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?
 A: Your equation for the plane that passes through $\{A, B, C\}$ should be
$\displaystyle \frac12\sqrt{\frac12} x + \frac12 z = -\frac18 \sqrt{ \frac12 }$.
Namely, you were off by one negative sign starting from there.
As for the integrals, guessing your intention from the way you set them up, it looks like you also need the planes that pass through $\{A,C,D\}$ and $\{B,C,D\}$.
Currently you have only the equations for the two lower planes. You also need the two upper planes.
That is, even if you plug in the correct equation for the $A$-$B$-$C$ plane and get zero (something like $\frac{-1}{768}+\frac1{768}+\frac{-1}{768}+\frac1{768}$), that would be a coincidence.

The formula of taking the arithmetic mean of the four points is valid here. Even though your tetrahedron is not a regular tetrahedron (which all four faces are equilateral triangles), it still has the relevant symmetry to make the centroid be at the "average".
A: 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.
