Compute the coordinates of the centre of mass for the region
$$K:\{x^2+y^2+z^2\leq R^2, \quad x\geq 0, \quad z\geq 0\}.$$
This problem should be easy, but I can't see where I make the error. Drawing out $K$, I find it's just a $1/4-$th sphere, cut by the $xy$ and $yz-$axis. Due to symmetry we have that $y_{cm}=0$ and $x_{cm}=z_{cm}.$ $D$ is the semi-circle on the $xy-$plane. The formula for the $x$ (and $z$) coordinate of the centre of mass is
$$x_{cm}=z_{cm}=\frac{\iiint_K x \ dxdydz}{\iiint_K \ dxdydz}=\frac{1}{m}\iiint_K x \ dxdydz=\frac{I_{x,z}}{m}.$$
So,
$$m=\iint_D \sqrt{R^2-(x^2+y^2)} \ dx dy=\iint_Er\sqrt{R^2-r^2} \ drd\theta=\pi\int_0^Rr\sqrt{R^2-r^2} \ dr=\frac{\pi R^3}{3}.$$
And
$$I_{x,z} = \iint_Dx\sqrt{R^2-(x^2+y^2)} \ dxdy=\iint_Er^2\cos{\theta}\sqrt{R^2-r^2} \ dr d\theta = 2\int_0^Rr^2\sqrt{R^2-r^2} \ dr.$$
This is where I'm stuck. That last integral is quite unpleasant. Any suggestions?
HINT
Let use a change of variable
$$r=R\cos \alpha \implies 2\int_0^Rr^2\sqrt{R^2-r^2} \ dr =2\int_{\frac{\pi}2}^0 R^2\cos^2 \alpha\sqrt{R^2-R^2\cos^2 \alpha} (-R\sin \alpha)\ d\alpha =2R^4\int_0^{\frac{\pi}2} \cos^2\alpha \sin^2 \alpha \ d\alpha =\frac{\pi R^4}8$$