# 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

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

∫(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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