The volume formed when region of circle x²+y²= 16 revolved about its diameter
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Answer:
Correct option is
A
3
46
π
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
=π∫
0
2
x
2
dy=π∫
0
2
(9−y
2
)dy
=π[9y−
3
1
y
3
]
0
2
=π[9×2−
3
1
(2)
3
−(9.0−
3
1
.0)]
=π(18−
3
8
)=
3
46
πcu units.
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