Centroid of the region bounded by y = f(x) and x = f(y) curve How do i find the centroid of the region if $x = f(y)$ is involved ?
For example : find the centroid of the region bounded by $y = x^2$ and $x = y^2$
And how do i find it if there is curve intersection that is below x axis ?
For example : find the centroid of the region bounded by $y = x^3$ and $x = y^2-1$
The formula I found in the book was
$$x = \frac 1{\mathcal{A}}\int_a^b x[f(x)-g(x)] \,dx$$
$$y = \frac 1{\mathcal{A}}\int_a^b \frac 12 (f(x)^2-g(x)^2) \,dx$$
$\mathcal{A}$ is the area of the region bounded by the curve.

In this case you can first try to sketch the region, as shown below.

From the sketch we learn that it suffice to consider the first quadrant where $ x, y \geq 0 $.
Thus the equation $ x = y^2 $ can be converted to $ y = \sqrt{x} $.
Put $ f(x) = \sqrt{x} $, $ g(x) = x^2 $, $ a = 0 $, $ b = 1 $, and using the formulae you have given yield
$$ A = \int_0^1 \sqrt{x} - x^2 \operatorname{d}\!x = \left. \frac23 x^{3/2} - \frac13 x^3 \right|^1_0 = \frac13,  $$
$$ x = \frac1A \int_0^1 x(\sqrt{x} - x^2) \operatorname{d}\!x = 3 \left[ \frac25 x^{5/2} - \frac14 x^4 \right]_0^1 = 0.45, $$
and
$$ y = \frac1A \int_0^1 \frac12 [(\sqrt{x})^2 - (x^2)^2]\operatorname{d}\!x = \frac32 \left[ \frac12 x^2 - \frac15 x^5  \right]_0^1 = 0.45. $$
