Question

A lamina occupies the region inside the circle x2+y2=14y but outside the circle x2+y2=49. the density at each point is inversely proportional to its distance from the origin. where is the center of mass

Answer 1

Lamina occupies x² + y² = 14y. Outside circle is x² + y² = 49
To find the mass of lamina, integrate given density function over the region
m = ∫∫D P(x, y) dA
Subtitute x = r cosФ and y = r sinФ in x² + y² = 14y
and x² + y² = 49
x² + y² = 14y
(r cosФ)² + ( rsinФ)² = 14(rsinФ)
r² = 14r sin Ф
x² +y² = 49
r² = 49
r = 7
Cntre mass (-x. -y)
-x= i/m ∫∫D xp(x,y) dA = 1/m∫∫ (r cosФ) p( r, Ф)r
dr dФ
-y = 1/m∫∫D yp(x, y) dA = 1/m ∫∫D (r sinФ) p(r, Ф) r drФ
where m = ∫∫D p(x, y) dA

Answer 2

The function to that give the center of mass is mathematically given as

$m = \int \int D p(x, y) dA$

Where is the center of mass?

Question Parameters:

A lamina occupies the region inside the circle x2+y2=14y

but outside the circle x2+y2=49.

Generally, the equation for the  mass of lamina  is mathematically given as

$m = \int \intD P(x, y) dA$

Where

$r^2 = 14r sin \theta$

x^2 +y^2= 49

r = 7

Therefore,The center of mass is

$-x= i/m \int \intD xp(x,y) dA \\\\-x= 1/m \int \int (r cos\theta) p( r, \theta)rdrd\theta$

And

$-y = 1/m\int \int D yp(x, y) dA \\\\-y= 1/m \int \int D (r sin\theta) p(r, \theta) r dr\theta$

Therefore, the center of mass will reside at

$m = \int \int D p(x, y) dA$

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