Ask Question
11 November, 02:11

Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and above the region in the xy-plane bounded by the curves y = x, y = 0, and x = 1; ρ (x, y, z) = 8.

+4
Answers (1)
  1. 11 November, 02:26
    0
    The mass of the solid is 16 units.

    The center of mass of the solid lies at (0.6875, 0.3542, 2.021)

    Work:

    Density function: ρ (x, y, z) = 8

    x-bounds: [0, 1], y-bounds: [0, x], z-bounds: [0, x+y+3]

    The mass M of the solid is given by:

    M = ∫∫∫ρ (dV) = ∫∫∫ρ (dx) (dy) (dz) = ∫∫∫8 (dx) (dy) (dz)

    First integrate with respect to z:

    ∫∫8z (dx) (dy), evaluate z from 0 to x+y+3

    = ∫∫[8x+8y+24] (dx) (dy)

    Then integrate with respect to y:

    ∫[8xy+4y²+24y]dx, evaluate y from 0 to x

    = ∫[8x²+4x²+24x]dx

    Finally integrate with respect to x:

    [8x³/3+4x³/3+12x²], evaluate x from 0 to 1

    = 8/3+4/3+12

    = 16

    The mass of the solid is 16 units.

    Now we have to find the center of mass of the solid which requires calculating the center of mass in the x, y, and z dimensions.

    The z-coordinate of the center of mass Z is given by:

    Z = (1/M) ∫∫∫ρz (dV) = (1/16) ∫∫∫8z (dx) (dy) (dz)

    Calculate the integral then divide the result by 16.

    First integrate with respect to z:

    ∫∫4z² (dx) (dy), evaluate z from 0 to x+y+3

    = ∫∫[4 (x+y+3) ²] (dx) (dy)

    = ∫∫[4x²+24x+8xy+4y²+24y+36] (dx) (dy)

    Then integrate with respect to y:

    ∫[4x²y+24xy+4xy²+4y³/3+12y²+36y]dx, evaluate y from 0 to x

    = ∫[28x³/3+36x²+36x]dx

    Finally integrate with respect to x:

    [7x⁴/3+12x³+18x²], evaluate x from 0 to 1

    = 7/3+12+18

    Z = (7/3+12+18) / 16 = 2.021

    The y-coordinate of the center of mass Y is given by:

    Y = (1/M) ∫∫∫ρy (dV) = (1/16) ∫∫∫8y (dx) (dy) (dz)

    Calculate the integral then divide the result by 16.

    First integrate with respect to z:

    ∫∫8yz (dx) (dy), evaluate z from 0 to x+y+3

    = ∫∫[8xy+8y²+24y] (dx) (dy)

    Then integrate with respect to y:

    ∫[4xy²+8y³/3+12y²]dx, evaluate y from 0 to x

    = ∫[20x³/3+12x²]dx

    Finally integrate with respect to x:

    [5x⁴/3+4x³], evaluate x from 0 to 1

    = 5/3+4

    Y = (5/3+4) / 16 = 0.3542

    The x-coordinate of the center of mass X is given by:

    X = (1/M) ∫∫∫ρx (dV) = (1/16) ∫∫∫8x (dx) (dy) (dz)

    Calculate the integral then divide the result by 16.

    First integrate with respect to z:

    ∫∫8xz (dx) (dy), evaluate z from 0 to x+y+3

    = ∫∫[8x²+8xy+24x] (dx) (dy)

    Then integrate with respect to y:

    ∫[8x²y+4xy²+24xy]dx, evaluate y from 0 to x

    = ∫[12x³+24x²]dx

    Finally integrate with respect to x:

    [3x⁴+8x³], evaluate x from 0 to 1

    = 3+8 = 11

    X = 11/16 = 0.6875

    The center of mass of the solid lies at (0.6875, 0.3542, 2.021)
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Find the mass and center of mass of the solid E with the given density function ρ. E lies under the plane z = 3 + x + y and above the ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers