Definite Integration Question 309
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
[UPSEAT 1999]
Options:
A) $ \frac{46}{3}\pi $
B) $ 12\pi $
C) $ 16\pi $
D) $ 28\pi $
Show Answer
Answer:
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 [ 9y-\frac{1}{3}y^{3} ]_0^{2} $
$ =\pi [ 9\times 2-\frac{1}{3}{{(2)}^{3}}-(9.0-\frac{1}{3}.0) ] $
$ =\pi [ 18-\frac{8}{3} ]=\frac{46}{3}\pi $ cubic unit.
BETA
