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EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of mass of the lamina. ²+ y = a² SOLUTION Let's place the lamina as the upper half of the circle x? + y2 = a?. (See the figure.) Then the distance from a point (x, y) to the center of the circle (0.) (the origin) is Vx? + y². Therefore the density function is P(x, y) = KV + y? -a Video Example () where K is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then Vx2 + y2 = r and the region D is given by 0

EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of mass of the lamina. ²+ y = a² SOLUTION Let's place the lamina as the upper half of the circle x? + y2 = a?. (See the figure.) Then the distance from a point (x, y) to the center of the circle (0.) (the origin) is Vx? + y². Therefore the density function is P(x, y) = KV + y? -a Video Example () where K is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then Vx2 + y2 = r and the region D is given by 0

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.6: Variation
Problem 33E
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EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of mass of the lamina. ²+ y = a² SOLUTION Let's place the lamina as the upper half of the circle x? + y2 = a?. (See the figure.) Then the distance from a point (x, y) to the center of the circle (0.) (the origin) is Vx? + y². Therefore the density function is P(x, y) = KV + y? -a Video Example () where K is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then Vx2 + y2 = r and the region D is given by 0 <rs a, osesn. Thus the mass of the lamina is P(x, y) dA - ea KV + y? dA de 2 dr de Кп Both the lamina and the density function are symmetric with respect to the y-axis, so the center of mass must lie on the y-axis, that is x = . The y-coordinate is given by yρ(x, y ) dA r sin(0) (Kr)r dr de Кла 3 sin(e) de паз -cos(0) na The center of mass is located at the point (x, y) =
Transcribed Image Text:EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of mass of the lamina. ²+ y = a² SOLUTION Let's place the lamina as the upper half of the circle x? + y2 = a?. (See the figure.) Then the distance from a point (x, y) to the center of the circle (0.) (the origin) is Vx? + y². Therefore the density function is P(x, y) = KV + y? -a Video Example () where K is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then Vx2 + y2 = r and the region D is given by 0 <rs a, osesn. Thus the mass of the lamina is P(x, y) dA - ea KV + y? dA de 2 dr de Кп Both the lamina and the density function are symmetric with respect to the y-axis, so the center of mass must lie on the y-axis, that is x = . The y-coordinate is given by yρ(x, y ) dA r sin(0) (Kr)r dr de Кла 3 sin(e) de паз -cos(0) na The center of mass is located at the point (x, y) =
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