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Replacing these values into the original position equation, the position vector is

r_CO = (24)/(5)i + (18)/(5)j

b) This portion of the problem is resolved with the parallel axis theorem: I^O_zz = I^C_zz + md². Recalling that x² + y² = r², and assuming that the plane of the example figure is the mid plane of the cylinder, the triple integral that describes the moment of inertia can be formed and solved as:

I^O_zz = ^(T)/(2)⌠⌡_{− (T)/(2)}^2π⌠⌡₀^R⌠⌡₀(x² + y²)rρ₀⎛⎝1 − (r)/(2R)⎞⎠drdθdz + R²∭ρ₀dV = ^(T)/(2)⌠⌡_{− (T)/(2)}^2π⌠⌡₀^R⌠⌡₀ρ₀r³⎛⎝1 − (r)/(2R)⎞⎠drdθdz + R²^(T)/(2)⌠⌡_{− (T)/(2)}^2π⌠⌡₀^R⌠⌡₀ρ₀⎛⎝r − (r²)/(2R)⎞⎠drdθdz = T2πρ₀^R⌠⌡₀⎛⎝r³ − (r⁴)/(2R)⎞⎠dr + R²ρ₀2πT^R⌠⌡₀⎛⎝r − (r²)/(2R)⎞⎠dr = T2πρ₀⎡⎣(R⁴)/(4) − (R⁴)/(10) + (R⁴)/(2) − (R⁴)/(6)⎤⎦ = R⁴2πρ₀T⎡⎣(3)/(4) − (3)/(30) − (5)/(30)⎤⎦ = 2R⁴πρ₀T(29)/(60) = (29)/(30)R⁴πρ₀T

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