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)

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

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
$=\pi {\int }_0^2{x}^{2}dy=\pi {\int }_0^2(9-y^2)dy$
$=\pi {[9y-\frac{1}{3}{y}^3]}_0^2$
$=\pi [9\times 2-\frac{1}{3}(2{)}^3-\left(9.0-\frac{1}{3}.0\right)]$
$=\pi \left(18-\frac{8}{3}\right)=\frac{46}{3}\pi$cu units.