Deriving centroid of quarter circle.

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SUMMARY

The discussion focuses on calculating the centroid of a quarter circle defined by the equation x² + y² = a² in the first quadrant. The user initially attempted to derive the centroid using the formula y' = (I y dA)/(I dA) and calculated the area differential dA as dy.x dy(a² - y²)^(1/2). After some confusion regarding the integration results, the user realized the mistake in their calculations regarding the exponent of a², ultimately confirming that the correct centroid is derived from the integral yielding a^(3/2)/3.

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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 !
 
Last edited:
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