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To find the volume of the given solid below the plane x+y+z=6 and above the disk x2+y2≤1, you can set up a triple integral in cylindrical coordinates.

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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.

User Danfordham
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