Definite Integration Question 303
Question: The volume of spherical cap of height h cut off from a sphere of radius a is equal to
[UPSEAT 2004]
Options:
A) $ \frac{\pi }{3}h^{2}(3a-h) $
B) $ \pi (a-h)(2a^{2}-h^{2}-ah) $
C) $ \frac{4\pi }{3}h^{3} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
The required volume of the segment is generated by revolving the area ABCA of the circle $ x^{2}+y^{2}=a^{2} $ about the x-axis and for the arc BA, Here $ CA=h $ and $ OA=a $ , (given)
$ OC=OA-CA=a-h $ , $ x $ varies from $ a-h $ to a.
The required volume $ =\int _{a-h}^{a}{\pi y^{2}dx} $
$ =\pi \int _{a-h}^{a}{(a^{2}-x^{2})dx} $ , $ (\because x^{2}+y^{2}=a^{2}) $
$ =\pi [ a^{2}x-\frac{1}{3}x^{3} ] _{a-h}^{a}=\pi [ (a^{3}-\frac{1}{3}a^{3}-{ a^{2}(a-h)-\frac{1}{3}{{(a-h)}^{3}} } ] $
$ =\pi [ ( a^{3}-\frac{1}{3}a^{3} )-{ a^{3}-a^{2}h-\frac{1}{3}( a^{3}-3a^{2}h+3ah^{2}-h^{3} ) } ] $
$ =\pi [ a^{2}h-a^{2}h+ah^{2}-\frac{1}{3}h^{3} ] $
$ =\frac{1}{3}\pi h^{2}(3a-h) $ .
BETA
