Final answer:
The question involves calculating the volume under a surface in three dimensions by setting up and evaluating a double integral over a specified region in the first octant.
Step-by-step explanation:
The student has asked to integrate a function f(x,y,z)=x over a specified region in the first octant, bounded above by the surface z=8-2x²-y² and below by z=y². To solve this, we can set up the integral in terms of the bounding surfaces, and choose appropriate limits for x, y, and z.
Because we're integrating with respect to x and the function is simply x, this problem simplifies to finding the volume under the surface z=8-2x²-y² and above z=y² in the first octant. We need to project the region into the xy-plane and determine the bounds for integration. The projected region would be the area where the circle x²+y²=8 (from rearranging z=8-2x²-y² to find where z=0) intersects with the first octant.
To integrate, we perform a double integral where x varies from 0 to √(8-y²)/√2, and y varies from 0 to √8, integrating the function x between the bounds z=y² and z=8-2x²-y². The process of carrying out this double integral will yield the desired volume.