I Triple equation for integral on a graph

[NODARman]

TL;DR Summary

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Hi, so I'm trying to find the volume of a shape using integral, I found the equation of one plane in 3D space but the second one is something like that, which I cannot write in integral as a function: $\frac{2(2x-a)}{a}=-\frac{2(6y-a\sqrt3)}{a\sqrt3}=\frac{2z-a\sqrt3}{a\sqrt3}$
In the 3D viewer program, I wrote this and it made a line, but I needed a plane. Other points of that shape were connected, for example, z=3y and it was a plane, that can be written in integral. When I solve for x it wasn't right in the 3D coordinate system space.
Like I can write integral of (3y)dy but not integral of (2x=3y=4z)dxdydz... :(

 

[scottdave]

Is this homework?

Are there other surfaces that bound the volume?

How is the parameter "a" used?

 

[NODARman]

Is this homework?

Are there other surfaces that bound the volume?

How is the parameter "a" used?

Sorry, I wrote little information, so it's a regular tetrahedron.
I recalculated that and the function of the line (which is bisectriz of the BCD triangle/surface) is a little bit different. But the main problem is how to calculate the volume under that triangle. I've searched the formula of the line (in 3D space) on the internet btw.

"a" is the length of the edges.

P.S. Kinda homework-ish...

 

[scottdave]

You may benefit from reviewing these resources for integration with "funny limits":
Here is a YouTube video:



and Paul's online notes does a pretty good job of explaining it:
https://tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx