Calculate the center of mass for :
The area bounded by parabola $y = x^2/b$ and the line $y = b$.
I got the following integral I just need verification that my work is correct.
First I got $$\frac{ds}{dy} = \frac{2x}{b},$$ so $$ds = \frac{2x}{b} dy.$$ So we have the integration as follows:
$$\frac{\int_{0}^{b}2 y \frac{\sqrt{(by)}}{b} dy}{\int_{0}^{b}2\frac{\sqrt{(by)}}{b}dy}$$
the rest of the calculation is easy so if someone could verify this that would be great!
We assume constant density. By symmetry the $x$-coordinate of the centroid is $0$. Now we go after the $y$-coordinate. Assume $b\gt 0$.
The area of our region is $\int_0^b 2x\,dy$, that is, $\int_0^b 2\sqrt{by}\,dy$.
Next we calculate the moment of our region about the $x$-axis. Consider a thin horizontal sice "at" height $y$, of width "$dy$". The moment of this about the $x$-axis is $(y)(2x)\,dy$, where $x=\sqrt{by}$, the width of the slice at height $y$. "Add up" (integrate) from $y=0$ to $y=b$. We get $\int_0^b 2y\sqrt{by}\,dy$.
Divide as usual.
Remark: I would rather use the fact that the moment of the region below $y=f(x)$, from $x=c$ to $x=d$, about the $x$-axis is $\int_c^d \frac{1}{2}(f(x))^2 \,dx$.