lamina in the xy plane is bounded by y=0,x=0 and 2x + y=2 has mass density at (x,y) equal to the distance to the x-axis. find the mass of the lamina.
the part that is tripping me up is the mass density
To find the mass of the lamina, we need to integrate the mass density over the entire lamina.
The mass density is given as the distance to the x-axis. This means that at each point (x, y) on the lamina, the mass density is equal to y.
First, let's find the limits of integration for x and y based on the given boundaries:
- y = 0 represents the x-axis, which means y starts from 0.
- x = 0 represents the y-axis, which means x starts from 0.
- 2x + y = 2 is the equation of a line. Solve this equation for y to get: y = 2 - 2x. This line intersects the x-axis at y = 2, so the upper limit for y is 2.
Now, we can set up the integral to find the mass:
M = ∬(R) ρ(x, y) dA
Where M is the mass, ∬(R) represents a double integral over the region R, ρ(x, y) is the mass density function, and dA is an infinitesimal element of area.
Since the mass density is given by ρ(x, y) = y, the integral becomes:
M = ∬(R) y dA
To evaluate this double integral, we need to define the limits of integration for x and y. In this case, the region R is bounded by y = 0, x = 0, and 2x + y = 2.
The limits for x are from 0 to the x-coordinate of the intersection point between the line 2x + y = 2 and the x-axis. To find this intersection point, we set y = 0 in the equation:
2x + 0 = 2
2x = 2
x = 1
Therefore, the limits for x are from 0 to 1.
The limits for y are from 0 to the y-coordinate of the intersection point between the line 2x + y = 2 and the y-axis, which is 2.
Now we can set up the double integral:
M = ∫(0 to 1) ∫(0 to 2) y dy dx
Integrating with respect to y first:
M = ∫(0 to 1) [(1/2) y^2] (0 to 2) dx
M = ∫(0 to 1) (1/2) (2^2) dx
M = ∫(0 to 1) 2 dx
M = [2x] (0 to 1)
M = 2(1) - 2(0)
M = 2
Therefore, the mass of the lamina is 2 units.