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); SEE YA