Answer
$(0,\frac{3h}{4},0)$
Work Step by Step
Please see the attached image first.
Let's take the x, and y axes through the symmetric planes of the object.
Here we use the equation, $y_{cm}=\frac{\int\space y\space dm}{M}-(1)$
Where, $y_{cm} -$ y coordinate of the center of mass, M - Mass of the object, y - y coordinate of the center of mass of the particle object.
from the triangle, we can write,
$\frac{2R}{h}=\frac{a}{y}=\gt (\frac{2R}{h})y=a-(2)$
$dv=\pi r^{2}h=\pi(\frac{a}{2})^{2}dy=\frac{\pi}{4}a^{2}dy$
$(2)=\gt dv=\frac{\pi}{4}(\frac{2Ry}{h})^{2}dy=\frac{\pi R^{2}y^{2}}{h^{2}}dy$
The mass of the particle dm is the same fraction of the total mass M as its volume dv is of the total object's volume $V=\frac{1}{3}\pi R^{2}h$
$\frac{dm}{M}=\frac{dv}{V}=\frac{\pi R^{2}y^{2}dy}{h^{2}}\times\frac{1}{\frac{1}{3}\pi R^{2}h}=\frac{3y^{2}dy}{h^{2}}-(3)$
$(3)=\gt(1)$,
$y_{cm}=\frac{1}{M}{\int_{0}^{h}\space y\space \frac{3y^{2}Mdy}{h^{3}}}=\frac{3}{h^{3}}\int_{0}^{h}y^{3}dy$
$y_{cm}=\frac{3}{h^{3}}\times\frac{y^{4}}{4}\biggr|_{0}^{h}=\frac{3h^{4}}{4}=\frac{3h}{4}$
Location of the center of the mass $=(0,\frac{3h}{4},0)$
(x,y,z)
