   Chapter 15.6, Problem 39E

Chapter
Section
Textbook Problem

Find the mass and center of mass of the solid E with the given density function ρ.39. E lies above the xy-plane and below the paraboloid z = 1 – x2 – y2; ρ (x, y, z) = 3

To determine

To find: The mass and center of mass of the solid E with density ρ .

Explanation

Given:

E lies above the xy-plane and below the paraboloid z=1x2y2 and the density is ρ(x,y,z)=3 .

Calculation:

Mass of the solid , mass(m)=density×volume .

The mass of the solidis, m=liml,m,ni=1lj=1mk=1nρ(xij*,yij*,zij*)ΔA=Eρ(x,y,z)dA .

Here, the density function is given by ρ(x,y,z) and E is the region that is occupied by the solid.

Covert the boundary of z into polar form. Substitute x=rcosθ and y=rsinθ in the above paraboloid. It gives z varies from 0 to 1r2 . Here the density is given by ρ(x,y,z)=3 .

m=A[01r23dz]dA=A[3z]01r2dA=3A[1r20]dA=3A[1r2]dA

The value of r varies from 0 to 1 and θ varies from 0 to 2π .

m=302π01(1r2)rdrdθm=02π01(rr3)drdθ

Integrate the above integral with respect to r and apply the limit of it.

m=302π[r22r44]01dθ=302π[((1)22(1)44)((0)22(0)44)]01dθ=302π[(1214)]01dθ=302π14dθ

Integrate the above integral with respect to θ and apply the limit of it.

m=34[θ]02π=34[2π0]=3π2

Thus, the mass of the solid is 3π2 .

For finding the center of mass,

Center of mass located at x¯=Myzm=1mExρ(x,y,z)dV .

x¯=Eρ(x,y,z)xdEm=E3xdEm=02π0101r23(rcosθ)dzrdrdθm

x¯=02π0101r23r2(cosθ)dzdrdθm (1)

Integrate the numerator part with respect to z and apply the limit of it.

Myz=302π01r2(cosθ)[z]01r2drdθ=302π01r2(cosθ)[1r20]drdθ=302π01(cosθ)[r2r4]drdθ

Integrate the above integral with respect to r and apply the limit of it.

Myz=302π(cosθ)[r33r55]01dθ=302π(cosθ)[((1)33(1)55)((0)33(0)55)]dθ=302π(cosθ)[(1315)(00)]dθ=2502π(cosθ)dθ

Integrate the above integral with respect to θ and apply the limit of it.

Myz=25[sinθ]02π=25[sin(2π)sin(0)]=25=0

Substitute the value in equation (1),

x¯=Myzm=0(3π2)=0

Center of mass located at y¯=Mzxm=1mEyρ(x,y,z)dV .

x¯=Eρ(x,y,z)ydEm=E3ydEm=02π0101r23(rsinθ)dzrdrdθm

x¯=02π0101r23r2(cosθ)dzdrdθm (2)

Integrate the numerator part with respect to z and apply the limit of it

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