A hemispherical metal bowl has internal radius 5 cm and the metal is 1 cm thick everywhere. Find the total surface area of the bowl — its outside, its inside and the flat ring around the rim — in ${\text{cm}}^2$, in terms of π.

The outer radius is 5 + 1 = 6 cm. Outer curved surface = 2π$(6^2)=72$π; inner curved surface = 2π$(5^2)=50$π; the rim is a flat ring of area π$(6^2-5^2)=11$π. Total = 72π + 50π + 11π = 133π ${\text{cm}}^2$. (122π forgets the rim, but real metal has a thickness you can see; 72π counts only the outside; 108π uses 3π$r^2$ for a SOLID hemisphere of radius 6, which has no hollow at all. Note π$(6^2-5^2)=11$π, not π$(6-5{)}^2$ = π.)