M x = 1/2 p ∫ ( f ( x )² - g ( x )² ) d x
f ( x ) = √( 1 - x²), g ( x ) = - 2
\(M x = 1/2 * 5 \int\limits^1_{-1} (- x^{2} -3)\, dx= \\ =-5/2 * [ x^{3}/3 + 3 x]^1 _{-1} = \\ =-5/2 * (1/3+3+1/3+3)=\)
= - 50/3
My = p ∫ x ( f ( x ) ) dx
My = p ∫ x ( √(1+x²)) dx
Substitution: 1 - x² = u, x dx = - du/2
\(M y = 5 * \int\limits^0_0 { \sqrt{u} } \, du = 0\)
M x = - 50/3, M y = 0
\(M = 5 * \int\limits^1_{-1} { (\sqrt{1+ x^{2} }+2)} \, dx = \\
=5* [1/2 \sqrt{1+ x^{2} } *x + 2 x + 1/2 *sinh ^{-1} x]^1_{-1}\)
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 )
