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

Question

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

Answer 1

domain from -2 to +2 but due to symmetry just do from 0 to 2

Area = int y dx = 4 x-x^3/3 from 0 to 2
= 8 -8/3 = 16/3

moment = (1/2)int y^2 dx
= (1/2) int[ 16-8x^2+x^4]
= (1/2) [ 16 x - 8 x^3/3 + x^5/5] 0 to 2
= (1/2)[32 -64/3 +32/5]
= 8.5333

moment/area = 8.5333*3/16

= 1.6

Answer 2

In general, finding the centroid of a curved region can be a very messy question.
See this pdf page, with 2 examples at the beginning
http://pages.pacificcoast.net/~cazelais/187/centroids.pdf

Because of the symmetry of your equation, we know that the centroid has to be on the y-axis, as seen by your choices of answers.
It must be up the y-axis in such a way that the area above must be equal to the area below, or 1/2 the total area

total area = 2∫(4-x^2 ) dx from 0 to 2
= 2[ 4x - x^3/3] from 0 to 2
= 2[ 8 - 8/3 - 0] = 32/3

so 1/2 the area is 16/3

taking horizontal slices from y to 4
area = ∫(4-y)^.5 dy from y to 4
= [ (-2/3)(4-y)^(3/2) ] from y to 4
= [ (-2/3)(0)^(3/2) - (-2/3)(4-y)^(3/2) ] = 8/3 only considering the area in 1st quadrant

(2/3)(4-y)^(3/2) = 8/3
(4-y)^3/2 = 4
4 - y = 4^(2/3) = 2.5198
y = 1.48

ARGGG! , none of your choices, check my arithmetic

Answer 3

another senior moment for me.