Find the center of mass in the 1st quadrant of the circle x^2+y^2=4 where rho(x,y)=k(x^2+y^2)

Thanks.

To find the center of mass in the first quadrant of the given circle, we need to calculate the centroid of the mass distribution.

First, let's define the region in the first quadrant enclosed by the circle. Since we are only considering the first quadrant, we can restrict our attention to the values of x and y that satisfy x ≥ 0 and y ≥ 0.

To find the center of mass, we need to calculate the weighted average of the coordinates (x, y) with the weight function ρ(x, y) = k(x^2 + y^2), where k is a constant.

Let's start by finding the total mass (M) of the region. This can be obtained by integrating ρ(x, y) over the given region.

M = ∫∫ ρ(x, y) dA

To convert to polar coordinates, we can make the substitution x = rcos(θ) and y = rsin(θ).

dA = dx dy = (r dr) (dθ)

The limits of integration in polar coordinates become:

0 ≤ r ≤ 2 (since the circle has a radius of 2)
0 ≤ θ ≤ π/2 (since we are only considering the first quadrant)

Now, let's calculate the total mass:

M = ∫∫ k(r^2) r dr dθ
= k ∫₀^(π/2) ∫₀² r^3 dr dθ

Integrating with respect to r, we get:

M = k ∫₀^(π/2) [r^4/4]₀² dθ
= k ∫₀^(π/2) (16/4) dθ
= k ∫₀^(π/2) 4 dθ
= 4k [θ]₀^(π/2)
= 2kπ

Now, let's find the coordinates (x̄, ȳ) of the center of mass using the formulas:

x̄ = ∫∫ x ρ(x, y) dA / M
ȳ = ∫∫ y ρ(x, y) dA / M

Since we have already converted to polar coordinates, we can substitute x = rcos(θ) and y = rsin(θ) in these formulas.

x̄ = k ∫₀^(π/2) ∫₀² (rcos(θ))(r^2) r dr dθ / (2kπ)
= 1/2 ∫₀^(π/2) cos(θ) ∫₀² r^4 dr dθ / π
= π/10 ∫₀^(π/2) cos(θ) (r^5/5)₀² dθ
= π/10 ∫₀^(π/2) cos(θ) (32/5) dθ
= 32π/50 ∫₀^(π/2) cos(θ) dθ
= 16π/50 [sin(θ)]₀^(π/2)
= 16π/50 (1 - 0)
= 8π/50
= 4π/25

Similarly,

ȳ = k ∫₀^(π/2) ∫₀² (rsin(θ))(r^2) r dr dθ / (2kπ)
= 1/2 ∫₀^(π/2) sin(θ) ∫₀² r^4 dr dθ / π
= π/10 ∫₀^(π/2) sin(θ) (r^5/5)₀² dθ
= π/10 ∫₀^(π/2) sin(θ) (32/5) dθ
= 32π/50 ∫₀^(π/2) sin(θ) dθ
= -32π/50 [cos(θ)]₀^(π/2)
= -32π/50 (0 - (-1))
= 32π/50
= 16π/25

Therefore, the center of mass in the first quadrant of the given circle is (x̄, ȳ) = (4π/25, 16π/25).