{\frac{1}{2}}}z dz dy dx $$
and the other part
$$ \int_{0}^{\frac{1}{2}}\int_{x+\frac{1}{2}}^{\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{\frac{1}{4}\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}}} zdz dy dx dx+\int_{\frac{1}{2}}^{0}\int_{0}^{-x-\frac{1}{2}}\int_{\frac{1}{4}\sqrt{\frac{1}{2}}}^{\frac{1}{4}\sqrt{\frac{1}{2}}-x\sqrt{\frac{1}{2}}}z dz dy dx $$
I don't get 0.
I can't see why. Is the formula I use not valid?
 [/INST] Few days ago, using vectog calculus, I have proved an analogous formula for centroid of a triangle. To appply this method here, two things are needed to check.


*

*The centroid is a common point of four lines, each of them joins a vertex with the centroid of the opposite face.

*The centroid divides this face in the ratio 3:1.


With this knowledge our task is trivial.
</s>