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.
To evaluate the triple integral of xz over the given region, we need to set up the bounds for each variable. Since you have a tetrahedron defined by its vertices, we can establish the boundaries using a combination of equations and inequalities.
Let's start by looking at the x-coordinate. We can see that it ranges from 0 to 1 as the tetrahedron extends along the x-axis. Therefore, the bounds for x are:
0 ≤ x ≤ 1
Next, let's consider the y-coordinate. Looking at the vertices, we see that y is always equal to 1 within the tetrahedron. So the bounds for y are:
y = 1
Finally, we need to determine the bounds for the z-coordinate. The z-value changes as we move within the tetrahedron along the y-axis. To find the equation defining this boundary, we can start by looking at the line connecting (0,1,0) and (0,1,1). This line can be parametrized as:
z = t, where t ranges from 0 to 1
To figure out the bounds of z, we need to consider the range of t that corresponds to the region of the tetrahedron we are interested in. We want z to lie within the corresponding values of y and x. Start by examining the coordinates of the vertices (0,0,0) and (0,1,0). Notice that for y = 0, z is always 0. So we need to find the range of t that makes z stay between 0 and 1 as y varies from 0 to 1 and x remains 0.
By comparing the coordinates, we can conclude that when y varies from 0 to 1, z varies from 0 to t along the line (0,t,0). Therefore, the range of t is:
0 ≤ t ≤ y
Combining all the bounds, we have:
0 ≤ x ≤ 1
y = 1
0 ≤ z ≤ y
Now, to evaluate the triple integral of xz over this tetrahedral region, you only need to set up the integral using these bounds:
∫∫∫ xz dV
Where the outermost integral is with respect to x, the middle integral is with respect to y, and the innermost integral is with respect to z.