   Chapter 16.7, Problem 39E

Chapter
Section
Textbook Problem

Find the center of mass of the hemisphere x2 + y2 + z2 = a2, z ≥ 0, if it has constant density.

To determine

To find: The center of mass of the hemisphere.

Explanation

Given data:

x2+y2+z2=a2 (1)

Formula used:

SFdS=DF|rϕ×rθ|dA (2)

rϕ=xϕi+yϕj+zϕk (3)

rθ=xθi+yθj+zθk (4)

Mass of the hemisphere (m) is,

m=SKdS=K4π(12a2)=2πa2K

Rearrange the equation.

K=m2πa2

Due to symmetry Mxz=Myz=0 that is x¯=y¯=0 .

Therefore,

Mxy=SzKdS {z¯=Mxy}

By using the spherical coordinates to parameterize the sphere, consider r(ϕ,θ)=asinϕcosθi+asinϕsinθj+acosϕk with 0θ2π and 0ϕπ2 .

Find rϕ .

Substitute asinϕcosθ for x , asinϕsinθ for y and acosϕ for z in equation (3),

rϕ=ϕ(asinϕcosθ)i+ϕ(asinϕsinθ)j+ϕ(acosϕ)k=(acosθ)ϕ(sinϕ)i+(asinθ)ϕ(sinϕ)j+(a)ϕ(cosϕ)k=acosθcosϕi+asinθcosϕjasinϕk

Find rθ .

Substitute asinϕcosθ for x , asinϕsinθ for y and acosϕ for z in equation (4),

rθ=θ(asinϕcosθ)i+θ(asinϕsinθ)j+θ(acosϕ)k=(asinϕ)θ(cosθ)i+(asinϕ)θ(sinθ)j+(acosϕ)θ(1)k=(asinϕ)(sinθ)i+(asinϕ)(cosθ)j+(acosϕ)(0)k=asinϕsinθi+asinϕcosθj

Find rϕ×rθ .

rϕ×rθ=(acosθcosϕi+asinθcosϕjasinϕk)×(asinϕsinθi+asinϕcosθj)=|ijkacosθcosϕasinθcosϕasinϕasinϕsinθasinϕcosθ0|={(0+a2sin2ϕcosθ)i+(a2sin2ϕsinθ0)j+(a2cos2θcosϕsinϕ+a2sin2θcosϕsinϕ)k}={a2sin2ϕcosθi+a2sin2ϕsinθj+a2cosϕsinϕ(cos2θ+sin2θ)k}={a2sin2ϕcosθi+a2sin2ϕsinθj+a2cosϕsinϕk} {cos2θ+sin2θ=1}

Find |rϕ×rθ|

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