Question

Find the mass of the solid cylinder D{(r,θ,z):0≤r≤3, 0≤z≤8} with density p(r,θ,z)= 1 + z/2.

Answer 1

The mass of the solid cylinder D with density p(r,θ,z) = 1 +
`(z)/(2)\)` where
`\(0 ≤ r ≤ 3\)` and
`\(0 ≤ z ≤ 8\)` is M = 132π.

Step-by-step explanation:

To find the mass of the solid cylinder, we'll integrate the density function p(r,θ,z) over the given region D. Using cylindrical coordinates, the volume element dV = r , dr , dθ , dz. The density function p(r,θ,z) = 1 +
`(z)/(2)\)`. Integrating over D, the mass M is given by:

M =
`\iiint_D p(r,θ,z) \, dV = \int_0^(2π) \int_0^3 \int_0^8 \left(1 + (z)/(2)\right) r \, dz \, dr \, dθ\]`

First integrating with respect to z, we get:

M =
`\int_0^(2π) \int_0^3 \left[(zr)/(2) + (z^2)/(4)\right]_0^8 \, dr \, dθ\]`

M =
`\int_0^(2π) \int_0^3 \left(4r + 16\right) \, dr \, dθ\]`

M =
`\int_0^(2π) \left[2r^2 + 16r\right]_0^3 \, dθ\]`

M =
`\int_0^(2π) \left(18 + 48\right) \, dθ\]`

M =
`\int_0^(2π) 66 \, dθ = 66 * 2π = 132π\]`

Thus, the mass of the solid cylinder D with the given density function is M = 132π.