M x = 1/2 p ∫ ( f ( x )² – g ( x )² ) d x
f ( x ) = √( 1 – x²),  g ( x ) = – 2
M x = 1/2 * 5  intlimits^1_{-1} (- x^{2} -3), dx= \ =-5/2 * [ x^{3}/3 + 3 x]^1 _{-1} = \ =-5/2 * (1/3+3+1/3+3)=
= – 50/3
My = p ∫ x * ( f ( x ) ) dx
My = p ∫ x ( √(1+x²)) dx
Substitution: 1 – x² = u,  x dx = – du/2
M y = 5 *  intlimits^0_0 { sqrt{u} } , du = 0
M x = – 50/3, M y = 0
M = 5 *  intlimits^1_{-1}  { (sqrt{1+ x^{2} }+2)} , dx  = \ 
=5* [1/2  sqrt{1+ x^{2} }  *x + 2 x + 1/2 *sinh ^{-1} x]^1_{-1}
M ≈ 5 * 6.3 ≈ 31.2  
x = M y / M = 0 / 31.5 = 0
y = M x / M =  -50/3 : 31.5 ≈ – 0.529
The center of mass is ( 0, -0.529 )