**1.** Consider . What is the volume of the region

(1)

The first thing we do: consider the rectangle and form a partition of into segments and into segments. This gives us a mesh of rectangles as specified by the following diagram:

Observe the area of is

(2)

We can approximate the volume of by a sort of Riemann sum, picking points in and taking

(3)

However, as with Riemann sums, we recover the exact volume when we take the limits :

(4)

if the limit exists.

**2. Definition.** We define the **“Double Integral”** of over the integral as

(5)

if this limit exists.

Note this sum is called a *double Riemann sum*, just as a for the single integral we had a *Riemann Sum*.

**3. Properties of Integrals.** We won’t prove, but note, there are three important properties double integrals satisfy. Two can be grouped together as linearity:

(6)

and

(7)

where and are continuous on , and is a constant.

If for all , then

(8)

**4. Problem:** How do we compute ?

### Iterated Integrals

**5. Fubini’s Theorem.** If is continuous on the integral , then

(9)

We won’t prove it (that’s what real analysis discusses!), but the parenthetic terms are functions of a single variable and describe the area of a “sheet”. By stacking “sheets” we can compute the volume of the region.

**6.** This is fine for rectangular regions, but over arbitrary regions what do we do? We bound the region by a pair of curves :

In these situations, we write

(10)

Please note the order of integration, and corresponding boundary of integration!

**7.** We can likewise bound the region by curves , for example:

The trick lies with writing these integrals as

(11)

Again, note the order of integration!

Example 1.Lets consider the domain

(12)

Evaluate the integral

(13)

Solution:\quad\ignorespaces We first doodle this domain, shade it in red:

Now we use Fubini’s theorem writing

(14)

Observe the order of integration, and the bounds of integration are determined by specifying and for the quantity. Similarly, since , we determine the bounds of integration for the integral.Using linearity, we can write

(15)

where

(16)

and

(17)

Now we can evaluate each of these separately.We see for , there are no expressions, so we have

(18)

Performing this inner integral

(19)

We plug this back in:

(20)

This can be solved quickly as

(21)

Thus

(22)

Moreover this implies .

What to do about ? Recall

We first perform the inner integral

(23)

We plug this back in

(24)

We can calculate this out directly as

(25)

But we forgot the coefficient in front of the integral! Thus our integral is

(26)

giving us our final result

(27)

which concludes our example.