find the center of mass of a cone of height 8cm and base diameter 16cm?

I suggest taking the centre of the base as origin, and the axis of the cone as the y axis.

Regard the cone as made up of "many" circular discs, each "very thin".

Integrate the mass of each disc multiplied by its distance from the origin, and finally divide by the mass of the cone. This gives the height of the centre of mass above the origin.

Let the density of the cone be ρ ( units of mass per cubic cm)

Then a disc of thickness dy, and at a height y above the origin,

has radius (8-y) (because height and radius of the cone are equal)

and mass ρπ(8-y)² dy

Mass of cone

= ∫ ρπ(8-y)² dy from y = 0 to 8

= -π(ρ/3)(8-y)³ from y = 0 to 8

= 512 π ρ/3 as we knew already.

Now to calculate ∫ y*ρπ(8-y)² dy

use y = 8 - (8-y) so that this integral is equal to

∫ 8*ρπ(8-y)² dy - ∫(8- y)*ρπ(8-y)² dy from y=0 to 8

= 8*512 π ρ/3 - ∫ρπ(8-y)³ dy

= 8*512 π ρ/3 + (ρπ/4)(8-y)⁴ dy

= 8*512 π ρ/3 - (ρπ/4)*8⁴

= (1024 π ρ / 3)

Hence height of centre of mass

= (1024 / 3) / (512 / 3)

= 2

The centre of mass is on the axis of the cone ( of course)

2 cm above the base.