Locate the bar y of the centroid of the area Solutiony 1 x

Locate the bar y of the centroid of the area.

y = 1 - x^2 /4

x = sqrt (4 - 4y)

x = 2* sqrt (1-y)

At x = 0, we have y = 1

Taking a strip parallel to x axis of width dy

Area of strip dA = 2x dy

dA = 2*2 *sqrt (1-y) dy

dA = 4* sqrt(1-y) dy

Yc = Integral (y dA) / Integral dA

Integral dA = Integral 4* sqrt(1-y) dy

= 4* (-2/3) *(1-y)^(3/2)

Varying it from y = 0 to y = 1 we get,

Integral dA = (-8/3)* [0 - 1]

= 8/3

Integral y dA = Integral 4* y*sqrt(1-y) dy

= 4* (-2/15)*(1-y)^(3/2) * (3y+2)

Varying it from y = 0 to y = 1 we get,

Integral y dA = (-8/15) [0 - 1* (3*0+2)]

= 16/15

Thus, Yc = (16/15) / (8/3)

= 2/5

= 0.4 m