Given a surface S or volume V and an integrand f(x, y, z), defined and reasonably
well behaved in S or V, we can form the analogue of Riemann sums in this context.

For volume: break V up into small pieces and sum a value of f in each piece
times its volume; take the limit as the volumes of all pieces goes to zero
in every possible way. If it exists this limit is the *multiple integral of
f over V*. The limit can be shown to exist when f is *continuous *and
V is *compact (*bounded and closed*)*.* *It often exists otherwise
as well.

Exercise

**Iterated integrals**

Any product which contains two or more integrals in it can be considered
an *iterated integral. *

Examples

Notation