Mass centrum for $y=\sqrt{\ln (x+1)}$ The region below 
$$y = \sqrt{\ln (x+1)}$$
and above the $x$-axis, $ 0 \leq x \leq 1$, is rotated about the $x$-axis. 
So I want to find the mass centrum for this solid. We know that for the x-coordinate $$ \frac{1}{m}\int_K x dm $$
So calculating $dm$ gave me $$dm =\ln (x+1) \pi \rho$$
And the $x$-coordinate is found after some calculations to be $$ \frac{1}{4(2\ln 2 -1)} $$
However how do I find the $y$ and $z$ coordinates?
 A: By symmetry, the centre of mass of the solid lies on the $x$-axis. So its $y$ and $z$ coordinates are $0$. This answers the question as asked. But for fun we compute the $x$-coordinate, which is correctly computed by the OP. This is done in order to introduce a little integration trick.
Without loss of generality we may assume that the density of the solid is $1$. Then the moment of the solid about the $y$-$z$ plane is
$$\int_0^1 \pi x\ln(x+1)\,dx.$$
We use integration by parts. Let $u=\ln(x+1)$ and let $dv=x\,dx$. Then $du=\frac{1}{x+1}\,dx$ and we can take $v=\frac{1}{2}(x^2-1)$. (That was the trick.)
Since $uv$ vanishes at $0$ and at $1$, our integral is
$$-\int_0^1 \frac{\pi}{2}\cdot\frac{x^2-1}{x+1}\,dx.$$
There is cancellation, and we get $\frac{\pi}{4}$.
We need to divide by the volume, which is
$$\int_0^1\pi \ln(x+1)\,dx.$$
We use a similar integration by parts trick, letting $u=\ln(x+1)$ and $dv=dx$. then $du=\frac{1}{x+1}\,dx$ and we can take $v=x+1$. The volume turns out to be $\pi(2\ln 2-1)$.
