Ex: Find the center of mass of the solid whose density &z and bounded above by the plane z=y, below by the xy-plane and laterally by the cylinder x² + y² = 4. Sol.: M = SSS 8dv π 2 rsine =[[[ 000 2 =]] //² 00 Mxy zrdzdrd0 = r³ sin² 0 drd0 = 10 π 2 rsine 00 0 π 2 00 T π 2 00 1/16 [+][0-sin20] =(1-0) (1 - 0) = π rsine ¹0 zdrdzdrd0 = drdo r³(1-cos20) drde ·SSS 000 π 2 rsine z²rdzdrde zy x²+²=4 x-0 # J=12 -2 2 r=2 r=2
Ex: Find the center of mass of the solid whose density &z and bounded above by the plane z=y, below by the xy-plane and laterally by the cylinder x² + y² = 4. Sol.: M = SSS 8dv π 2 rsine =[[[ 000 2 =]] //² 00 Mxy zrdzdrd0 = r³ sin² 0 drd0 = 10 π 2 rsine 00 0 π 2 00 T π 2 00 1/16 [+][0-sin20] =(1-0) (1 - 0) = π rsine ¹0 zdrdzdrd0 = drdo r³(1-cos20) drde ·SSS 000 π 2 rsine z²rdzdrde zy x²+²=4 x-0 # J=12 -2 2 r=2 r=2
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
Chapter10: Analytic Geometry
Section10.1: The Rectangular Coordinate System
Problem 40E: Find the exact volume of the solid that results when the region bounded in quadrant I by the axes...
Related questions
Question
![Ex: Find the center of mass of the solid whose density &-z and bounded above by the plane z-y,
below by the xy-plane and laterally by the cylinder x² + y² = 4.
Sol.:
M =
• SIS
π 2 rsine
8dv
= S S S
000
Mxy
2
-]³in²
0 0
zrdzdrd0 =
=
2
2 rsine
z²r
11:00
2
00
000
2
¡įra
00
r³ sin² 0 drd0 =
TC
1/16
= [] [0-sin²0 = (¹ - 0) (π – 0) = π
20
10
4 4
¹0
π 2 rsine
zdrdzdrd0 =
drdo
r³(1- cos20)drdo
π 2 rsine
!!!
000
z²rdzdrde
x²+²=4
x=0
J=±2
r=2
r=2
-2
zy
2
$
How is the angle (0)
equal π ?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94aa169f-469d-4cfa-9752-84c25fa011d4%2F188e20a7-a9a1-474e-b2bd-ce84e3e97e71%2Fyd2lgr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Ex: Find the center of mass of the solid whose density &-z and bounded above by the plane z-y,
below by the xy-plane and laterally by the cylinder x² + y² = 4.
Sol.:
M =
• SIS
π 2 rsine
8dv
= S S S
000
Mxy
2
-]³in²
0 0
zrdzdrd0 =
=
2
2 rsine
z²r
11:00
2
00
000
2
¡įra
00
r³ sin² 0 drd0 =
TC
1/16
= [] [0-sin²0 = (¹ - 0) (π – 0) = π
20
10
4 4
¹0
π 2 rsine
zdrdzdrd0 =
drdo
r³(1- cos20)drdo
π 2 rsine
!!!
000
z²rdzdrde
x²+²=4
x=0
J=±2
r=2
r=2
-2
zy
2
$
How is the angle (0)
equal π ?
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