Final answer:
The volume of the solid can be found by setting up a triple integral in cylindrical coordinates, where the limits for r are 0 to 1, for θ are 0 to 2π, and z is from 0 to 6 - r(cos(θ) + sin(θ)).
Step-by-step explanation:
To find the volume of the solid below the plane x+y+z=6 and above the disk x^2+y^2≤1, we need to set up a triple integral in cylindrical coordinates. In cylindrical coordinates, the disk has a simple representation because it corresponds to the region where the radial distance r is less than or equal to 1, and the angle θ can vary from 0 to 2π.
The conversion from Cartesian to cylindrical coordinates gives us:
- x = r cos(θ)
- y = r sin(θ)
- z remains the same.
The equation of the plane x+y+z=6 in cylindrical coordinates becomes r cos(θ) + r sin(θ) + z = 6, which simplifies to z = 6 - r(cos(θ) + sin(θ)).
Considering the bounds for r and θ, the triple integral takes the form:
∫ ∫ ∫ (6 - r(cos(θ) + sin(θ))) r dr dθ dz
The bounds of integration for r are from 0 to 1, for θ from 0 to 2π, and for z from 0 to 6 - r(cos(θ) + sin(θ)). This will yield the volume of the specified solid.
Integrating this will give us the desired volume.