Find center of mass of the lamina that occupies the region D and has the given density function ρ.
D is the triangular region with vertices (0, 0), (2, 1), (0, 3); ρ(x, y) = 5(x + y)
mass=30
Find center of mass
If you found the mass, you're halfway there.
M = ∫ ρ(x,y) dA
The x-coordinate of the center of mass is just
Mx = ∫xρ(x,y) dA / M
and similarly for My.
To find the center of mass of a lamina in the region D with a given density function ρ, we can follow these steps:
1. Start by calculating the total mass of the lamina. In this case, the mass is given as 30.
2. Divide the total mass by the area of the lamina to find the average density. In this case, the area of the triangular region D can be found using the formula for the area of a triangle: A = (1/2) * base * height. In our case, the base is 2 and the height is 3, so the area is (1/2) * 2 * 3 = 3.
3. Divide the average density by the total mass to find the density function. In this case, the average density is 30/3 = 10. So the density function is ρ(x, y) = 10.
4. To find the center of mass (x_c, y_c) of the lamina, we need to calculate the double integral of the product of the density function ρ(x, y) and the coordinates (x, y) over the region D, divided by the total mass.
x_c = (1/mass) * ∬(D) ρ(x, y) * x dA
y_c = (1/mass) * ∬(D) ρ(x, y) * y dA
5. Substitute the given density function ρ(x, y) = 10 and the mass = 30 into the center of mass formulas to solve for x_c and y_c.
x_c = (1/30) * ∬(D) 10x dA
y_c = (1/30) * ∬(D) 10y dA
Now, we need to evaluate the double integrals over the triangular region D:
∬(D) x dA = ∬(D) x dy dx
∬(D) y dA = ∬(D) y dy dx
To set up the limits of integration for the double integrals, we can express the region D using the parameters x and y:
0 ≤ x ≤ 2
0 ≤ y ≤ 3 - (3/2)x
Now, we can substitute these limits of integration into the expressions for x_c and y_c and compute the double integrals to find the center of mass (x_c, y_c).