PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 10.1, Problem 90E

a.

To determine

To write: an equation for the sphere, the center of the bearing ball corresponds to the origin of the three-dimensional coordinate system.

a.

Expert Solution
Check Mark

Answer to Problem 90E

The equation of the sphere is, x2+y2+z2=36.

Explanation of Solution

Given information: Ball bearings allow machinery to move at extremely high speeds.

and carry large loads. A bearing ball is in the shape of a sphere with a volume of 288π cubic millimeters.

Formula used:

The standard equation of a sphere with center ( h, k, j ) and radius r is given by:

  (x−h)2+(y−k)2+(z−j)2=r2.

Calculation:

Let r is the radius of the sphere (ball bearing), and the center of the bearing ball corresponds to the origin of the three-dimensional coordinate system.

Therefore, ( h, k, j ) = (0, 0, 0).

  Volume of the sphere(Ball​  bearing) =288π43πr3=288πr3=288×34=216r=2163r=6

Hence, the equation of the sphere:

  h=k=j=0,r=6. (x−0) 2 + (y−0) 2 + (z−0) 2 = 6 2 x2+y2+z2=36.

Therefore, the equation of the sphere is, x2+y2+z2=36.

b.

To determine

To graph: the sphere using the three-dimensional graphing utility.

b.

Expert Solution
Check Mark

Explanation of Solution

Given information: The equation of the sphere is, x2+y2+z2=36.

Calculation:

The graph of the sphere using the three-dimensional graphing utility is shown below.

  PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS), Chapter 10.1, Problem 90E

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Chapter 10.1, Problem 89E
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Chapter 10.1, Problem 91E

Chapter 10 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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