A) \[\frac{46}{3}\pi \]
B) \[12\pi \]
C) \[16\pi \]
D) \[28\pi \]
Correct Answer: A
Solution :
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 \[=\pi \int_{0}^{2}{{{x}^{2}}dy}\]\[=\pi \int_{0}^{2}{(9-{{y}^{2}})dy}\] \[=\pi \left[ 9y-\frac{1}{3}{{y}^{3}} \right]_{0}^{2}\]\[=\pi \left[ 9\times 2-\frac{1}{3}{{(2)}^{3}}-(9.0-\frac{1}{3}.0) \right]\] \[=\pi \left[ 18-\frac{8}{3} \right]=\frac{46}{3}\pi \]cubic unit.
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