From a solid right circular cylinder with height $h$ and radius of the base $r$ , a right circular cone of the same height and same base is removed. Find the volume of the remaining solid.
$\frac{2}{3} π r^{2} h$
$π r^{2} h$
$2 π r^{2} h$
$\frac{1}{3} π r^{2} h$

Question

From a solid right circular cylinder with height $h$ and radius of the base $r$ , a right circular cone of the same height and same base is removed. Find the volume of the remaining solid.
$\frac{2}{3} π r^{2} h$
$π r^{2} h$
$2 π r^{2} h$
$\frac{1}{3} π r^{2} h$

Toppr Admin · Accepted Answer

Volume-xA0-of the remaining solid - Volume of cylindrical section -x2212- Volume of-xA0-conical shape-So- Volume of the material removed -x3C0-r2h-x2212-13-x3C0-r2h-23-x3C0-r2h

From a solid right circular cylinder with height h and radius of the base r, a right circular cone of the same height and same base is removed. Find the volume of the remaining solid.23πr2hπr2h2πr2h13πr2h

Volume of the remaining solid = Volume of cylindrical section − Volume of conical shape.So, Volume of the material removed =πr2h−13πr2h=23πr2h

From a solid right circular cylinder with height $h$ and radius of the base $r$ , a right circular cone of the same height and same base is removed. Find the volume of the remaining solid.
$\frac{2}{3} π r^{2} h$
$π r^{2} h$
$2 π r^{2} h$
$\frac{1}{3} π r^{2} h$

Volume of the remaining solid $=$ Volume of cylindrical section $-$ Volume of conical shape.

So, Volume of the material removed $= π r^{2} h - \frac{1}{3} π r^{2} h = \frac{2}{3} π r^{2} h$