Final answer:
The volume of the pyramid with specified vertices can be found using a triple integral, with limits from 0 to 22 for both x and y, and 0 to a linear function dependent on x for z, which accounts for the tapering of the pyramid.
Step-by-step explanation:
The volume of the pyramid with vertices at (0,0,0), (22,0,0), (22,22,0), (0,22,0), and (0,0,44) can be calculated using a triple integral. We should aim to set it up such that it is aligned with the pyramid's base and height for efficiency. To find the limits of integration for such a pyramid, consider the pyramid's base in the xy-plane, with vertices at (0,0), (22,0), (22,22), and (0,22). The height extends along the z-axis to (0,0,44).
Since we know the pyramid is symmetrical and has a square base, we can use symmetry to our advantage by setting the limits for x and y from 0 to 22. For the z-axis, because the height of the pyramid is linearly related to x and y, we can set the limit for z from 0 to a linear function that starts at z = 44 when x=0 (on the line y=0) and goes to z=0 as x reaches 22.
Thus, the triple integral to find the volume V would be:
\[ V = \int_{0}^{22} \int_{0}^{22} \int_{0}^{44(1 - (x/22))} dz\, dy\, dx \]