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?
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?
area = 4πr^2 = 16π
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
Yes, your explanation is correct.
For the volume calculation:
The volume of the original candle is given as πr^2h = 112π.
The volume of the hemisphere on top of the candle is 2/3πr^3 = 16π/3.
So, the volume of the remaining candle after removing the hemisphere is obtained by subtracting the volume of the hemisphere from the original volume of the candle:
Remaining volume = Original volume - Hemisphere volume
= 112π - 16π/3
For the surface area calculation:
The surface area of the original candle is given as 2πr^2 + 2πrh = 88π.
The surface area of the hemisphere is 4πr^2 = 16π.
So, the surface area of the remaining candle after removing the hemisphere is obtained by subtracting the surface area of the hemisphere from the original surface area of the candle:
Remaining area = Original area - Hemisphere area
= (2πr^2 + 2πrh) + (16π - 4π)
Therefore, your analysis of calculating the remaining volume and area is correct based on the given formulas and calculations.