EXAMPLE 3 A wire takes the shape of a semicircle x² + y2 = 49, y 0, and is thicker near its base than near the top. Find the center of mass of the wire if the linear density at any point is proportional to its distance from the line y = 8.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
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EXAMPLE 3
A wire takes the shape of a semicircle x² + y2 = 49, y 2 0, and is thicker near its base than
near the top. Find the center of mass of the wire if the linear density at any point is proportional to its distance
from the line y = 8.
SOLUTION
As in this example we use the parametrization x = 7 cos(t),
y = 7 sin(t), 0st<n, and find
that ds = 7 dt. The linear density is
p(x, y) = k(8 – y)
where k is a constant, and so the mass of the wire is
TT
k(8 – y) ds
8
(7) dt
m =
%3D
TT
= 7k| 8t +
From these equations we have
y
УР(х, у) ds
m
1
yk(8 – y) ds
%3D
7k(8t – 14)
1
- 49 sin?(t)) dt
8Tt
14
49 t +
1
TT
49
sin(2t)
4
8Tt
14
By symmetry, we see that x = 0, so the center of mass is
[
(X, y) =
Transcribed Image Text:EXAMPLE 3 A wire takes the shape of a semicircle x² + y2 = 49, y 2 0, and is thicker near its base than near the top. Find the center of mass of the wire if the linear density at any point is proportional to its distance from the line y = 8. SOLUTION As in this example we use the parametrization x = 7 cos(t), y = 7 sin(t), 0st<n, and find that ds = 7 dt. The linear density is p(x, y) = k(8 – y) where k is a constant, and so the mass of the wire is TT k(8 – y) ds 8 (7) dt m = %3D TT = 7k| 8t + From these equations we have y УР(х, у) ds m 1 yk(8 – y) ds %3D 7k(8t – 14) 1 - 49 sin?(t)) dt 8Tt 14 49 t + 1 TT 49 sin(2t) 4 8Tt 14 By symmetry, we see that x = 0, so the center of mass is [ (X, y) =
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Author:
Swokowski
Publisher:
Cengage