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.

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

Oh, not sure about that last line, have to do the integral out at every height because x is in the integrand.

To set up the bounds for evaluating the triple integral, we need to determine the range for each variable (x, y, and z) within the given tetrahedron.

Let's start by considering the bounds for x. Looking at the vertices of the tetrahedron, we can see that x varies from 0 to 1, which means the bounds for x are x = 0 to x = 1.

Next, let's determine the bounds for y. By inspecting the vertices, we find that y ranges from 0 to 1. However, we need to express the bounds in terms of x to capture the triangular shape of the tetrahedron.

Inspecting the x-coordinates of the vertices, we can see that x = 0 acts as the base of the triangle and extends to x = 1. So, for each value of x within the range we found earlier, y should vary from y = 0 to y = x. Therefore, the bounds for y are y = 0 to y = x.

Finally, let's determine the bounds for z. By examining the vertices again, we observe that z varies from 0 to 1. However, similar to y, we need to express the bounds in terms of both x and y to account for the triangular shape.

Inspecting the coordinates of the vertices, we notice that z = 0 forms a triangular plane, and as we move along that plane from x = 0 to x = 1, the corresponding z values change linearly from z = 0 to z = 1. Within each line segment from y = 0 to y = x, z should vary linearly from z = 0 to z = x. Thus, the bounds for z should be z = 0 to z = x.

To summarize, the bounds for the triple integral of xz over the given tetrahedron are:
- x ranges from 0 to 1
- y ranges from 0 to x
- z ranges from 0 to x

Now that we have established the bounds, you can proceed with the integration of the function xz with respect to x, y, and z over their respective ranges to evaluate the triple integral.