Double Integrals Over General Regions

Evaluate double integrals over regions bounded by curves rather than straight edges — Type I and Type II regions.

Every double integral so far has lived over a rectangle: fixed limits on both variables, the same box no matter what x or y happens to be. Most real regions aren't rectangles — a wedge, a lens shape between two curves, the footprint of an actual part. The rectangle was training wheels.

Ready to put it into practice?

Start practicing

A Type I region: x runs over a fixed interval, and for each x, y is trapped between two curves that depend on x.

D = \{(x,y) \mid a \le x \le b,\; g_1(x) \le y \le g_2(x)\}

Type I region

$a, b$ Fixed numbers — the outer x-bounds, the same for every x in the region.
$g_1(x)$ The lower boundary curve: where the region starts in y, at a given x.
$g_2(x)$ The upper boundary curve: where the region ends in y, at a given x.

The integral over it, written as an iterated integral:

\iint_D f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx

Iterated integral over a Type I region

$\int_a^b$ Outer integral: sweep x over its fixed interval, exactly like a rectangle's outer bound.
$\int_{g_1(x)}^{g_2(x)}$ Inner integral: for the current x, sweep y from the lower curve up to the upper curve.
$dy\,dx$ Integrate with respect to y first, then with respect to x.

Compare to the rectangle version: the outer bounds a, b are still plain numbers — that never changes. What's different is the inner bounds. They used to be plain numbers c and d. Now they're functions of x, because the strip's top and bottom slide as x moves. Picture vertical strips of width dx, each one running from the lower curve up to the upper curve, then add the strips from x = a to x = b.

Turn the strips 90 degrees and you get a Type II region: y runs over a fixed interval, and for each y, x is trapped between two curves.

D = \{(x,y) \mid c \le y \le d,\; h_1(y) \le x \le h_2(y)\}, \qquad \iint_D f\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy

Type II region

$h_1(y)$ The left boundary curve: where the region starts in x, at a given y.
$h_2(y)$ The right boundary curve: where the region ends in x, at a given y.

Same idea, axes swapped: horizontal strips instead of vertical ones. Some regions are only Type I, some are only Type II, and plenty are both — you just pick whichever direction makes the description, and the resulting integral, easier. The practice problems below are all Type I.

The intersection points of the two boundary curves are a and b — solve g_1(x) = g_2(x) for x and you've found the outer limits. This is the step students skip, and then wonder why their bounds are wrong. No intersection points, no valid Type I region.

Order of integration is no longer a free swap the way it was for a rectangle. Fubini still holds, but switching from Type I to Type II means re-deriving the bounds from scratch — you can't just relabel dx and dy and keep the same numbers.

Worked example

\iint_D (x+y)\,dA, \quad D: -1 \le x \le 2,\; x^2 \le y \le x+2

$x^2 = x+2 \;\Longrightarrow\; x^2 - x - 2 = 0 = (x-2)(x+1) \;\Longrightarrow\; x = -1,\, 2$
Find where the parabola and the line cross — those two x-values become the fixed outer bounds. Checking x = 0: the line gives y = 2, the parabola gives y = 0, so the line sits on top through the whole region.
$\int_{x^2}^{x+2} (x+y)\,dy = \left[xy + \dfrac{y^2}{2}\right]_{x^2}^{x+2} = -\dfrac{x^4}{2} - x^3 + \dfrac{3}{2}x^2 + 4x + 2$
Inner integral, with respect to y, from the lower curve to the upper curve. x is frozen throughout, so it rides along as a constant just like it did over a rectangle.
$\int_{-1}^{2} \left(-\dfrac{x^4}{2} - x^3 + \dfrac{3}{2}x^2 + 4x + 2\right) dx = \dfrac{189}{20}$
Outer integral, over the fixed interval found in step 1. This part is exactly like every rectangle problem already done.

Answer:

\dfrac{189}{20}

How to solve one of these

1.
Sketch the region. Even a rough sketch shows immediately whether $y$ or $x$ is the variable trapped between two curves.
2.
Find the intersection points of the boundary curves. Set the two curve expressions equal and solve. These $x$ -values (Type I) or $y$ -values (Type II) become the fixed outer bounds.
3.
Identify the lower and upper curve. Plug in one point inside the interval to check which curve is on top — don't guess from the equations alone.
4.
Write the iterated integral with matched bounds. Inner differential pairs with the curve bounds; outer differential pairs with the two intersection values.
5.
Evaluate the inner integral. Treat the outer variable as a constant, substitute the curve expressions as the limits, and simplify — the result is a plain polynomial in the outer variable.
6.
Evaluate the outer integral. A single-variable integral over fixed numeric limits — nothing new here.
7.
Check the shape. If the final polynomial in the outer variable has a wildly high degree or an unexpected sign, recheck which curve was called upper vs. lower.

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

Sweep vertical strips from x = a to x = b; each strip runs from the lower curve g_1(x) up to the upper curve g_2(x).

Ready to put it into practice?

Start practicing