Answer:
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)