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

Answer 1

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