The centre of mass of a solid hemisphere of radius $$8\,\text{cm}$$ is $$x\,\text{cm}$$ from the centre of the flat surface. Then value of $$x$$ is___

We take the solid hemisphere of radius $$R=8\,\text{cm}$$. Because of symmetry, its centre of mass will lie on the central vertical axis passing through the centre of the flat circular face. We measure every distance $$z$$ from this flat face, so $$z=0$$ at the plane of the face and $$z=R$$ at the pole of the hemisphere.

To locate the centre of mass we use the continuous form of the centre-of-mass formula for volume distributions:

$$z_{\text{cm}}=\dfrac{\displaystyle\int z\,dV}{\displaystyle\int dV}.$$

We slice the hemisphere into thin circular discs perpendicular to the axis. A slice at distance $$z$$ from the base has thickness $$dz$$ and radius $$r$$. From the equation of a sphere we have

$$r^{2}+z^{2}=R^{2}\;\Longrightarrow\;r^{2}=R^{2}-z^{2}.$$

The volume of this thin disc is

$$dV=\pi r^{2}\,dz=\pi\bigl(R^{2}-z^{2}\bigr)\,dz.$$

Now we write the numerator of the centre-of-mass expression:

$$\int_{0}^{R} z\,dV=\int_{0}^{R} z\,\pi\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\int_{0}^{R}\bigl(R^{2}z - z^{3}\bigr)\,dz.$$

Integrating term by term,

$$\pi\left[\frac{R^{2}z^{2}}{2}-\frac{z^{4}}{4}\right]_{0}^{R} =\pi\left(\frac{R^{2}R^{2}}{2}-\frac{R^{4}}{4}\right) =\pi\left(\frac{R^{4}}{2}-\frac{R^{4}}{4}\right) =\pi\,\frac{R^{4}}{4}.$$

Next we find the denominator (the total volume of the hemisphere):

$$\int_{0}^{R} dV=\int_{0}^{R}\pi\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\int_{0}^{R}\bigl(R^{2}-z^{2}\bigr)\,dz =\pi\left[R^{2}z-\frac{z^{3}}{3}\right]_{0}^{R} =\pi\left(R^{3}-\frac{R^{3}}{3}\right) =\pi\,\frac{2R^{3}}{3}.$$

Substituting these two results into the centre-of-mass formula,

$$z_{\text{cm}}=\frac{\pi\,\dfrac{R^{4}}{4}} {\pi\,\dfrac{2R^{3}}{3}} =\frac{R^{4}}{4}\times\frac{3}{2R^{3}} =R\,\frac{3}{8} =\frac{3R}{8}.$$

Putting $$R=8\,\text{cm}$$, we obtain

$$z_{\text{cm}}=\frac{3\times8}{8}\,\text{cm}=3\,\text{cm}.$$

So, the answer is $$3\,\text{cm}$$.