Deriving centroid of quarter circle.

[txpc]

Homework Statement

Find the centroid of the region cut from the first quadrant by the circle x^2+y^2=a^2

Homework Equations

I know that y' = (I y dA)/(I dA)

The Attempt at a Solution

Taking a strip dy with length x, I obtain dA = dy.x dy(a^2-y^2)^(1/2)

So I y DA = I y*(a^2-y^2)^(1/2) dy from a->0

This integral leads to a^(3/2)/3

Now, I dA is simply 1/4 the area of a circle radius a = a^2.pi/4

By dividing these 2 values, I obtain 4/3.pi.sqrt(a) instead of 4a/3pi which means I'm off by a sqrt a somewhere which I can't seem to figure out the error, I think the best bet would be the definite integral giving a^(3/2)/3 but even after plugging it in a few integrators, it comes out the same.

Any help is greatly appreciated.

EDIT :

Oh dear me, I'm so silly. Forgot it was (a^2)^3/2 !

 

[kuruman]

It's a good thing you figured it out on your own.