Find the moment of inertia I2 for the solid above the xy-plane bounded by the paraboloid z = x² + y² = 9, assuming the mean density to be constant C. please solve it..it's urgent..
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Consider a disc of radius r and thickness dz at a height of z above the x-y plane. Its radius r is given by the bounding paraboloid r² = x² + y².
So r² = z.
Mass dm of the disc = dm = C* π r² * dz = C π z dz
Total mass = M = Integral z = 0 to Z dm = πC Z² / 2
We know that MOI of the thin disc = dm r² / 2
= (C π z dz) z / 2 = (C π/2) z² dz
So MOI = integral from 0 to Z { Cπ/2 z² dz}
= Cπ/6 * Z³ , in terms of Z
= M * Z/3 in terms of mass
Given Z = 9, So MOI = πC 9³ /6 = 243 C π /2
So r² = z.
Mass dm of the disc = dm = C* π r² * dz = C π z dz
Total mass = M = Integral z = 0 to Z dm = πC Z² / 2
We know that MOI of the thin disc = dm r² / 2
= (C π z dz) z / 2 = (C π/2) z² dz
So MOI = integral from 0 to Z { Cπ/2 z² dz}
= Cπ/6 * Z³ , in terms of Z
= M * Z/3 in terms of mass
Given Z = 9, So MOI = πC 9³ /6 = 243 C π /2
kvnmurty:
click on red heart thanks above pls
This is for the solid that is bounded by z =0, z = 9, z = x^2+y^2.. the conical shaped paraboloid solid..
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