Answer
$\frac{3R}{8}$
Work Step by Step
Please see the attached image first.
Let's take the x,y planes through the symmaetric planes of the hemisphere.
To find the location of the center of the mass we use the following equation.
$x_{cm}=\frac{\int xdm}{M}-(A)$
Let's assume this is an uniform hemisphere with the density $\rho$
We can write, $dm=\rho\times\pi (\frac{h}{2})^{2}dx-(1)$
We know that the,
$R^{2}=x^{2}+(\frac{h}{2})^{2}$
$4(R^{2}-x^{2})=h^{2}-(2)$
Where R - radius of hemisphere.
$(2)=\gt (1)$
$dm=\frac{4\pi \rho}{4}(R^{2}-x^{2})dx-(3)$
$(3)=\gt(A)$
$x_{cm}=\frac{\int_{0}^{R} x\frac{4\pi \rho}{4}(R^{2}-x^{2})dx}{\frac{2}{3}\pi R^{3}\rho}$
$x_{cm}=\frac{3}{8R^{3}}[R^{2}\int_{0}^{R}xdx-\int_{0}^{R}x^{3}dx]$
$x_{cm}=\frac{3}{2R^{3}}[R^{2}\times \frac{x^{2}}{2}\biggr|_{0}^{R}-\frac{x^{4}}{4}\biggr|_{0}^{R}]=\frac{3}{2R^{3}}(\frac{R^{4}}{2}-\frac{R^{4}}{4})$
$x_{cm}=\frac{3R}{8}$
