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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621

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Calculus (MindTap Course List)

8th Edition
James Stewart
ISBN: 9781285740621
Textbook Problem

If E is the solid of Exercise 18 with density function ρ ( x , y , z ) = x 2 + y 2 , find the following quantities, correct to three decimal places.

(a) The mass

(b) The center of mass

(c) The moment of inertia about the z-axis

To determine

To find:

The exact value of the mass for E using computer algebra system.

Explanation

1) Concept:

If the density function of a solid object that occupies the region E is ρ (x,y,z), in units of mass per unit volume, at any given point (x , y, z), then its mass is

m=E ρ(x,y,z) dV

2) Given:

E is a solid in the first octant bounded by the cylinder y2+z2=9 and the planes x=0

y=3x and z=0

The density function

ρx,y,z=x2+y2

3) Calculation:

We have,

y2+z2=9

Consider z=0

y2+02=9

Thus, y=±3

But region is first octant, therefore consider positive 3.

y2+z2=9z2=9-y2

z=9-y2

Therefore, we get,

E= x,y,z  0x1,   3xy3,   0z9-y2 }

Thus, using  ρx,y,z=x2+y2 into,

m=E ρ(x,y,z)dV gives,

m=E x2+y2 dV

m= 013x309-y2x2+y2 dz dy dx

Evaluating this integration using the Computer Algebra System,

Write following command on Mathematica,

Integratex2+y2,x,0,1,y,3x,3,z,0,9-y2

This gives the output as,

565

This can also be rewritten as 11.2 in decimal form.

Conclusion:

m=565= 11.2

To find:

The exact value of the center of mass for E using computer algebra system.

Solution:

The center of mass is

x-=0.375,   y-= 2.209,   z-=0.9375

1) Concept:

The center of mass located at the point x-,y-,z-, where

x-=Myzm   y-=Mxzm  z-=Mxym

And its moments about three coordinate planes are,

Myz=E x ρx, y,zdV,  Mxz=E y ρx, y,zdV,  Mxy=E z ρx, y,zdV

2) Given:

E is a solid in the first octant bounded by the cylinder y2+z2=9 and the planes x=0,

y=3x and z=0

The density function

ρx,y,z=x2+y2

3) Calculation:

We have,

y2+z2=9 y2+02=9

Thus, y=3

y2+z2=9z2=9-y2

z=9-y2

Therefore, we get E= x,y,z  0x1,   3xy3,   0z9-y2 }

Thus, using  ρx,y,z=x2+y2 into

x-=1m Myz

Myz=E x ρ(x,y,z) dV

Therefore, x-=1mE x ρ(x,y,z) dV

x-=1m013x309-y2x x2+y2 dz dy dx

x-=m-1013x309-y2x x2+y2 dz dy dx

Evaluating this integration using the Computer Algebra System,

Since m=565

1m=m-1=556

Write following command on Mathematica

Integrate5/56xx2+y2,x,0,1,y,3x,3,z,0,9-y2

This gives the output as,

855π7168

This approximates to

0

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