Find the mass of the triangular region with vertices (0, 0), (6, 0), and (0, 2), with density function ρ(x,y)=x^2+y^2
To find the mass of the triangular region, we need to integrate the density function over the region.
The triangular region is bounded by the lines y = 0, y = x/3, and y = 2. So, we can integrate the density function ρ(x,y) = x^2 + y^2 over this region.
We need to split the triangular region into two parts: the region bounded by the lines y = 0, y = x/3, and y = 2 and the region bounded by the lines y = 0, y = 2, and x = 6 (which is a rectangle).
For the first part of the triangular region, we can integrate ρ(x,y) = x^2 + y^2 with respect to x from x = 0 to x = 6. For each value of x, y varies from y = 0 to y = x/3. So, the integral for the first part is:
∫[0 to 6] ∫[0 to x/3] (x^2 + y^2) dy dx
For the second part of the triangular region (the rectangle), we can integrate ρ(x,y) = x^2 + y^2 with respect to x from x = 0 to x = 6. For each value of x, y varies from y = x/3 to y = 2. So, the integral for the second part is:
∫[0 to 6] ∫[x/3 to 2] (x^2 + y^2) dy dx
To find the mass of the entire triangular region, we need to evaluate the sum of these two integrals:
Mass = ∫[0 to 6] ∫[0 to x/3] (x^2 + y^2) dy dx + ∫[0 to 6] ∫[x/3 to 2] (x^2 + y^2) dy dx
To find the mass of the triangular region, we need to integrate the density function ρ(x, y) over the region.
The triangular region is defined by the vertices (0, 0), (6, 0), and (0, 2). We can set up the integral as follows:
M = ∫∫R ρ(x, y) dA
where R is the triangular region.
To find the limits of integration, we need to determine the boundaries of the triangular region. The left boundary is the line y = 0, the lower boundary is the line x = 0, and the upper boundary is the line y = 2 - (2/6)x.
Let's integrate the density function over the triangular region step by step:
1. Integrate with respect to y:
∫[0 to 2 - (2/6)x] ρ(x, y) dy
= ∫[0 to 2 - (2/6)x] (x^2 + y^2) dy
= [x^2y + (y^3/3)] [0 to 2 - (2/6)x]
= [(x^2(2 - (2/6)x) + ((2 - (2/6)x)^3)/3]
2. Integrate the result from step 1 with respect to x:
∫[0 to 6] [(x^2(2 - (2/6)x) + ((2 - (2/6)x)^3)/3] dx
Now, we can simplify and evaluate this integral to find the mass of the triangular region.