Find the centroid of the area bounded by the parabola y=4x^2 and the xaxis
A.(0,1.6)
B.(0,1.7)
C.(0,1.8)
D.(0,1.9)
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3 answers

domain from 2 to +2 but due to symmetry just do from 0 to 2
Area = int y dx = 4 xx^3/3 from 0 to 2
= 8 8/3 = 16/3
moment = (1/2)int y^2 dx
= (1/2) int[ 168x^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 👍
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answered by Damon 
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 yaxis, as seen by your choices of answers.
It must be up the yaxis in such a way that the area above must be equal to the area below, or 1/2 the total area
total area = 2∫(4x^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 = ∫(4y)^.5 dy from y to 4
= [ (2/3)(4y)^(3/2) ] from y to 4
= [ (2/3)(0)^(3/2)  (2/3)(4y)^(3/2) ] = 8/3 only considering the area in 1st quadrant
(2/3)(4y)^(3/2) = 8/3
(4y)^3/2 = 4
4  y = 4^(2/3) = 2.5198
y = 1.48
ARGGG! , none of your choices, check my arithmetic 👍
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answered by Reiny 
another senior moment for me.
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answered by Reiny
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