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A solid E lies within the cylinder x2 + y2 = 9, below the plane z = 21, and above the paraboloid z = 9 - x² - y². (See the figure above.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E. Solution In cylindrical coordinates, the cylinder is r = E = = {(r, 0, z)| 0 ≤ 0 ≤ 2π, 0 ≤ r ≤ 3,9-r² ≤ z ≤ 21} Since the density at (x, y, z) is proportional to the distance from the z-axis, the density function is f(x, y, z) = K√ x² + y² = Kr m = where K is the proportionality constant. Therefore, from this formula, the mass of E is -√√√ K√ x² + y ² dv *2πt 3 21 = · [²1²³1² ( JO 3 = 62²h 6 °K ²² [1 Kr² = K[2h de [² (12r² + rª) dr = 2nK [ I. 11 and the paraboloid is z = r dz dr de dr de , so we can write

A solid E lies within the cylinder x2 + y2 = 9, below the plane z = 21, and above the paraboloid z = 9 - x² - y². (See the figure above.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E. Solution In cylindrical coordinates, the cylinder is r = E = = {(r, 0, z)| 0 ≤ 0 ≤ 2π, 0 ≤ r ≤ 3,9-r² ≤ z ≤ 21} Since the density at (x, y, z) is proportional to the distance from the z-axis, the density function is f(x, y, z) = K√ x² + y² = Kr m = where K is the proportionality constant. Therefore, from this formula, the mass of E is -√√√ K√ x² + y ² dv *2πt 3 21 = · [²1²³1² ( JO 3 = 62²h 6 °K ²² [1 Kr² = K[2h de [² (12r² + rª) dr = 2nK [ I. 11 and the paraboloid is z = r dz dr de dr de , so we can write

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
A solid E lies within the cylinder 
x2 + y2 = 9,
 below the plane 
z = 21,
 and above the paraboloid 
z = 9 − x2 − y2.
 (See the figure above.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E.
Solution
In cylindrical coordinates, the cylinder is 
r = 
 and the paraboloid is 
z = 
 
 
 
 ,
 so we can write
E = 
 
(r, ?, z)
 
 0 ≤ ? ≤ 2?, 0 ≤ r ≤ 3, 9 − r2 ≤ z ≤ 21
 
Since the density at 
(x, y, z)
 is proportional to the distance from the z-axis, the density function is
f(x, y, z) = K
  x2 + y2
 = Kr
where K is the proportionality constant. Therefore, from this formula, the mass of E is
m
=
 
 
E
K
  x2 + y2
 dV
=
2?
 
0
3
 
0
21
 
9 − r2
 
 
 
 
 
 
 
r dz dr d?
=
2?
 
0
3
 
0
Kr2
 
 
 
 
 
 
 
 dr d?
=
K
2?
 
0
d?
3
 
0
(12r2 + r4) dr
=
2?K
 
 
 
 
 
 
  3
0
=
 
 
 
 .
Example Video Example A solid E lies within the cylinder x² + y² = 9, below the plane z = 21, and above the paraboloid z = 9 - x² - y². (See the figure above.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E. Solution In cylindrical coordinates, the cylinder is r = E = {(r, 0, 2) | 0 ≤ 0 ≤ 2π, 0 ≤ r ≤ 3,9-² ≤ z ≤ 21} Since the density at (x, y, z) is proportional to the distance from the z-axis, the density function is f(x, y, z) = K√√x² + y² = Kr where K is the proportionality constant. Therefore, from this formula, the mass of E is • √√√ K √ x² + y² dv JE m = = 2π 3 21 ITT ( '2π 3 = 62²th 6 °K²³² [[ 2πt - * [*²00 [²(12²³² +1²) 0 K de dr = 2πK|| I 10 Dra and the paraboloid is z = r dz dr de dr de so we can write I
Transcribed Image Text:Example Video Example A solid E lies within the cylinder x² + y² = 9, below the plane z = 21, and above the paraboloid z = 9 - x² - y². (See the figure above.) The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E. Solution In cylindrical coordinates, the cylinder is r = E = {(r, 0, 2) | 0 ≤ 0 ≤ 2π, 0 ≤ r ≤ 3,9-² ≤ z ≤ 21} Since the density at (x, y, z) is proportional to the distance from the z-axis, the density function is f(x, y, z) = K√√x² + y² = Kr where K is the proportionality constant. Therefore, from this formula, the mass of E is • √√√ K √ x² + y² dv JE m = = 2π 3 21 ITT ( '2π 3 = 62²th 6 °K²³² [[ 2πt - * [*²00 [²(12²³² +1²) 0 K de dr = 2πK|| I 10 Dra and the paraboloid is z = r dz dr de dr de so we can write I
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