Asked by Durva
A cylindrical candle of diameter 8cm and height 7cm is lit. After burning, a hemispherical
depression of diameter 4cm is left in the candle. Find the volume and surface area of this
partially burnt candle.
depression of diameter 4cm is left in the candle. Find the volume and surface area of this
partially burnt candle.
Answers
Answered by
oobleck
cylinder:
volume = πr^2 h = 112π
area = 2πr^2 + 2πrh = 88π
hemisphere:
volume = 2/3 πr^3 = 16π/3
area = 4πr^2 = 16π
so the remaining candle has
volume = 112π - 16π/3
original volume minus the hemisphere
area = πr^2 + 2πrh + 2πr^2 + (16π-4π)
that is, the original bottom + lateral + (top - hole)
volume = πr^2 h = 112π
area = 2πr^2 + 2πrh = 88π
hemisphere:
volume = 2/3 πr^3 = 16π/3
area = 4πr^2 = 16π
so the remaining candle has
volume = 112π - 16π/3
original volume minus the hemisphere
area = πr^2 + 2πrh + 2πr^2 + (16π-4π)
that is, the original bottom + lateral + (top - hole)
Answered by
Yoe
volume = πr^2 h = 112π
area = 2πr^2 + 2πrh = 88π
hemisphere:
volume = 2/3 πr^3 = 16π/3
area = 4πr^2 = 16π
so the remaining candle has
volume = 112π - 16π/3
original volume minus the hemisphere
area = πr^2 + 2πrh + 2πr^2 + (16π-4π)
that is, the original bottom + lateral + (top - hole)
area = 2πr^2 + 2πrh = 88π
hemisphere:
volume = 2/3 πr^3 = 16π/3
area = 4πr^2 = 16π
so the remaining candle has
volume = 112π - 16π/3
original volume minus the hemisphere
area = πr^2 + 2πrh + 2πr^2 + (16π-4π)
that is, the original bottom + lateral + (top - hole)
Answered by
Anonymous
Wtf
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