Question

The part of circle $y = 2$ is revolved about y-axis. The volume of generating solid will be

• A. $\frac { 46 } { 3 } \pi$
• B. $12 \pi$
• C. $16 \pi$
• D. $28 \pi$

Answer 1

Answer: (A)

$\frac { 46 } { 3 } \pi$

Answer 2

The part of circle $x ^ { 2 } + y ^ { 2 } = 9$ in between $y = 0$ and $y = 2$ is revolved about y- axis. Then a frustum of sphere will be formed.

The volume of this frustum $= \pi \int _ { 0 } ^ { 2 } x ^ { 2 } d y = \pi \int _ { 0 } ^ { 2 } \left( 9 - y ^ { 2 } \right) d y$ $= \pi \left[ 9 y - \frac { 1 } { 3 } y ^ { 3 } \right] _ { 0 } ^ { 2 }$ $= \pi \left[ 9 \times 2 - \frac { 1 } { 3 } ( 2 ) ^ { 3 } - \left( 9.0 - \frac { 1 } { 3 } .0 \right) \right]$

$= \frac { 46 } { 3 } \pi$ cubic unit.