A double integral accumulates a scalar quantity over a two-dimensional region $D$:

$$\iint_D f(x,y),dA.$$

If $f\ge0$, it can be interpreted as the volume below the surface $z=f(x,y)$ and above the region $D$.

Vertically simple regions

If

$$D={(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)},$$

then

$$\iint_D f,dA

\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y),dy,dx.$$

The inner integral accumulates along a vertical slice, and the outer integral combines those slices across the region.

Example

For the triangle

$$D={(x,y):0\le x\le1,\ 0\le y\le x},$$

the area is

$$\iint_D1,dA

\int_0^1\int_0^x1,dy,dx

\int_0^1x,dx

\frac12.$$

Reversing the order

The same region can often be described using horizontal rather than vertical slices. Changing the order from $dy,dx$ to $dx,dy$ does not change the geometric integral, but it changes the bounds and may make the calculation much easier.

A successful setup begins with the region: sketch or understand $D$ first, then choose the order whose slices describe it most simply.