There must be an enclosed region, because it’s not possible to find the centroid of an open region. Therefore, I’m assuming that the y-axis forms the required boundary. x=y³ can be written y=∛x.
To find the limits of integration we need to solve ∛x=x-3, that is, x=(x-3)³.
x³-9x²+27x-27-x=x³-9x²+26x-27=0.
To solve this cubic, we can use Newton’s Method where f(x)=(x-3)³-x, f'(x)=3(x-3)²-1 and the iterative formula applies:
x=x-f(x)/f'(x), where we start with x=x₀ an arbitrary value in the expression on the right-hand side to arrive at the next iteration x₁ on the left-hand side.
If x₀=5, x₁=4.7272..., x₂=4.6736..., x₃=4.6717..., x₄=4.6716998817...
From this, x=4.6717 approx (irrational number which cannot be easily expressed otherwise). The lower limit is x=0 (y-axis).
Area A of the region=∫(∛x-x+3)dx[0,4.6717].
A=[¾x^(4/3)-x²/2+3x]⁴˙⁶⁷¹⁷₀=8.96 approx.
We now use the x-moment to find Ax̄, where x̄ is the x value of the centroid:
Ax̄=∫x(∛x-x+3)dx[0,4.6717].
Ax̄=[(3/7)x^(7/3)-x³/3+3x²/2]⁴˙⁶⁷¹⁷₀=14.3871 approx.
x̄=1.6057 approx.
The y-moment is Aȳ=½∫(x^⅔-(x-3)²)dx[0,4.6717].
Aȳ=½[⅗x^(5/3)-x³/3+3x²-9x]⁴˙⁶⁷¹⁷₀=-1.362 approx.
ȳ=-0.152 approx. Centroid is (1.6057,-0.1520).