Find the mass of the triangular region with vertices (0, 0), (6, 0), and (0, 2), with density function ρ(x,y)=x2+y2
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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}.
To find the mass of the triangular region, we need to integrate the product of the density function and the area element over the region.
The area element in Cartesian coordinates is given by dA = dx dy.
The mass can be calculated as follows:
Mass = ∫∫D ρ(x,y) dA
where D represents the triangular region.
The equation for the density function is given as ρ(x,y) = x^2 + y^2.
Now, let's find the limits of integration for x and y:
For x, the limits range from 0 to 6 as given by the x-coordinates of the vertices.
For y, the limits range from 0 to the y-coordinate of the upper vertex, which is 2.
Thus, the mass of the triangular region is:
Mass = ∫₀² ∫₀⁶ (x^2 + y^2) dx dy
Now, let's integrate with respect to x first, then with respect to y:
Mass = ∫₀² [∫₀⁶ (x^2 + y^2) dx] dy
Integrating with respect to x, we get:
Mass = ∫₀² [(1/3)x^3 + y^2x] from 0 to 6 dy
Mass = ∫₀² [(1/3)(6)^3 + y^2(6) - (1/3)(0)^3 - y^2(0)] dy
Mass = ∫₀² (72 + 6y^2) dy
Integrating with respect to y, we get:
Mass = [72y + 2y^3/3] from 0 to 2
Mass = [72(2) + 2(2)^3/3] - [72(0) + 2(0)^3/3]
Mass = [144 + 16/3] - [0]
Mass = 144 + 16/3
Mass = 432/3 + 16/3
Mass = 448/3
Therefore, the mass of the triangular region with the given density function is 448/3.