To find the mass, we need to integrate the density function ρ(x,y) over the triangular region.
The triangular region is bounded by the x-axis, the line y = 2, and the line x = 6. In the first quadrant, it is bounded by the x-axis, the line y = 2, and the line x = 6. We can split the region into two parts: a rectangle and a right triangle.
The rectangle has base 6 and height 2. The mass of the rectangle is given by the integral:
∫∫(rectangle) ρ(x,y) dA
= ∫(0 to 6) ∫(0 to 2) (x^2 + y^2) dy dx
= ∫(0 to 6) (x^2y + (1/3)y^3) evaluated from 0 to 2 dx
= ∫(0 to 6) (2x^2 + (1/3)8) dx
= ∫(0 to 6) (2x^2 + (8/3)) dx
= [ (2/3)x^3 + (8/3)x]^6_0
= (2/3)(6^3) + (8/3)(6)
= (2/3)(216) + (8/3)(6)
= 144 + 16
= 160.
Now, let's find the mass of the right triangle. The right triangle has base 6 and height 2. The mass of the triangle is given by the integral:
∫∫(triangle) ρ(x,y) dA
= ∫(0 to 2) ∫(0 to (3/2)y) (x^2 + y^2) dx dy
= ∫(0 to 2) [(1/3)x^3 + y^2x] evaluated from 0 to (3/2)y dy
= ∫(0 to 2) [(1/3)(3/2)y^3 + y^2(3/2)y] dy
= ∫(0 to 2) [(1/2)y^3 + (3/2)y^3] dy
= ∫(0 to 2) [(2/2)y^3 + (6/2)y^3] dy
= ∫(0 to 2) (8/2)y^3 dy
= (8/2) [(1/4)y^4] evaluated from 0 to 2
= (8/2)(1/4)(2^4 - 0)
= (8/2)(1/4)(16)
= 8.
Therefore, the total mass is 160 + 8 = <<160+8=168>>168. Answer: \boxed{168}.
Find the mass of the triangular region with vertices (0, 0), (6, 0), and (0, 2), with density function ρ(x,y)=x2+y2
.
1 answer