Final answer:
To evaluate the given triple integral, set up the integral with proper bounds for z, y, and x, and then perform integration in the order dz, dy, dx considering the bounds given by the planes and the equation y²z=1.
Step-by-step explanation:
The question asks to evaluate the triple integral of the function y within the region bounded by the planes x=0, y=0, z=0, and y²z=1. To solve this, we must first establish the limits of integration for each variable. Since we have the equation y²z=1, we can express z as z=1/y². Therefore, the limits for z are from 0 to 1/y², considering the plane z=0 as the lower bound and y²z=1 as the upper bound.
For the y-integral, there are no explicit bounds given for y other than it must be greater than 0 due to the y=0 plane. But since y² cannot be negative, we can infer that y must also be bounded above, possibly by infinity. Lastly, since there are no restrictions on x apart from x being non-negative (x=0), we can also assume x ranges from 0 to infinity.
Once the limits of integration are established, the triple integral can be set up as:
∫∫∫ y dz dy dx with bounds for z from 0 to 1/y², bounds for y from 0 to infinity, and bounds for x from 0 to infinity.