Question

The part of circle $x^2+y^2=9$ in between $y=0$ and $y=2$ 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)

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
$=\pi \underset{0}{\overset{2}{\int }}{x}^{2}dy=\pi \underset{0}{\overset{2}{\int }}\left(9-y^2\right)dy$
$=\pi {\left[9y-\frac{1}{3}{y}^3\right]}_0^2$
$=\pi \left[9\times 2-\frac{1}{3}(2{)}^3-\left(9\cdot 0-\frac{1}{3}\cdot 0\right)\right]$
$=\pi \left(18-\frac{8}{3}\right)=\frac{46}{3}\pi$ cu units