Find the mass m, the moments mx and my, and the center of mass (x, y) of the uniform flat plate of density rho = 1 bounded by the curves y = x2 ln x and y = -x2 ln x and the lines x = 0 and x = 1.

m = 2/9, M(x) = 0, M(y) = 1/8, COM = (9/16, 0)

Step-by-step explanation:

Given that the density of the uniform plate is 1 and it is bound by the curves y = x²lnx and y = -x²lnx and the lines x = 0 and  x = 1

Mass of the given lamina is

m = ∫∫ ρ(x,y) dA

m = ∫(0,1)∫(x²lnx,-x²lnx) 1 dy.dx

m = ∫(0,1) [y]x²lnx, -x²lnx dx

m = ∫(0,1) (-x²lnx – x²lnx) dx

m = ∫(0,1) (-2x²lnx) dx

m = 2∫(0,1) (-x²lnx) dx

m = 2[-((x³lnx)/³) + x³/9 ](0,1)

m = 2[(1³/9)ln1 + 1/9 - 0]

m = 2/9

Now, M(x) =∫∫ y.ρ(x,y) dA and M(y) =∫∫ x.ρ(x,y) dA

M(x) = ∫∫(D) y.ρ(x,y) dA

M(x) = ∫(0,1)∫(x²lnx,-x²lnx) y dy.dx

M(x) = ∫(0,1) [y²/2] x²lnx,-x²lnx dx

M(x) = ½ ∫(0,1) {(-x²lnx)² – (x²lnx)²} dx

M(x) = ½ ∫(0,1) (x⁴ (lnx)² – x⁴(lnx)²) dx

M(x) = 0

M(y) =  ∫(0,1)∫(x²lnx,-x²lnx) x dy.dx

M(y) = ∫(0,1) x[y] x²lnx,-x²lnx dx

M(y) = ∫(0,1) x(-x²lnx -x²lnx) dx

M(y) = 2∫(0,1) (-x³lnx) dx

M(y) = 2[-x⁴lnx/4 + x⁴/16]0,1

M(y) = 2[-ln1/4 + 1/16 - 0]

M(y) = 2/16 = 1/8

x’ = M(y)/m = (1/8) / (2/9) = 9/16

y’ = M(x)/m = (0/2) / (2/9) = 0

Hence the center of mass is at the point (x’,y’) = (9/16 , 0)

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step-by-step explanation:

i believe the answer is the polygon, heptagon.