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91Ó°ÊÓ

Find the specified quantities for the solids described below:

The mass of the region from Exercise 51, assuming that the density at every point is proportional to the square of the point’s distance from the xy-plane.

Short Answer

Expert verified

The mass is given bym=7kπ1024

Step by step solution

01

Given Information

The density at every point is proportional to the square of the point’s distance from the xy-plane.

The region bounded above by the plane is given by equationz=x and bounded below by the paraboloid by equationz=x2+y2.

02

Evaluation of limits

Relation between rectangular and cylindrical coordinates is given by

r=x2+y2,tanθ=yx,z=z

And

x=rcosθ,y=rsinθ,z=z

Rectangular coordinates are z=xandz=x2+y2

Cylindrical coordinates are z=rcosθand z=r2

Cartesian limits are x2+y2≤z≤x

x=rcosθ⇒z=x⇒z=rcosθ

r2=x2+y2⇒z=x2+y2⇒z=r2

⇒r2≤z≤rcosθ

x2+y2=xgivesr2=rcosθ⇒r=0,r=cosθ→0≤r≤cosθ

Also

r=0givescosθ=0⇒-π/2≤θ≤π/2

Cylindrical limits are

role="math" localid="1652385337856" r2≤z≤rcosθ,0≤r≤cosθ,-π/2≤θ≤π/2

03

Evaluation of Mass

The density at every point is proportional to the square of the point’s distance from the xyplane.

⇒ÒÏ=kz2

Required mass is m=∭EÒÏdxdydz

m=∭Ekz2dxdydz

m=∫θ=-π/2π/2∫r=0cosθ∫z=r2cosθkz2rdzdrdθ

m=∫θ=-π/2π/2∫r=0cosθ(kr)z33z=r2z-rcosθdrdθ

m=∫θ=-π/2π/2∫r=0cosθkr3r3cos3θ-r6drdθ

Solving further yields

m=∫θ=-π/2π/2k3r55cos3θ-r88r=0cosθdθ

m=k3θ=-π/2π/2cos8θ5-cos8θ8dθ

m=2k3340∫θ=0π/2cos8θdθ

Solve using integration

m=k20∫θ=0π/2cos8θdθ

m=k2078×56×34×12×π2

m=7kπ1024 (required mass)

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