Center of Mass

Question

a. Find the coordinates of the center of mass (centroid) of the portion of the cardioid r=1+cosθ above the x-axis; that is, the region described by R= {(r,θ) : 0 ≤ θ ≤ π, 0 ≤ r ≤ 1+cosθ}. 

 

b. A solid is bounded above by the paraboloid z=4-x²-y² and below by the cone z=√(x²+y²). The density of the solid satisfies 

ρ(x,y,z)=z. Find the coordinates of the center of mass.

 

Thank you

Answer 1

For part a)

 

The x coordinate of the center of mass is   

 

x_cm =  ∫∫ dθ r dr  r cos(θ)  /   M    where M = ∫∫ dθ r dr     

 

The y coordinate of the center of mass is 

 

y_cm =   ∫∫ dθ r dr  r sin(θ)   / M

 

The double integrals should be regarded as concatenated forms with the r integration being the inner integral with 

limits 0 to ( 1+ cos(θ)).    

 

The r integration is elementary, leaving  trigonometric integrals  over θ .   These trigonometric integrals (over 0 to pi) can be evaluated in closed form.     The results are

 

M = 3 pi/4  and  

 

x_cm = ( 5 pi/8) / M   =  5/6   and

 

y_cn = (4/3) /M =  (16/9 )/pi  ~  .5659

 

For b) it appears that the problem statement is incomplete.   The paraboloid  allows z to be negative , thus generating a negative density,

Answer 2

a) First, the area should be found:

A=(1/2)∫₀^π(1+cosθ)²dθ=3π/4.

Now we need to find the two coordinates (r, θ) as

r=(1/A)∫∫_RrdA=(4/3π)∫₀^π∫₀^1+cosθrdrdθ=3π/4.

θ=(1/A)∫∫_RθdA=1/2(π²-4)

 

b) Mass=∫∫∫ρ(x,y,z)dxdydz.

 

M_yz=∫∫∫xρ(x,y,z)dxdydz=. 

M_xz=∫∫∫yρ(x,y,z)dxdydz= 

M_xy=∫∫∫zρ(x,y,z)dxdydz=. 

(x, y, z)=(Myz/M, Mxz/M, Mxy/M).

 

But it seems to me it is easier to use polar coordinates. Try.