Mass and center of mass Let S be a surface that represents a thin shell with density ρ. The moments about the coordinate planes ( see Section 13.6 ) are M y z = ∬ S x ρ ( x , y , z ) d S , M x z = ∬ S y ρ ( x , y , z ) d S , and M x y = ∬ S z ρ ( x , y , z ) d S . The coordinates of the center of mass of the shell are x ¯ = M y z m , y ¯ = M x z m , z ¯ = M x y m , where m is the mass of the shell. Find the mass and center of mass of the following shells. Use symmetry whenever possible . 67. The constant-density cone with radius a , height h , and base in the xy -plane
Mass and center of mass Let S be a surface that represents a thin shell with density ρ. The moments about the coordinate planes ( see Section 13.6 ) are M y z = ∬ S x ρ ( x , y , z ) d S , M x z = ∬ S y ρ ( x , y , z ) d S , and M x y = ∬ S z ρ ( x , y , z ) d S . The coordinates of the center of mass of the shell are x ¯ = M y z m , y ¯ = M x z m , z ¯ = M x y m , where m is the mass of the shell. Find the mass and center of mass of the following shells. Use symmetry whenever possible . 67. The constant-density cone with radius a , height h , and base in the xy -plane
Solution Summary: The author explains the formula used to find the mass and center of mass of a cone with constant density.
Mass and center of massLet S be a surface that represents a thin shell with density ρ. The moments about the coordinate planes (see Section 13.6) are
M
y
z
=
∬
S
x
ρ
(
x
,
y
,
z
)
d
S
,
M
x
z
=
∬
S
y
ρ
(
x
,
y
,
z
)
d
S
, and
M
x
y
=
∬
S
z
ρ
(
x
,
y
,
z
)
d
S
. The coordinates of the center of mass of the shell are
x
¯
=
M
y
z
m
,
y
¯
=
M
x
z
m
,
z
¯
=
M
x
y
m
, where m is the mass of the shell. Find the mass and center of mass of the following shells. Use symmetry whenever possible.
67. The constant-density cone with radius a, height h, and base in the xy-plane
Write the given third order linear equation as an equivalent system of first order equations with initial values.
Use
Y1 = Y, Y2 = y', and y3 = y".
-
-
√ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t²
\y(3) = 1, y′(3) = −2, y″(3) = −3
(8) - (888) -
with initial values
Y
=
If you don't get this in 3 tries, you can get a hint.
Question 2
1 pts
Let A be the value of the triple integral
SSS.
(x³ y² z) dV where D is the region
D
bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0.
Then the value of sin(3A) is
-0.003
0.496
-0.408
-0.420
0.384
-0.162
0.367
0.364
Chapter 14 Solutions
Student Solutions Manual, Single Variable for Calculus: Early Transcendentals
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Area Between The Curve Problem No 1 - Applications Of Definite Integration - Diploma Maths II; Author: Ekeeda;https://www.youtube.com/watch?v=q3ZU0GnGaxA;License: Standard YouTube License, CC-BY