Find the mass and center of mass of the lamina that occupies the reg.pdf

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The triangular region D bounded by the lines x=0, y=x, and 2x+y=6 has a non-uniform mass density of ρ(x,y)=x^2. The total mass is calculated to be 44 and the center of mass is calculated to be (6/5, 12/5) by taking moments about the x and y axes.

Formation

Find the mass and center of mass of the lamina that occupies the region D where D is the
triangular region enclosed by the lines x=0, y=x, and 2x+y=6, the density function of the lamina
is (x,y)=x2
Answer: center of mass: (x bar,Y bar) = (6/5,12/5)
Solution
The mass is given by m = ?[0,2] ?[x,6 - 2x] 11x² dy dx = ?[0,2] 11x²y |[x, 6 - 2x] dx
?[0,2] 11x²(6 - 2x) - 11x²(x) dx ?[0,2] (66x² - 33x³) dx 33 ?[0,2] (2x² - x³) dx 33((2/3)x³ - (1/4)x4)
|[0,2] 11(2)(8) - (33/4)(16) - 0 176 - 132 = 44 For the center of mass you need the moments
about x and y. The moment about the x-axis is Mx = ?? 11x²y dA and the moment about the y-
axis is My = ?? 11x³ dA dA represents the same region we've been working with so the integrals
are very similar to the mass's integral. The final center of mass will be (x¯, y¯) = (My/44, Mx/44)

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