Question

A cone of height H with a base of radius r is cut by a plane parallel to and h units above the base where h < H. Find the volume of the solid (frustum of a cone) below the plane.

Answer 1

Answer:

$\Delta V = \frac{1}{3}\times \pi \frac{r^{2}}{H^{2}}\left ( H^{3}-(H-h)^{3} \right )$

Step-by-step explanation:

radius of cone = r

height of cone = H

Let the radius of upper cone is r' and the height of plane from the base is h. So, the height of upper cone = H - h

According to the diagram

$\frac{r'}{r}=\frac{H - h }{H}$

$r'=r\frac{H - h }{H}$     .... (1)

Volume of complete cone

$V = \frac{1}{3} \pi r^{2}H$  

Volume of upper cone

$V' = \frac{1}{3} \pi r'^{2}(H - h)$

$V' = \frac{1}{3} \pi \frac{r^{2}(H - h)^{3}}{H^{2}}$

The volume of frustum of cone below the plane is

ΔV = V - V'

$\Delta V = \frac{1}{3}\times \pi r^{2} \left ( H - \frac{\left (H-h \right )^{3}}{H^{2}} \right )$

$\Delta V = \frac{1}{3}\times \pi \frac{r^{2}}{H^{2}}\left ( H^{3}-(H-h)^{3} \right )$