Calculus (double check )

Find the centroid of the area bonded by the parabola y =4-x^2 and the x-axis?

A.(0,1.6) <<<< correct?
B.(0,1.7)
C.(0,1.8)
D.(0,1.9)

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  1. Let f(x)=4-x^2
    f(x)=0 at x=±2
    So integration is done from x=-2 to +2.

    Note that dA = f(x)dx

    Area, A
    = ∫dA
    = ∫f(x)dx
    = ∫(4-x^2)dx
    = [4x-x^3/3] x=-2 to +2
    = [8-8/3 -(-8) -(-8/3)]
    = 16-16/3
    = 32/3

    Moments about y-axis:
    ∫xdA
    =∫x*f(x)dx
    = 0 (by symmetry, or you can do the integration)
    so x-centroid = 0

    Moments about x-axis
    ∫(y/2)dA
    =∫(f(x)/2)*f(x)dA
    =256/15

    So y-centroid
    = (256/15) / A
    = (256/15) / (32/3)
    = 8/5
    = 1.6

    So the centroid is at (0,1.6), so A is correct.

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