Did the problem say the solid was a tetrahedron with those corners? I will have to assume that to be the case.
Now look at that thing.
It has a base in the xy plane. Labeling those points as a, b, c and d then a b and d have z = 0
SO
every layer up in z is a triangle parallel to the base in the x,y plane.
You can find the x and y of the corners of that triangle at every z because each edge is a straight line.
For example when z = 0
x goes from 0 to 1 while y goes from 0 to 1 or y = x
then x goes from 0 to 1 while y goes from 1 to 1 or y = 1
so the integral at that z would be from x = 0 to x = 1
and from y = x to y = 1
of course that contribution would be zero because z = 0 at this level
however you can define the limits at every height z as the sides of the triangle at that height.
so what you want is x z times the area of that triangle at each z
Hi, I've been working on this problem:
Evaluate the triple integral of xz with the vertices (0,0,0), (0,1,0), (0,1,1), and (1,1,0).
I drew out the tetrahedron, but I can't set up the bounds. Usually, they give you an equation as a bound. Here, I'm just completely lost.
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Oh, not sure about that last line, have to do the integral out at every height because x is in the integrand.
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