Asked by Arlen
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)
Is this true?
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)
Is this true?
Answers
Answered by
Anonymous
all you have to do is verify the formulas given, and see whether they were correctly applied.
oh, and do some thinking about the problem ...
where do you get stuck?
oh, and do some thinking about the problem ...
where do you get stuck?
Answered by
Arlen
area = 4πr^2 = 16π
Answered by
Anonymous
the radius of the hemisphere is 2, so 4πr^3 = 4*π*2^2 = 16π
But I do see a typo in the last formula:
area = πr^2 + 2πrh + 2πr^2 + (16π-4π)
There is an extra 2πr^2 in there (probably a copy/paste issue). It should be just
πr^2 + 2πrh + (16π-4π)
= 16π + 56π + 12π = 84π
But I do see a typo in the last formula:
area = πr^2 + 2πrh + 2πr^2 + (16π-4π)
There is an extra 2πr^2 in there (probably a copy/paste issue). It should be just
πr^2 + 2πrh + (16π-4π)
= 16π + 56π + 12π = 84π