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)

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right