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